The Greeks

An option pricing model is a function that takes the price of the underlying asset, or spot price, $S$ and other market inputs $\{z_1, \dots, z_k\}$ and then outputs an option’s fair value $V$:

$V = f(S; z_1, \dots, z_k). \tag{1}$

An important way to understand a particular option-pricing model is to think about the output price’s sensitivity to the model’s inputs. For example, what happens to the option price $V$ when a particular parameter $z_i$ changes? To do this, we can take partial derivatives of $f$ with respect to the inputs. For example, arguably the most important sensitivity to understand is how the option price changes as a function of spot price or

$\text{sensitivity of option to spot} = \frac{\partial V}{\partial S}. \tag{2}$

We can visualize this sensitivity as the slope at a point along a price curve, defined by a particular option pricing model $f$. In an instant in time, when spot is valued at $S$, this partial derivative evaluated at $S$ is a line representing the rate of change of the option price as a function of the spot price (Figure $1$). Clearly, a steeper slope means the option is more sensitive to the underlying’s price.

In the options-pricing literature, these partial derivatives are called the Greeks, so-named because they are denoted with Greek letters. For example, the sensitivity in Equation $2$ is called delta ($\Delta$).

However, without assuming a particular option pricing model, we cannot specify delta or the other Greeks further. We could have a high-level discussion about the Greeks without defining $f$, but any practical discussion requires that we do so. The most common choice for $f$ is the Black–Scholes equation (Black & Scholes, 1973), which gives the fair price of a European-style option under assumptions such as the stock price following an Itô process and that the stock does not pay dividends. Inputs to Black–Scholes are the spot price $S$, the option’s strike price $K$, the option’s time to expiry $T$, the stock’s volatility $\sigma$, and the risk-free interest-rate $r$. Thus, throughout this post, I will refer to various Greeks with the understanding that these are the Black–Scholes Greeks. However, it is important to keep in mind that the basic idea is more general: an option’s price is sensitive to the inputs of its pricing model, and this model is not necessarily Black–Scholes. One could, for example, use the Bachelier model (Bachelier, 1900) or the Black model (Black, 1976) or the Cox-Ross-Rubinstein model (Cox et al., 1979; Rendleman, 1979) to price options.

We will stick with Black–Scholes. Now let $C$ denote the price of a call option. The Black–Scholes price of a European-style call is

$\begin{aligned} C &= S \Phi(d_1) - K e^{-r T} \Phi(d_2), \\ d_1 &= \frac{1}{\sigma \sqrt{T}} \left[ \log(S/K) + \left[ r + (1/2) \sigma^2 \right] T \right], \\ d_2 &= d_1 - \sigma \sqrt{T}, \end{aligned} \tag{3}$

where $\Phi(x)$ is the cumulative distribution function (CDF) of the standard normal distribution:

$\Phi(x) := \frac{1}{\sqrt{2\pi}} \int_{-\infty}^x e^{-z^2 / 2} dz. \tag{4}$

Since $\Phi(x)$ is the CDF of the standard normal distribution, its derivative $\Phi^{\prime}(x)$ is the probability density function (PDF) of the standard normal distribution. I’ll denote this with $\varphi(x)$ or

$\varphi(x) := \Phi^{\prime}(x). \tag{5}$

As a preliminary, see A1 for computing this derivative when the input to $\Phi$ is itself a function of $x$, i.e. for computing $\Phi^{\prime}(g(x))$ for some function $g(x)$. This fact will be used repeatedly when deriving the Greeks.

The intuition for Equation $3$ is that the Black–Scholes formula is the weighted difference between the spot $S$ and strike $K$ (Figure $2$, left). The weights account for the time-value of money and the probability of the option ending in-the-money under the risk-neutral measure. See my post on the binomial options-pricing model for a detailed yet intuitive explanation of Black–Scholes using a simpler yet analogous model. Note that we assume the stock does not pay dividends, and therefore there is no adjustment to $S$. In a separate paper, (Merton, 1973) derived Black–Scholes with stochastic interest rates, dividend-paying stocks, and American options; and thus Black–Scholes is sometimes called Black–Scholes–Merton to give credit to Merton.

The Black–Scholes formula for a European-style put with price $P$ is

$P = K e^{-r T} \Phi(-d_2) - S \Phi(-d_1). \tag{6}$

Here, we can see that the payoff is roughly flipped (Figure $2$, right). We’ll dig into some more subtle differences later.

However, for deriving the Greeks, we will find it useful to represent a call in terms of a put via put-call parity:

$C + K e^{-r T} = P + S. \tag{7}$

Intuitively, this equation says: if you have the right to buy a stock plus the money to do so, you’re in the same financial position as if you had the right to sell the stock short plus you own the stock to sell. If you’re unconvinced, simply write down two portfolios, one for the left-hand side containing a call and a zero-coupon bond paying $K$ at time $T$ and one for the right-hand side with a put and a stock; then convince yourselves that these have the same payout when $S_T \geq K$ or $S_T \lt K$.

Throughout this post, I’ll always derive the Black–Scholes Greek for a call option and then use put-call parity to more easily derive the Greek for a put. This approach is especially nice in light of the linearity of differentiation. Why does this work? Put-call parity is a principle of efficient markets, and thus an option pricing formula that admits no arbitrage should adhere to put-call parity. And we can convince ourselves that Black–Scholes adheres to put-call parity by deriving the Greeks for puts using both Equations $6$ and $7$ and observing that we get the same result either way. I will do this in appendices.

Now let’s dig into five of the most important Greeks: delta ($\Delta$), gamma ($\Gamma$), rho ($\rho$), theta ($\theta$), and vega ($\mathcal{V}$). See (Hull, 1993) and (Cox & Rubinstein, 1985) for textbook discussions of this topic.

Delta

As we saw above, delta ($\Delta$) captures the sensitivity, or more precisely the instantaneous rate of change, of the option price to the spot price. The Black–Scholes delta for a call option is

$\Delta(C) = \frac{\partial C}{\partial S} = \Phi(d_1), \tag{8}$

where $\Phi(x)$ and $d_1$ are defined in Equations $3$ and $4$. See A2 for a key lemma that will be used throughout the derivations of the Greeks and see A3 for a derivation of $\Delta$.

To understand delta, let’s start with an example. Suppose we have a stock with price $S = $100$, a call option with a price of $C = $10$, and a delta between the two of $0.7$. (The prices do not actually matter here, but they will in later examples.) Now imagine that we sell calls for $3000$ shares of this stock, and that we want to hedge our resulting short position. We can use delta to compute the appropriate size of our hedge! We should buy $0.7 \times 3000 = 2100$ shares of the underlying stock. Why? This is just a direct application of Equation $8$. First, let’s replace the differentials (e.g. $\partial C$) with small moves in the assets (e.g. $dC$). This is a fine thing to do because Equation $8$ says that $\Delta$ is the linear approximation of how the Black–Scholes price of our calls will change. We can write this as

$dC = \Delta \times dS. \tag{9}$

Now let $w$ denote the number of shares of the stock associated with the calls we sold (here $w = 3000$). Then by Equation $9$, we have

$dC \times w = \Delta \times dS \times w. \tag{10}$

Our proposed hedge above is $\Delta \times w = 2100$, and this works because if the stock price changes by $dS = $1$, then the call moves by $dC = $0.7$, and we will make $$2100$ on our hedge and lose $$0.7 \times 3000 = $2100$ on our calls. In practice, this calculation is made slightly more tedious because options are typically sold with a contract multiplier indicating the number of shares per contract, but this is just a matter of bookkeeping. Thus, while the options we sold have a delta of $0.7$, our actual portfolio position has a delta of zero. We say that our portfolio is delta neutral, and this means that if spot changes a little, the value of our options position does not change significantly.

Dollar delta

In the previous example, the raw values of the stock and call did not matter, and we had no sense of our total notional exposure. Thus, in practice, mathematical delta (Equation $8$) is often converted to dollar delta, sometimes called notional delta, because this quantity tells us how much notional exposure we have to the underlying. Dollar delta is just delta times spot or

$$\Delta = \Delta S. \tag{11}$

For example, our notional exposure in the previous example was the number of shares times dollar delta or

$\begin{aligned} $\Delta \times w &= \Delta \times S \times w \\ &= 0.7 \times $100 \times 3000 \\ &= $210,000. \end{aligned} \tag{12}$

To hedge, we should buy $$210,000$ of stock, which is obviously the same as saying we should buy $2100$ shares of stock at $S = $100$.

One benefit of thinking in a notional amount is that now we can multiply dollar delta by a percent change in the stock. In contrast, since delta represents a dollar change in stock move per dollar change in option move, we cannot directly multiply mathematical delta by a relative move such as a return or percent return. To see this, consider the returns in our running example. Let $S_1$ denote the stock price after move $dS$. Then the return on $S$ is

$R_S = \frac{S_1 - S}{S} = \frac{101 - 100}{100} = 0.01, \tag{13}$

while the return on $C$ to a new price $C_1$ is

$R_C = \frac{C_1 - C}{C} = \frac{\Delta dS}{C} = \frac{0.7}{10} = 0.07. \tag{14}$

And clearly

$\Delta R_S = 0.7 \times 0.01 = 0.007 \neq 0.07 = R_C. \tag{15}$

So it is nonsensical to multiply mathematical delta by a relative move. We must operate on it in a way that makes sense given Equation $8$.

However, we can multiply dollar delta by a relative move since

$\begin{aligned} $ \Delta \times R_S &= \Delta \times S \times \left( \frac{S_1 - S}{S} \right) \\ &= \Delta \times dS \\ &= dC. \end{aligned} \tag{16}$

So dollar delta is nice because it tells us both our notional exposure to the underlying and because we can multiply it by a relative move in the underlying to get a dollar change in the option.

Portfolio delta

Imagine we had a portfolio of $n$ options. We can represent the value of our portfolio, $V_p$, as a weighted sum of these $n$ assets,

$V_p = \sum_{i=1}^n w_i V_i, \tag{17}$

where the portfolio weights represent the quantity of each option, and where $V_i$ is the value of the $i$-th call or put. By the linearity of differentiation, we can easily compute the delta of this portfolio:

$\begin{aligned} \Delta (V_p) &= \frac{\partial V_p}{\partial S} \\ &= \sum_{i=1}^n w_i \frac{\partial V_i}{\partial S} \\ &= \sum_{i=1}^n w_i \Delta (V_i). \end{aligned} \tag{18}$

Thus, the delta of a portfolio can be computed by adding up the deltas of the individual options.

For example, recall our short position in call options from the example above. Now imagine that we additionally have a long position in call options worth of $2000$ shares of the same underlying, where the delta of each option is $0.6$. Then the delta of our portfolio in shares of stock is

$\begin{aligned} -3000(0.7) + 2000(0.6) &= -2100 + 1200 \\ &= -900. \end{aligned} \tag{19}$

Thus, our portfolio has a dollar delta of $-$90,000$, which we can make delta-neutral by buying $$90,000$ of stock. This makes sense. Before we had to notional hedge of $$210,000$. Now we have a notional hedge that is roughly half that, given we have a long position in calls that behave roughly like half a stock.

Delta hedging

In our leading example, we hedged our delta from our option position, a short position in calls. Now that we understand dollar delta and portfolio delta, let’s think a little bit about a more general way to formulate the problem. Imagine we have a portfolio with a delta of $\Delta_{\Pi}$. And imagine that another asset has a delta of $\Delta_i$. Then to be delta neutral, we want to buy $w_i$ of this asset such that

$w_i \Delta_i + \Delta_{\Pi} = 0. \tag{20}$

Solving for $w_i$, we get

$w_i = -\frac{\Delta_{\Pi}}{\Delta_i}. \tag{21}$

But what asset should we use here? Well, we could find an option with a delta of $\Delta_i$, but a simpler thing to do would be to buy the underlying asset, which has a delta of one ($\Delta_i = 1$). So our hedge in shares or dollars should be proportional to $w_i = \Delta_{\Pi}$, which is precisely what we see in Equation $10$. We can easily think about this in terms of shares or dollars by multiplying both sides of Equation $21$ as desired.

Functional form

Now that we have some sense of how delta is used, let’s dig into the precise definition of the Black–Scholes delta (Equation $8$). This is a wonderful result, but it may not be immediately obvious what it means. I’ll discuss a common interpretation in a moment. But first, note that since $\Phi(x)$ is the CDF of a symmetric distribution, we can immediately guess that shape of delta is roughly a sigmoid function, which it is (Figure $3$)! This also makes sense if we visualize the slope of the tangent line changing as we move along the Black–Scholes price curve in Figure $1$. Clearly, delta for a call must initially be nearly zero when $S \ll K$, must increase as $S$ approaches and passes $K$, and finally must flatten off to nearly one when $S \gg K$.

Notice that an at-the-money (ATM) option has a delta of slightly greater than $0.5$. This is because while $\Phi(x)$ is symmetric with respect to its input $d_1$, the variable $d_1$ is a logarithmic function of $S$. Confusingly, some people will claim that the delta of an ATM option is $0.5$. For example, Investopedia says

An ATM option has a delta of $±0.50$, positive if it is a call, negative for a put.

But this is demonstrably false. Just consider the full definition of delta for a call when $S = K$:

$\Delta(C) = \Phi\!\left( \frac{\left[ r + (1/2) \sigma^2 \right] \sqrt{T}}{\sigma} \right). \tag{22}$

Since $\Phi(x)$ is the CDF of the standard normal distribution, we can see that $\Delta(C) = 0.5$ only when $T = 0$, assuming interest rates and volatility are nonzero. But in my mind, it does not make much sense to talk about the delta of an ATM option at expiry. I’ll discuss this a bit more when discussing delta’s sensitivities to other parameters. Anyway, in most realistic scenarios (positive interest rates, nonzero volatility, and nonzero time to expiry), the term $d_1$ is greater than zero when $S=K$, and thus the Black–Scholes delta of an ATM option is slightly greater than $0.5$.

A deep in-the-money (ITM) option has a delta close to one, meaning that the option behaves like spot. And a deep out-of-the-money (OTM) option has a delta close to zero, meaning that the option is insensitive to the underlying’s price changes. I think it’s worth observing that many of these points should hold true for any option pricing model. It does not make sense to have an option pricing model whose delta is zero for an ITM option, for example. So the coherency of Black–Scholes Greeks with reality is one justification for using this particular model.

As a practical matter, we often refer to an option’s delta in percentage terms, so an option with a delta of $0.5$ is sometimes called a “$50$-delta option”. And of course, since a $50$-delta option is often considered ATM, this phrase is often used to describe ATM options.

It makes sense that the delta for a call option is in the range $[0, 1]$, since $\Delta$ it is the slope of the relationship between the spot and option (Figure $2$, left). Using put-call parity, we can immediately derive the delta for a put:

$\begin{aligned} \Delta(P) &= \frac{\partial}{\partial S} P \\ &= \frac{\partial}{\partial S} \left(C + Ke^{-rT} - S\right) \\ &= \Delta(C) - 1. \end{aligned} \tag{23}$

Since $\Delta(C)$ ranges in $[0, 1]$, clearly $\Delta(P)$ is in the range $[-1, 0]$. And this makes sense given the Black–Scholes price curve for a put (Figure $2$, right). So the delta for a call and put have the exact same shape for all input parameters, but are just shifted by one. See A3 for a derivation.

Moneyness

At this point, we have enough of an understanding of delta to investigate a common interpretation of delta, namely that it captures the moneyness of an option, how likely it is that the option ends ITM.

To understand this, we need to understand $\Phi(d_1)$ and $\Phi(d_2)$ from the Black–Scholes equation. $\Phi(d_2)$ has a clean interpretation. It is the probability that the option ends ITM in a risk-neutral world. If this is not clear to you, then I would recommend interpreting $\Phi(d_2)$ by mapping it onto the appropriate term in the binomial options-pricing formula, which is discrete-time analog of Black–Scholes (see the right term in brackets in Equation $26$ in this post).

But since the moneyness interpretation is true of $\Phi(d_2)$, it is only true of $\Phi(d_1)$ to the extent that $d_1 \approx d_2$. How are these terms different? They differ by the volatility of spot:

$d_2 = d_1 - \sigma \sqrt{T}. \tag{24}$

What this reasoning suggests is that the moneyness interpretation of delta is approximately true when volatility or time to expiry are low but is less true when volatility or time to expiry are high. When we compare $\Phi(d_1)$ and $\Phi(d_2)$ for varying volatilities, this is precisely the relationship we see (Figure $4$). For example, the delta $\Phi(d_1)$ and moneyness $\Phi(d_2)$ of a $\sigma=0.1$ option are closer in value than the delta and moneyness of a $\sigma=0.5$ option.

Delta sensitivities

As we have seen already, the precise shape of delta changes based on the other Black–Scholes inputs. Just as we can understand Black–Scholes by fixing all parameters but one, we can understand a single Greek such as delta by fixing all other parameters but one.

First, let’s look at how delta changes while holding all other parameters fixed as the time to expiry $T$ decreases (Figure $5$). I like to plot this with $T$ decreasing since this is means the graph can be read left-to-right as physical time moves forward. We can see that all options, regardless of moneyness, have more similar deltas when the expiration date is far enough out. Then as expiration approaches, ITM, ATM, and OTM options converge to deltas of $1$, $0.5$, and $0$ respectively for calls and $0$, $-0.5$, and $-1$ respectively for puts. In Figure $5$, I have only plotted these curves for $\Delta(C)$, since the graph for $\Delta(P)$ is identical except for a shift in the $y$-axis.

Another way to see this is to visualize how the Black–Scholes price curve (for calls) changes as a function of time. In the left subplot of Figure $2$, I plotted the Black–Scholes price curve for a call as the time to expiry $T$ decreases to zero. As we can see, when $T$ is large (here, $10$ years), the option’s delta is roughly $0.5$ over a large range of stock values. However, as the time to expiry approaches, the price curve becomes more hockey-stick shaped until the moment before expiry, at which point it is essentially a piecewise linear function with a discontinuity at the strike price. What does this tell us? Long before expiry, OTM and ITM calls have a similar delta, but these two deltas will converge to zero and one by expiry (all other values being fixed). And what happens to the delta of an ATM call? Well, its delta becomes increasingly sensitive to the precise value of the stock.

We can also think about how delta changes with respect to volatility (Figure $6$). When volatility decreases, OTM calls become less sensitive to the underlying, while ITM calls become more sensitive to the underlying.

This makes sense because in the absence of volatility, the world is predictable, and the delta function becomes a step function. Either the stock is worth more than the strike, in which case the option moves like a stock, or the stock is worth less than the strike, in which case the option is insensitive to the stock price. And in a world without volatility, it does not make sense to talk about the delta of ATM options, and we can see from Equation $22$ that this value is undefined. The only reason the discontinuity is not at $S=K$ is because of a nonzero interest rate. As with Figure $5$, the graph for puts is identical except for a shift in the $y$-axis.

Changing delta

Finally, it is important to remember that delta is a derivative and is therefore the slope of the tangent line to the Black–Scholes price curve at a particular spot price! When spot changes in value from $S$ to $S_1$, our original estimate of delta is no longer correct. It should be a new value $\Delta_1$, representing the slope of the line tangent to the Black–Scholes price curve at $S_1$ (Figure $7$).

Thus, hedging an options position by trading the underlying requires continuous rebalancing of the hedged position, due to this continuously accumulating hedging error. One way to visualize this ever-changing error is by visualizing the gap between the linear equations $\Delta S_1 + b$ and $\Delta_1 S_1 + b_1$, where $b$ and $b_1$ are the $y$-intercepts their respective deltas (Figure $8$, pink region). What would increase this gap between the two points? More curvature to the Black–Scholes price curve! Thus, in order to account for this hedging error, we could add a term that accounts for the curvature of the Black–Schole’s model a particular point. To do this, we use a second-order Greek called gamma.

Gamma

The Greek gamma ($\Gamma$) is an option’s sensitivity to delta. In other words, delta is a first-order Greek, while gamma is a second-order Greek, since gamma is a second partial derivative:

$\Gamma = \frac{\partial \Delta}{\partial S} = \frac{\partial \left( \frac{\partial V}{\partial S} \right)}{\partial S} = \frac{\partial^2 V}{\partial S^2}. \tag{25}$

The Black–Scholes gamma for a call option is

$\Gamma(C) = \frac{\partial \Delta(C)}{\partial S} = \frac{\varphi(d_1)}{S \sigma \sqrt{T}}, \tag{26}$

where $\varphi(x)$ is defined in Equation $5$. See A4 for a derivation.

Using our running example so far, the gamma of a call with inputs $S=$100$, $K=$100$, $\sigma=15\%$, $T=1$, and $r=4\%$ is roughly $0.025$. This means that delta changes by $0.025$ for every $$1$ change in spot. As with delta, we can replace the differentials above with moves in spot and delta to get our linear approximation of curvature:

$d\Delta = \Gamma \times dS. \tag{27}$

For example, in Figure $7$ spot changed by $dS=$20$. So our estimate of the change in delta using gamma estimated at $S$ is $d\Delta = $0.025 \times 20 = 0.5$. We can see that the actual change in delta was $0.94 - 0.63 = 0.31$. We overestimated the change in delta because the curvature at point $S$ is greater than the curvature at point $S_1$.

Dollar gamma

Much like delta, gamma is often converted to a notional value, called dollar gamma. While dollar delta is easy—it is just delta times spot or Equation $11$—dollar gamma is made complicated by the fact that there are many different definitions of it. See this Stack Exchange for an example of this confusion. I am going to use this definition:

$$\Gamma = \Gamma S^2. \tag{28}$

I like Equation $28$ because, as mentioned in that Stack Exchange post, it matches the definition of dollar gamma used by Peter Carr—see around Equation $9$ in (Carr & Madan, 2001)—and because it is analogous to dollar delta. An alternative definition of dollar gamma is to halve Equation $28$, so

$$\Gamma = \frac{1}{2} \Gamma S^2. \tag{29}$

In my mind, this is also a reasonable definition.

To see why these definitions make sense, let’s consider an analogy with physics. If we drive a car $50$ miles per hour for $2$ hours, we have gone $100$ miles. But what if we start at exactly $50$ miles per hour while accelerating at a constant $5$ mile per hours squared? This means that our velocity is increasing by $5$ miles per hour per hour. So after two hours, our total distance traveled is not $100$ miles but rather $110$ miles. Our change in position (distance) from baseline velocity is still just $100$ miles, but we also changed position due to acceleration. And acceleration times hours gives us our instantaneous velocity due to acceleration at each time point. If we multiply this by time again, we get the distance traveled due to these instantenous velocities. Of course, this relationship is captured in a classic equation of motion that most people learned in high school physics:

$p = vt + \frac{1}{2} at^2, \tag{30}$

where $p$ is position, $t$ is time, $v$ is velocity, and $a$ is acceleration. My argument here is that $p$ is analogous to the dollar change in our options position, $vt$ is analogous to dollar delta ($t = S$ and $v = \Delta$), and finally $at^2$ is analogous to dollar gamma. The former captures how much the notional value of our position has changed due to a first-order change in the underlying, while the latter captures how much the notional value of our position has changed due a second-order change in the underlying:

$\text{notional change in portfolio} = $ \Delta + \frac{1}{2} $\Gamma. \tag{31}$

And if we substitute our definitions of dollar delta and dollar gamma into Equation $29$, we get

$\text{notional change in portfolio} = \Delta S + \frac{1}{2} \Gamma S^2. \tag{32}$

And we can visualize this nicely. In Figure $8$, dollar delta is the blue region representing how much our portfolio has changed due to some notional change in spot price, while dollar gamma is the pink region representing how much our portfolio has changed due to the same notional change in spot price. If we add these two regions together, we get the total notional change in our portfolio.

You might be asking: aren’t we just doing a second-order Taylor approximation? Yes! To see this formally, let $f(x)$ denote the Black–Scholes pricing function. If we abuse notation a little bit and let $\Delta(S)$ and $\Gamma(S)$ denote the first and second partial derivatives of $f(x)$ evaluated at $S$, the first-order Taylor approximation of $f(x)$ around the point $S$ is

$f(x) \approx f(S) + \Delta(S)(x-S) + \frac{\Gamma(S)(x-S)^2}{2}. \tag{33}$

I visualize this in Figure $9$. To me this, really justifies Carr’s definition of $\Gamma S^2$ as dollar gamma. The $1/2$ is only from the Taylor expansion. If this section doesn’t make sense, see this excellent 3Blue1Brown video on the Taylor series, particularly around the $14\!:\!25$ mark.

Finally, yet another alternative definition of dollar gamma is to divide Equation $28$ by $100$, so

$$\Gamma = \frac{\Gamma S^2}{100}. \tag{34}$

(Note that yet another definition of dollar gamma I have seen is dividing it by $100^2$; see here.) There is a motivation for this definition, and I think it’s worth understanding. Consider dollar delta divided by $100$:

$\% \Delta = \frac{$ \Delta}{100}. \tag{35}$

Since dollar delta has units of dollars per return in $S$, the units of Equation $35$ are dollars per percent. For example, if we have a $$1000$ notional long position in $55$-delta calls, then we have a dollar delta of $$550$ and a dollar-delta-per-percent of $$5.5$ per $1\%$. What this represents is a change in dollar delta per a $1\%$ change in stock. Put differently, for every $1\%$ change in the underlying stock price, we earn $$5.5$ in our option position. For example, if the stock changed by $4\%$, then our dollar delta would change by $$22$. Particularly if we scale these values by the number of contracts to get our portfolio’s notional position, we can see that Equation $35$ is very useful for quickly estimating our notional risk from a percent change in the underlying.

The gamma analog to Equation $35$ is thus

$\% \Gamma = \frac{\Gamma S}{100}. \tag{36}$

This tells us how much our mathematical delta (Equation $8$) changes a percent change in the underlying. But if we’re interested in dollar delta, then we would multiply Equation $36$ by $S$, giving us the definition of dollar gamma in Equation $34$. In other words, Equation $34$ captures how dollar delta changes per a percent change in the underlying. In my mind, this definition is useful from a trading perspective but is less mathematically coherent. So I prefer to refer to Equation $28$ as dollar gamma, although Equation $34$ is useful in its own way.

Gamma hedging

As with delta, we may want to hedge our gamma exposure or make our portfolio gamma neutral. This means our position would be insensitive to large price shocks. However, it may not be immediately obvious how to do this, since the underlying asset has zero gamma:

$\frac{\partial^2 S}{\partial S^2} = \frac{\partial}{\partial S} 1 = 0. \tag{37}$

So unlike delta, we cannot simply buy some amount of the underlying to make our portfolio gamma neutral.

What we need instead is to gamma hedge with options. Imagine that our portfolio is delta neutral but has a gamma exposure of $\Gamma_{\Pi}$, and that an option $i$ has a gamma of $\Gamma_i$. We would like to buy $w_i$ of the option such that

$w_i \Gamma_i + \Gamma_{\Pi} = 0. \tag{38}$

This means we want to buy

$w_i = -\frac{\Gamma_{\Pi}}{\Gamma_i} \tag{39}$

of the option. After we do this, however, our delta hedge will be wrong. That’s because we just bought $w_i$ of an option, which means we have new delta exposure to the underlying. Since deltas add (see Equation $18$), the total unhedged delta of our portfolio is

$\Delta_{\Pi} + \Delta_i. \tag{40}$

We have already hedged $\Delta_{\Pi}$, and can be delta neutral if we hedge $\Delta_i$. This means we buy $w_i = -\Delta_i$ of the underlying (Equation $21$). Since we hedge our delta with spot and spot has no gamma exposure, we’re done.

Functional form

Now that we have some sense of how gamma can be used, let’s look at its Black–Scholes definition in detail.

Based on Equation $25$, we can immediately infer that gamma take values in $[0, 1]$, should be roughly symmetric with respect to $S$, and should decrease as $S$, $\sigma$ or $T$ increase. Again, I say “roughly” because while the normal distribution is symmetric with respect to its input, here the input to the PDF is not the stock price but $d_1$, which is logarithmic with respect to $S$. So much like delta, which is a slightly asymmetric sigmoid function, gamma is slightly asymmetric normal distribution (Figure $10$).

We can derive the gamma for a put option using put-call parity:

$\Gamma(P) = \frac{\partial}{\partial S} \left[ \Phi(d_1) - 1 \right] = \frac{\partial}{\partial S} \Phi(d_1) = \Gamma(C), \tag{41}$

which matches gamma for a put if we derive it directly from Black–Scholes (see A4). So gamma is the same for puts and calls!

I think there’s a nice way to visualize this equivalence. Consider Figure $11$. I have plotted the Black–Scholes price curve $f(x)$ (solid black line) for calls (left) and puts (right) as well as the delta approximation at a given point $S=$110$ (blue line). (I chose an off-center point to underscore the point being made here.) We can visualize the curvature of $f(x)$ by thinking about the gap between the blue tangent line and $f(x)$. We can see that for calls and puts, this gap is the same, suggesting that the curvature is the same. It is not a proof, and the gap itself is not gamma, but might give some sense for why gamma might be the same for calls and puts.

Finally, based on the lemma in A2, we could also write Equation $26$ as

$\Gamma = K e^{-rT} \frac{\varphi(d_2)}{S^2 \sigma \sqrt{T}}. \tag{42}$

This is useful to know since sometimes different authors write gamma in different ways.

Gamma sensitivities

As with delta, it is useful to visualize gamma changing as we vary one other parameter. As we already mentioned, it should be clear that gamma is small when an option is ITM or OTM, and it is bigger when the option is ATM. We can observe the first fact in Figure $10$ (gamma decreases as $S$ increases). This makes intuitive sense, as the option will become less sensitive to moves in the underlying the more ITM the option is. This is a different way of saying that delta “levels off” as $S$ increases, as it does in Figure $2$.

And as with delta, we can see how gamma changes as the time to expiry $T$ decreases (Figure $12$). We can see that the gammas for ITM and OTM options ultimately trend to zero. This makes sense, as these options are increasingly less sensitive to changes in delta. But ATM options become very sensitive to changes in delta as they near expiry. This also makes sense, as the Black–Scholes price curve becomes a hockey stick with a kink at $S=K$ (Figure $2$), and gamma is measuring this increasing curvature. In fact, gamma becomes so high close to expiry that I stopped plotting the ATM curve in Figure $12$ well before $T=0$. Otherwise, the plot would be unreadable.

Lastly, we can see how gamma changes as $\sigma$ changes (Figure $13$). In my mind, this result is fairly intuitive. If volatility increases, then for the same stock price, the option’s sensitivity to changes in the stock price is lower.

Rho

At this point, we have a good understanding of how an option’s price changes with respect to spot. Now let’s change tack and look at first-order Greeks with respect to other parameters. In the remainder of this post, we’ll look at theta ($\Theta$), rho ($\rho$), and vega ($\mathcal{V}$). I’ll start with two easier ones, rho and theta, and leave vega—and option’s sensitivity to volatility—for last.

The Greek rho ($\rho$) measures an option’s sensitivity to interest rates. The Black–Scholes rhos for calls and puts are

$\begin{aligned} \rho(C) &= \frac{\partial C}{\partial r} = T K e^{-r T} \Phi(d_2), \\ \rho(P) &= \frac{\partial P}{\partial r} = - T K e^{-rT} \Phi(-d_2). \end{aligned} \tag{43}$

See A5 for derivations. I placed the two definitions of $\rho$ above side-by-side because it is clear that they have opposite signs, and I think it’s useful to think about why this might be the case. Recall our justification for put-call parity. A call option plus a zero-coupon bond paying $K$ at time $T$—and is thus equal to $K e^{-rT}$ today—is equivalent to a put option plus the underlying stock. The key thing to think about here is how the bond is subject to interest rates. As interest rates increase, the present value of the strike (bond) $K e^{-rT}$ decreases—we need less money today in order for it to be worth the same amount at time $T$—and thus $C$ becomes relatively more expensive to $S$. Thus, with zero interest rates, ATM puts and calls with the same inputs are worth the amount (Figure $14$, left). With zero interest rates and with $S \neq K$, the difference between calls and puts is equal to $S - K$ (Figure $14$, middle and right). In my mind, this is why calls and puts have sensitivities to rates that are the opposite sign. As interest rates increase, the fair price of a call increases, while the fair price of a put decreases.

Now that we have some understanding for how calls and puts change with respect to interest rates, let’s look at an example of using the Greek rho. Let’s use our rolling example of an ATM call with $S=K=$100$, time to expiry of $T=1$ years, annualized volatility of $\sigma=15\%$, and an annualized interest rate of $r=4\%$. Using these values, the price of a Black–Scholes call option is roughly $$8$ (you can sanity check this number with Figure $14$). If we plug these values into Equation $43$ above, we get

$\rho(C) = \frac{\partial C}{\partial r} \approx $55 \text{ per $1\%$ change in annualized rates}. \tag{44}$

That’s a pretty big number, but what does it mean? Well, since Black–Scholes interest rates are annualized, then this means that a $1\%$ move in annualized interest rates will result in a $0.01 \times 55 = $0.55$ move in our call option. So rather than the fair price being $$8$, it will now be $$8.55$. And if this example was for a put option, the logic would be the same but the sign of rho would flip: rho for a put with the same inputs would be $\rho \approx -40$. This tracks with what we see in Figure $14$.

We can sanity check these changes in option values with an example. Imagine we bought $100$ shares of the underlying stock at its current price $S = $100$. So we need to pay $$10,000$ today. However, alternatively, we could buy calls for $100$ shares of stock, meaning we get exposure to $100$ shares of stock for $$8 \times 100 = $800$. Thus, as a buyer of calls, we have the same potential upside as the buyer of stock but we have $10,000 - 800 = $9200$ surplus dollars that we can put into zero-coupon bonds. At $4\%$ annualized interest, we would earn $$368$ in one year. If interest rates increased by $1\%$, then we would earn $$92$ more dollars. So to a first approximation, the value of the calls going up $$0.55 \times 100 = $55$ seems reasonable.

Functional form

Finally, let’s look at the functional form of rho with respect to spot price $S$ (Figure $15$). We can see that rho for a call looks a lot like delta (Figure $3$), which makes sense since rho for a call is the CDF of standard normal $\Phi(d_2)$, scaled by $T K e^{-r T}$. And the graph for a put would be similar but shifted along the $y$-axis.

Given this similarity to delta, it seems straightforward to quickly approximate rho for an ATM option. For example, imagine that an option expires in one week or $T = 7/365 \approx 0.02$, that $K=$100$ and that $r=4\%$. Then $-r T \approx -0.0008$, which means that $e^{-rT} \approx 0.9992$—think about the first-order Taylor approximation if this does not make sense—and so $\rho \approx 2 \times 1 \times 0.5$. So $\rho$ should be close to unity. The true value is approximately $0.98$. This value is significantly smaller than our previous example (Equation $44$) due a much smaller time to expiry.

Rho hedging

To my knowledge, rates hedging is less widely discussed in the derivatives-pricing literature. I suspect this is because are interest rates are relatively constant and because changes in rates are predictable in the sense that they happen after central bank announcements. Thus, an options dealer simply needs to re-hedge their rates exposure around these events. Furthermore, one can hedge rates without options via fixed-income assets such as bonds.

As with gamma, we cannot hedge our rates exposure with the underlying asset, which has zero rho. But we need to find an asset $i$ with rates exposure $\rho_i$ such that

$w_i \rho_i + \rho_{\Pi} = 0. \tag{45}$

We could hedge our rates exposure with an option, but another idea would be to use a fixed-income asset. This underscores an important idea, which is that we can think of other types of assets such as futures or bonds as having Greeks as well. Here, we can think of the rho of a bond as a linear approximation of how the bond price changes per a change in rates.

Theta

The Greek theta ($\Theta$) captures the sensitivity of the option price to the passage of time. The Black–Scholes theta for a call option is

$\Theta(C) = -\frac{\partial C}{\partial T} = -\frac{S \sigma \varphi(d_1)}{2 \sqrt{T}} - r K e^{-r T} \Phi(d_2). \tag{46}$

See A6 for a derivation. This Greek is often referred to as time decay because an option’s value typically (but not always!) decreases over time if all other variables are held constant. Why? My intuition is this: imagine that an option is fairly priced today but then time passes without any change to the interest rate, the underlying, and so forth. Then the option is worth less because the optionality, which was priced in, has decreased. For example, a call option is a bet on the stock going up, and now the stock has less time for the stock to go up.

One easy way to verify this is to simply plot the Black–Scholes price of an option as $T$ approaches $0$ (Figure $16$). We can see that the value of calls is decreasing for both OTM and ITM options, but it decays to zero for OTM options and to nonzero values for ITM options. To be clear, theta is measuring how fast each option’s value is decaying, so options that start at a higher value but still end in zero must have a faster time decay.

As before, let’s start with an example. Imagine an ATM call option has a spot and strike price of $$100$, has an implied volatility of $15\%$, and expires in a quarter of the year ($T=0.25$). And imagine that the annual interest rate is $4\%$. Then the option’s theta is

$\frac{\partial C}{\partial T} \approx -$8.03 \text{ per year}. \tag{47}$

How do we interpret this number? As with the other Greeks, we can multiply it by a change in time to get a linear approximation of the change in our option’s value. For example, to approximate how our option has changed in value after one week has passed, we multiply theta by a change in time (annualized):

$-8.03 \times \left( \frac{7}{365} \right) \approx -$0.15 \text{ per week}. \tag{48}$

So all else being constant, this call option’s value will decay by $$0.15$ per week. Note that we can convert theta to a daily value by simply dividing by the number of days. So here, the daily theta is

$-\frac{8.03}{365} \approx $-0.02 \text{ per day}. \tag{49}$

Now we can simply multiply this quantity by a number of days to get how our option’s value decays in notional terms over that period.

Negative sign

Before looking at the functional form, let’s ask a more basic question. Why does theta have a negative sign in Equation $46$? In this post so far, we have denoted the time to expiry as $T$, and we imagine that $T$ is decreasing. But we typically think of time as increasing. So we need a negative sign to capture this inversion between the derivative and our model of time. If you’re unconvinced, you could write the time to expiry as

$T - t \tag{50}$

where $T$ is a fixed constant, and $t$ is physical time, which is now increasing. So before, $T$ was the time to expiry but decreased, and now $T-t$ is the time to expiry and decreases as $t$ increases. To compute theta here, we take partial derivatives with respect to $t$, and the chain rule gives us the negative sign, e.g.

$\frac{\partial}{\partial t} \sqrt{T-t} = -\frac{1}{2}\left(T-t\right)^{-1/2}. \tag{51}$

So the negative sign is a convention due to how we think about time, and it is coherent across different ways of modeling the problem.

Functional form

Now let’s dig into the Black–Scholes functional form of theta.

Broadly, we might expect theta to be often negative, but at least for me, Equation $46$ is a bit more complicated to intuitive than the other Greeks. So let’s just plot theta as a function of the stock price $S$ and see what we get (Figure $17$). What we find is that theta for a call is non-positive for OTM, ATM, and ITM options, with a small negative peak around the strike. And I think this makes sense. A deep OTM option is worthless and must decay to zero at expiry; therefore, its time decay is close to zero. An ITM call has a nonzero value at expiry, and thus will have a smaller time decay. Finally, an ATM call is worth somewhere between an OTM and ITM call today, but its value at expiry is zero. Thus, an ATM call has the fastest time decay. As with the other Greeks, the peak is not precisely when $S = K$. We can also see this in Figure $16$.

As before, we can use put-call parity to derive theta for a put:

$-\frac{\partial}{\partial T} P = -\frac{S \sigma \varphi(d_1)}{2 \sqrt{T}} + r K e^{-rT} \Phi(-d_2). \tag{52}$

Again, see A6 for a derivation. At a high level, the key idea is that the partial derivative with respect to $T$ results in a nonzero interest-rate term:

$-\frac{\partial C}{\partial T} + r K e^{-rT} = -\frac{\partial P}{\partial T}. \tag{53}$

As we can see, theta for a put is equivalent to theta for a call, plus some non-negative offset $r K e^{-rT}$ assuming non-negative interest rates. In fact, this observation is why I put this section on theta after the section on rho. Equation $53$ does not make intuitive sense unless we first appreciate how calls and puts behave as interest rates change. As interest rates go up, calls become more expensive, and puts become cheaper. And if an option is worth more (less) today but has the same value at expiry, then it must have a faster (slower) time decay.

For calls with non-negative interest rates, theta is non-positive across OTM, ATM, and ITM options (Figure $17$). However, for puts with non-negative interest rates, theta is negative when the option is ATM or OTM, but it is positive when the option is ITM (Figure $18$). This means that for an ITM put, it’s value is increasing with time! This may be surprising but it is true (Figure $20$, right). Why? One way to think about it is that present value of the eventual payoff for a put is

$K e^{-rT} - S. \tag{54}$

So as the time to expiry $T$ decreases, the term $K e^{-rT}$ increases. So an ITM put is increasing in value with the passage of time. We see the opposite effect with calls. This effect only occurs when $r \gt 0$. When $r = 0$, then a deep ITM put is simply worth $K-S$ across time (Figure $19$, left).

Theta hedging

While $S$ is a random variable—and the other Black–Scholes parameters are also often modeled as random variables rather than market-specified constants—time to expiry $T$ is non-random. Therefore, while it’s important to understand theta conceptually, it is not a Greek that is typically hedged. We could hedge our time exposure, of course, by finding an asset $i$ with time decay $\Theta_i$ such that

$w_i \Theta_i + \Theta_{\Pi} = 0. \tag{55}$

However, this does not offer any utility, as our position is not subject to unpredictable “time shocks”.

Vega

Finally, the Greek vega ($\mathcal{V}$) measures an option’s sensitivity to the standard deviation $\sigma$ of the underlying asset, or the underlying’s volatility:

$\mathcal{V} = \frac{\partial V}{\partial \sigma} = S \varphi(d_1) \sqrt{T}. \tag{56}$

We can easily see via put-call parity that vega will be the same for calls and puts. See A7 for a derivation. As an aside, “vega” is not the name of a Greek letter, and I’m unsure of the etymology. See this Stack Exchange for a discussion.

As a caveat, Black–Scholes assumes that the volatility of the underlying stock is constant. So it is a bit conceptually inconsistent to use the Black–Scholes vega since the entire derivation of Black–Scholes assumes constant volatility. However, in practice, this is done as a useful first approximation. Since the volatility of the stock is unobservable, $\sigma$ is often referred to as the implied volatility, since we can observe only the market-implied volatility from the price of options.

Let’s start with an example. Again, we’ll use values $S=K=$100$, $\sigma=15\%$, $T=1$ years, and $r=4\%$. With these values, the Black–Scholes vega of either a call or a put is

$\frac{\partial V}{\partial \sigma} \approx $37.6 \text{ per $1\%$ change in annualized volatility}. \tag{57}$

So if the annualized implied volatility increases from $15\%$ to $16\%$, vega approximates the option will change by $37.6 \times 0.01 = $0.376$. The actual price change using Black–Scholes is $$0.377$.

Dollar vega

We multiplied delta by spot to get our notional exposure to spot. And we multiplied gamma by spot squared since it is the second-partial derivative with respect to spot. Using similar logic, we can compute a dollar vega by multiplying our vega by an implied volatility $\sigma$:

$$\mathcal{V} = \mathcal{V} \sigma. \tag{58}$

Here, $\sigma$ is he market-implied volatility using Black–Scholes. Alternatively, we could think about a percentage vega or

$\%\mathcal{V} = \mathcal{V} \times 1\%. \tag{59}$

This represents our notional vega per a $1\%$ move in annualized implied volatility. This is a more practical number than raw vega. For example, in our example above, the percent vega would be $$0.376$. This is interpretable since it says how much our position changes in dollar terms per a $1\%$ change in implied volatility. And since it is in notional terms, we can multiply it by shares to the percent vega in terms of our portfolio.

Vega hedging

As with gamma, rho, and theta, a position in spot has zero vega exposure, since

$\frac{\partial S}{\partial \sigma} = 0. \tag{60}$

What this means is that if our portfolio position has vega $\mathcal{V}_{\Pi}$, we cannot neutralize the risk by taking a position in the underlying. Again, we need to find an asset $i$ with vega $\mathcal{V}_{i}$ and achieve vega neutrality by taking a position $w_i$ in the asset such that

$w_i \mathcal{V}_i + \mathcal{V}_{\Pi} = 0. \tag{61}$

Again, we could use a options, or any other asset with appropriately matched vega, to construct a portfolio which is vega neutral. My understanding is that an options market maker will keep their book vega neutral by simply selling vega to other clients via selling particular options.

As with gamma hedging, however, vega hedging might re-introduce exposures that will need to be re-hedged. This should make intuitive sense. First, we need to find a linear combination of options such that our portfolio’s gamma and vega exposures are both zero. Then we can zero out our delta, since doing so does not effect the other Greeks. See section $19.8$ on vega in (Hull, 1993) for more details.

Functional form

The functional form of vega is, like gamma, a scaled PDF of the normal distribution (Figure $20$). Intuitively, an option’s sensitivity to changes in implied volatility should decrease as either the spot price or time to expiry decrease. It also makes sense that vega should peak around the ATM point, when $S \approx K$. Why? Imagine an option is deep OTM. Even if the stock has a very high realized volatility—imagine a meme stock that whose price is whipping up and down while still remaining far below the strike—the fair price of the option does not need to change. Thus, vega should be small for deep OTM or deep ITM options.

As with gamma, the lemma in A2 allows us to write vega in terms of the strike price today, $K e^{-rT}$, rather than in terms of the spot price $S$:

$\mathcal{V} = K e^{-rT} \varphi(d_2) \sqrt{T}. \tag{62}$

This equation is nice because it has a clean interpretation. Recall that $\varphi(d_2)$ is the probability that an option ends ITM under the risk-neutral measure. And $K e^{-rT}$ is the present value of the strike price (at time $T$) or perhaps the fair price of a zero-coupon bond. So we can see that vega is like an expectation! The option either ends ITM and has a fair price of $K e^{-rT}$ with probability $p := \varphi(d_2)$ or ends OTM and is worthless with probability $1 - p$, or

$\begin{aligned} \mathcal{V} &= (K e^{-rT} \sqrt{T}) \times p + 0 \times (1 - p) \\ &= K e^{-rT} \sqrt{T} \varphi(d_2). \end{aligned} \tag{63}$

I have not read anyone write about vega this way, but in my mind, it seems like one view of vega is a kind expected value of the option.

Finally, note that given Equations $26$ and $56$, we can easily express gamma in terms of vega or vice versa:

$\mathcal{V} = \Gamma S^2 \sigma T. \tag{64}$

So if we are an option’s trader and know our gamma risk and volatility, you can do some quick math to estimate vega.

Vega sensitivities

We can see from Equation $56$ that vega increases as either the spot price $S$ or time to expiry $T$ increase. This is pretty intuitive. For example, if an option expires in one year and the implied volatility increases, there is a larger range in final outcomes for the option than if it expired in one day (Figure $21$, left).

When implied volatility decreases, vega for a particular spot price will decrease (Figure $21$, right). This effect is a bit subtle, and I’ve drawn the Figure $21$ slightly differently to highlight this. And of course, it should be clear that vega is highest when an option is roughly ATM. As with other Greeks, this makes intuitive sense, since this is when the option’s value could go either way. In this case, the option is particularly sensitive to its inputs.

Unifying view

As we have seen, the Greeks are complex topic. It can be tricky to reason about how a high-dimensional, nonlinear price surface changes as its various inputs change. So in conclusion, let me try to unify this into a coherent story.

First, a mathematical story. Recall that Taylor’s theorem for univariate functions can be generalized to higher-dimensional functions. If we view our derivatives-pricing function $f(S; z_1, \dots, z_k)$ as one such multidimensional function—rather than as a univariate function of just spot—, then the Greeks are the partial derivatives in the multidimensional Taylor expansion! Using the Greeks in this post, we could write the Taylor expansion as

$dV = \Delta dS + \rho dr + \Theta dT + \mathcal{V} d\sigma + \frac{1}{2} \Gamma (dS)^2 + \dots. \tag{65}$

As we add higher-order derivatives for a particular input, our approximation along that dimension will get better provided higher-order derivatives are nonzero. And as the various inputs change, this multidimensional function representing the fair price of the option changes. If we want to hedge a particular sensitivity or Greek, then we are constantly rebalancing our portfolio to re-neutralize our position. So this is one coherent view of the Greeks.

Second, a financial story. One view of Equation $65$ is that $dV$ is really the profit and loss (P&L) from our option position. This means that each term in the approximation is a separate P&L: we have a delta P&L, a rho P&L, a vega P&L, and so forth. And since we can hedge individual or combinations of Greeks, we could use a portfolio of options to trade one of the inputs to Black–Scholes! We would hedge out all the Greeks except one, and when that input moved, our position in options would change in value, capturing the relevant P&L.

But why would you do this? First, most of the inputs to Black–Scholes can either be traded directly, such as delta via stocks or interest rates via fixed-income products like bonds, or they cannot be traded, such as time. However, there is one input to Black–Scholes that is unobservable, difficult to trade, and impacts our total P&L: implied volatility! And vega P&L is the key. We can construct a portfolio that has a vega exposure, allowing us to convert changes in implied volatility into P&L in our options. This is why volatility trading is so intimately tied to options trading (Carr & Madan, 2001).

But there’s a second problem with this approach. Imagine that we try to trade volatility by buying a basket of options that has limited exposure to all Greeks but vega. And now imagine that the stock price is extremely volatile. Do we make money? The answer, as we can see in Figure $20$, is that it depends. It depends because vega is only nonzero for options within a certain moneyness range. Far OTM and far ITM options have effectively zero vega, and therefore even though the stock price might be volatile, we do not necessarily capture a meaningful change in option prices if vega is negligible. So options via the Greeks allow us to trade the inputs to our derivative-pricing model, but they do not necessarily let us trade those inputs cleanly.

A natural evolution in this line of thinking is to construct derivatives that have constant exposure to vega. This is the essential idea of a variance swap, which is a financial contract designed to have exposure to pure volatility (Demeterfi et al., 1999). And if we want to trade variance swaps, it makes sense to want an index that tracks the fair price of such a swap. This is the idea behind the VIX (Whaley, 1993; Whaley, 2009), which is essentially the fair price of a volatility swap.

And so we can trace a thread from an old idea to a new one. Trading options is at least several hundred years old (Haug & Taleb, 2011), while trading volatility is a relatively modern idea, one which arguably started with the observation that options imply a volatility (Latane & Rendleman, 1976). And at the center of these two seemingly distinct ideas is the Greeks.

   

Appendix

A1. Deriving $\Phi^{\prime}(f(x))$

Let $F(x)$ denote the CDF of a random variable $X$ with PDF $f(x)$. We can write $F(x)$ in terms of $f(x)$:

$F(x) = \int_{-\infty}^x f(z) dz. \tag{A1.1}$

Using the fundamental theorem of calculus, it is easy to see that the CDF must be the anti-derivative of the PDF:

$\frac{d}{dx} F(x) = f(x). \tag{A1.2}$

But I was curious: what is the derivative of $F$ when $x$ is a function of another function $b$? Consider this case:

$F(b(x)) = \int_{-\infty}^{b(x)} f(z) dz. \tag{A1.3}$

To compute its derivative, we can decompose the integral into two parts:

$F(b(x)) = \int_{-\infty}^a f(z) dz + \int_{a}^{b(x)} f(z) dz \tag{A1.4}$

for some constant $a$. The derivative is thus

$\begin{aligned} \frac{d}{dx} F(b(x)) &= \frac{d}{dx} \int_{-\infty}^a f(z) dz + \frac{d}{dx} \int_{a}^{b(x)} f(z) dz \\ &= \frac{d}{dx} \int_{a}^{b(x)} f(z) dz. \end{aligned} \tag{A1.5}$

Note that the left integral is zero because the term is constant with respect to $x$. So we have

$\frac{d}{dx} F(b(x)) = \frac{d}{dx} \int_{a}^{b(x)} f(z) dz. \tag{A1.6}$

Let $a(x) = a$ be a constant. We can then invoke Leibniz integral rule to compute

$\frac{d}{dx} \int_{a(x)}^{b(x)} f(z) dz = f(b(x)) \frac{d}{dx} b(x) - f(a(x)) \frac{d}{dx} a(x) + \int_{a(x)}^{b(x)} \frac{\partial}{\partial x} f(z) dz. \tag{A1.7}$

But the middle and right terms are both zero because their respective derivatives are zero, since each function is constant with respect to $x$. This gives us

$\frac{d}{dx}F(b(x)) = \frac{d}{dx} \int_{-\infty}^{b(x)} f(z) dz = f(b(x)) \frac{d}{dx} b(x). \tag{A1.8}$

Mapping this onto Black–Scholes, we can see that we can take partial derivatives such as

$\frac{\partial}{\partial S} \Phi(d_1) = \varphi(d_1) \frac{\partial}{\partial S} d_1. \tag{A1.9}$

The reasoning above holds in general, i.e. for other partial derivatives.

A2. Risk neutrality

Under the risk-neutral measure, the expected following equality must hold:

$S \varphi(d_1) = K e^{-rT} \varphi(d_2). \tag{A2.1}$

Intuitively, this equation states that the stock price is equal to the strike price, while adjusting for factors such as volatility, interest rates, and the probability of the option ending in-the-money. We want to prove that this equation holds under Black–Scholes, as we will use this result repeatedly.

Observe that by the definition of $\varphi(x)$, the PDF of the standard normal distribution, we have

$\begin{aligned} \varphi(d_2) &= \frac{1}{\sqrt{2\pi}} \exp\left\{ -d_2^2 / 2\right\} \\ &= \frac{1}{\sqrt{2\pi}} \exp\left\{ -\left(d_1 - \sigma \sqrt{T} \right)^2 / 2\right\} \\ &= \frac{1}{\sqrt{2\pi}} \exp\left\{ -\left(d_1^2 + \sigma^2 T - 2 d_1 \sigma \sqrt{T} \right) / 2\right\} \\ &= \frac{1}{\sqrt{2\pi}} \exp\left\{ -d_1^2 / 2 \right\} \exp\left\{ \left( - \sigma^2 T + 2 d_1 \sigma \sqrt{T} \right) / 2\right\} \\ &= \varphi(d_1) \exp\left\{ \left( - \sigma^2 T + 2 d_1 \sigma \sqrt{T} \right) / 2\right\}. \end{aligned} \tag{A2.2}$

And by the definition of $d_1$, note that

$2 d_1 \sigma \sqrt{T} = 2 \log(S/K) + 2rT + \sigma^2 T. \tag{A2.3}$

Putting this together, we get

$\begin{aligned} \varphi(d_2) &= \varphi(d_1) \exp\left\{ \left( - \sigma^2 T + 2 \log(S/K) + 2rT + \sigma^2 T \right) / 2\right\} \\ &= \varphi(d_1) \exp\left\{ \log(S/K) + rT \right\} \\ &= \varphi(d_1) \frac{S}{K} e^{rT} \end{aligned} \tag{A2.4}$

or that

$K e^{-rT} \varphi(d_2) = S \varphi(d_1), \tag{A2.5}$

as desired.

Intuitively, this means that the Black–Scholes model adheres to the notional of risk-neutrality or that it satisfies the no-arbitrage condition.

As a final note, observe that since $\varphi(x)$ is the PDF of symmetric distribution, we know that $\varphi(x) = \varphi(-x)$. So we can write this lemma as do $K e^{-rT} \varphi(-d_2) = S \varphi(d_1), \tag{A2.6}$

If you would like, you can convince yourself that this is true by plugging in $-d_2$ rather than $d_2$ in Equation $\text{A}2.2$. You’ll see this symmetry is enforced by the square.

A3. Deriving delta

From A1, we know we need to compute the partial derivative of $d_1$ with respect to $S$. This is

$\begin{aligned} \frac{\partial}{\partial S} d_1 &= \frac{\partial}{\partial S} \left[ \frac{1}{\sigma \sqrt{T}} \left[ \log(S/K) + \left(r + \frac{\sigma^2}{2}\right) T \right] \right] \\ &= \frac{\partial}{\partial S} \left[ \frac{\log(S)}{\sigma \sqrt{T}} \right] \\ &= \frac{1}{S \sigma \sqrt{T}}. \end{aligned} \tag{A3.1}$

We can then use the definition $d_2$ to compute

$\frac{\partial}{\partial S} d_2 = \frac{\partial}{\partial S} (d_1 - \sigma \sqrt{t}) = \frac{\partial}{\partial S} d_1. \tag{A3.2}$

Now let us complete the derivation:

$\begin{aligned} \Delta(C) &= \frac{\partial C}{\partial S} \\ &= \frac{\partial}{\partial S} \left[ S\Phi(d_1) - Ke^{-rT} \Phi(d_2) \right] \\ &= \frac{\partial}{\partial S} S \Phi(d_1) - Ke^{-rT} \frac{\partial}{\partial S} \Phi(d_2) \\ &= S \varphi(d_1) \frac{\partial}{\partial S} d_1 + \Phi(d_1) - Ke^{-rT} \varphi(d_2) \frac{\partial}{\partial S} d_2 \\ &= \frac{\varphi(d_1)}{\sigma \sqrt{T}} + \Phi(d_1) - \frac{Ke^{-rT} \varphi(d_2)}{S \sigma \sqrt{T}}. \end{aligned} \tag{A3.3}$

Now applying the lemma in A2 to Equation $\text{A}3.3$, we get:

$\Delta(C) = \Phi(d_1), \tag{A3.4}$

as desired.

The derivation for delta for a put is essentially the same:

$\begin{aligned} \Delta(P) &= \frac{\partial P}{\partial S} \\ &= \frac{\partial}{\partial S} \left[ K e^{-rT} \Phi(-d_2) - S \Phi(-d_1) \right] \\ &= K e^{-rT} \frac{\partial}{\partial S} \Phi(-d_2) - \frac{\partial}{\partial S} \left[S \Phi(-d_1)\right] \\ &= K e^{-rT} \varphi(-d_2) \frac{\partial}{\partial S} (-d_2) - \left[\Phi(-d_1) + S \varphi(-d_1) \frac{\partial}{\partial S} (-d_1)\right] \\ &= -K e^{-rT} \varphi(-d_2) \frac{\partial}{\partial S} d_2 - \Phi(-d_1) + S \varphi(-d_1) \frac{\partial}{\partial S} d_1. \end{aligned} \tag{A3.5}$

Applying Equation $\text{A3.2}$ and the lemma in A2, we get

$\begin{aligned} \Delta(P) &= -S \varphi(-d_1) \frac{\partial}{\partial S} d_1 - \Phi(-d_1) + S \varphi(-d_1) \frac{\partial}{\partial S} d_1 \\ &= -\Phi(-d_1). \end{aligned} \tag{A3.6}$

Since the normal distribution is symmetric, we can write this as

$-\Phi(-d_1) = -\left[ 1 - \Phi(d_1) \right] = \Phi(d_1) - 1 \tag{A3.7}$

as desired.

A4. Deriving gamma

After deriving delta, deriving gamma is easy. It is

$\Gamma(C) = \frac{\partial}{\partial S} \Phi(d_1) = \Phi^{\prime}(d_1) \frac{\partial}{\partial S} d_1 = \frac{\varphi(d_1)}{S \sigma \sqrt{T}}, \tag{A4.1}$

where we have used the fact that

$\frac{\partial}{\partial S} d_1 = \frac{1}{S \sigma \sqrt{T}}, \tag{A4.2}$

from the derivation of delta. We can see from the definition of delta for a put (Equation $21$) that gamma for a put should be the same as gamma for a call:

$\Gamma(P) = \frac{\partial}{\partial S} \left[ \Phi(d_1) - 1 \right] = \Gamma(C). \tag{A4.3}$

Alternatively, we could use put-call parity to derive Equation $\text{A}4.3$.

A5. Deriving rho

Using the chain and product rules, we have

$\begin{aligned} \frac{\partial C}{\partial r} &= \frac{\partial}{\partial r} \left[ S \Phi(d_1) - K e^{-rT} \Phi(d_2) \right] \\ &= S \varphi(d_1) \frac{\partial}{\partial r} d_1 - \left[ -T K e^{-rT} \Phi(d_2) + K e^{-rT} \varphi(d_2) \frac{\partial}{\partial r} d_2 \right] \\ &= S \varphi(d_1) \frac{\partial}{\partial r} d_1 + T K e^{-rT} \Phi(d_2) - K e^{-rT} \varphi(d_2) \frac{\partial}{\partial r} d_1 \\ &\stackrel{\star}{=} T K e^{-rT} \Phi(d_2). \end{aligned} \tag{A5.1}$

Terms cancel in step $\star$ because of the lemma in A2 and since

$\frac{\partial}{\partial r} d_1 = \frac{\partial}{\partial r} d_2. \tag{A5.2}$

The derivation for a put is nearly identical:

$\begin{aligned} \frac{\partial P}{\partial r} &= \frac{\partial}{\partial r} \left[ K e^{-rT} \Phi(-d_2) - S \Phi(-d_1) \right] \\ &= -S \varphi(d_1) \frac{\partial}{\partial r} (-d_1) + \left[ -T K e^{-rT} \Phi(-d_2) + K e^{-rT} \varphi(-d_2) \frac{\partial}{\partial r} (-d_2) \right] \\ &= S \varphi(d_1) \frac{\partial}{\partial r} d_1 - T K e^{-rT} \Phi(-d_2) - K e^{-rT} \varphi(d_2) \frac{\partial}{\partial r} d_1 \\ &= - T K e^{-rT} \Phi(-d_2). \end{aligned} \tag{A5.3}$

Alternatively, we could use put-call parity:

$\begin{aligned} \frac{\partial}{\partial r} P &= \frac{\partial}{\partial r} C + \frac{\partial}{\partial r} K e^{-rT} - \frac{\partial}{\partial r} S \\ &= T Ke^{-rT} \left[ \Phi(d_2) - 1 \right] \\ &= -T K e^{-rT} \Phi(-d_2). \end{aligned} \tag{A5.4}$

In the last step, we have used the symmetry of the normal distribution:

$\Phi(x) - 1 = -\Phi(-x). \tag{A5.5}$

A6. Deriving theta

The derivation of theta is a bit tedious since we will need to use the product rule. So let us break it into parts. First, observe that:

$\frac{\partial}{\partial T} d_2 = \frac{\partial}{\partial T} d_1 - \frac{\sigma}{2 \sqrt{T}}. \tag{A6.1}$

And for brevity, let’s define a function $g(T)$ as

$g(T) := K e^{-rT}, \tag{A6.2}$

which implies that

$\frac{\partial g}{\partial T} = - r K e^{-r T}. \tag{A6.3}$

Finally, observe

$\begin{aligned} &\frac{\partial}{\partial T} g(T) \Phi(d_2) \\ &= g^{\prime}(T) \Phi(d_2) + g(T) \Phi^{\prime}(d_2) \frac{\partial}{\partial T} d_2 \\ &= - r K e^{-r T} \Phi(d_2) + K e^{-r T} \Phi^{\prime}(d_2) \frac{\partial}{\partial T} d_2. \end{aligned} \tag{A6.4}$

We can now put these observations together to derive $\Theta$:

$\begin{aligned} -\Theta(C) &= \frac{\partial C}{\partial T} \\ &= -\frac{\partial}{\partial T} \left[ S \Phi(d_1) - K e^{-r T} \Phi(d_2) \right] \\ &= S \Phi^{\prime}(d_1) \frac{\partial}{\partial T} d_1 - \frac{\partial}{\partial T} \left[ g(T) \Phi(d_2) \right] \\ &= S \varphi(d_1) \frac{\partial}{\partial T} d_1 - \left[ - r K e^{-r T} \Phi(d_2) + K e^{-r T} \Phi^{\prime}(d_2) \frac{\partial}{\partial T} d_2 \right] \\ &= S \varphi(d_1) \frac{\partial}{\partial T} d_1 + r K e^{-r T} \Phi(d_2) - K e^{-r T} \varphi(d_2) \frac{\partial}{\partial T} d_2 \\ &= S \varphi(d_1) \frac{\partial}{\partial T} d_1 - K e^{-r T} \Phi^{\prime}(d_2) \frac{\partial}{\partial T} d_2 + r K e^{-r T} \Phi(d_2) \\ &\stackrel{\star}{=} S \varphi(d_1) \frac{\partial}{\partial T} d_1 - S \varphi(d_1) \frac{\partial}{\partial T} d_2 + r K e^{-r T} \Phi(d_2) \\ &= S \varphi(d_1) \left[ \frac{\partial}{\partial T} d_1 - \frac{\partial}{\partial T} d_2 \right] + r K e^{-r T} \Phi(d_2) \\ &= S \varphi(d_1) \left[ \frac{\partial}{\partial T} d_1 - \left[\frac{\partial}{\partial T} d_1 - \frac{\sigma}{2 \sqrt{T}} \right] \right] + r K e^{-r T} \Phi(d_2) \\ &= S \varphi(d_1) \left[ \frac{\sigma}{2 \sqrt{T}} \right] + r K e^{-r T} \Phi(d_2) \\ &= \frac{S \sigma \varphi(d_1)}{2 \sqrt{T}} + r K e^{-r T} \Phi(d_2). \end{aligned} \tag{A6.5}$

In step $\star$, we use the lemma in A2. Negate the result above to get Equation $27$.

Now let’s derive theta for a put from the Equation $6$. As before, let’s address the term with the product rule separately. We will see that it is:

$\begin{aligned} \frac{\partial}{\partial T} g(T) \Phi(-d_2) &= g^{\prime}(T) \Phi(-d_2) + g(T) \varphi(-d_2) \frac{\partial}{\partial T} (-d_2) \\ &= - r K e^{-rT} \Phi(-d_2) - K e^{-rT}\varphi(-d_2) \frac{\partial}{\partial T} d_2. \end{aligned} \tag{A6.6}$

Putting this all together, we get

$\begin{aligned} -\Theta(P) &= \frac{\partial P}{\partial T} \\ &= \frac{\partial}{\partial T} \left[ K e^{-r T} \Phi(-d_2) - S \Phi(-d_1) \right] \\ &= - r K e^{-rT} \Phi(-d_2) - K e^{-rT}\varphi(-d_2) \frac{\partial}{\partial T} d_2 + S \varphi(-d_1) \frac{\partial}{\partial T} d_1 \\ &\stackrel{\star}{=} - r K e^{-rT} \Phi(-d_2) - S \varphi(d_1) \frac{\partial}{\partial T} d_2 + S \varphi(d_1) \frac{\partial}{\partial T} d_1 \\ &= - r K e^{-rT} \Phi(-d_2) - S \varphi(d_1) \left[ \frac{\partial}{\partial T} d_2 - \frac{\partial}{\partial T} d_1 \right] \\ &= - r K e^{-rT} \Phi(-d_2) - S \varphi(d_1) \left[ \frac{\partial}{\partial T} d_1 - \frac{\sigma}{2 \sqrt{T}} - \frac{\partial}{\partial T} d_1 \right] \\ &= - r K e^{-rT} \Phi(-d_2) + \frac{S \sigma \varphi(d_1)}{2 \sqrt{T}}. \end{aligned} \tag{A6.7}$

Step $\star$ is the subtle one. On this step, we both apply the lemma in A2 and use the fact that the standard normal distribution is symmetric and thus $\varphi(d_1) = \varphi(-d_1)$.

If we negate this and align terms with $\Theta(C)$, we get

$\Theta(P) = -\frac{S \sigma \varphi(d_1)}{2 \sqrt{T}} + r K e^{-rT} \Phi(-d_2). \tag{A6.8}$

We can verify that we get the same result if we derive theta for a put using put-call parity:

$\begin{aligned} \frac{\partial}{\partial T} C + \frac{\partial}{\partial T} K e^{-rT} &= \frac{\partial}{\partial T} P + \frac{\partial}{\partial T} S \\ \frac{\partial}{\partial T} P &= \frac{\partial}{\partial T} C + \frac{\partial}{\partial T} K e^{-rT} - \frac{\partial}{\partial T} S \\ &= \frac{\partial}{\partial T} C - r K e^{-rT}. \end{aligned} \tag{A6.9}$

Since theta for a put is negative as well, we have

$\begin{aligned} -\frac{\partial}{\partial T} P &= -\frac{S \sigma \varphi(d_1)}{2 \sqrt{T}} - r K e^{-rT} \Phi(d_2) + r K e^{-rT} \\ &= -\frac{S \sigma \varphi(d_1)}{2 \sqrt{T}} + r K e^{-rT} [1 - \Phi(d_2)] \\ &= -\frac{S \sigma \varphi(d_1)}{2 \sqrt{T}} + r K e^{-rT} \Phi(-d_2). \end{aligned} \tag{A6.10}$

In the last step, we used the symmetry of the normal distribution (Equation $\text{A}5.5$)

A7. Deriving vega

Observe

$\frac{\partial}{\partial \sigma} d_2 = \frac{\partial}{\partial \sigma} d_1 - \sqrt{T}. \tag{A7.1}$

Much like the delta derivation, we can use the lemma in A2 to ge

$\begin{aligned} \mathcal{V}(C) &= \frac{\partial C}{\partial \sigma} \\ &= S \varphi(d_1) \frac{\partial}{\partial \sigma} d_1 - K e^{-rT} \varphi(d_2) \frac{\partial}{\partial \sigma} d_2 \\ &= S \varphi(d_1) \frac{\partial}{\partial \sigma} d_1 - S \varphi(d_1) \frac{\partial}{\partial \sigma} d_2 \\ &= S \varphi(d_1) \frac{\partial}{\partial \sigma} d_1 - S \varphi(d_1) \left[ \frac{\partial}{\partial \sigma} d_1 - \sqrt{T} \right] \\ &= S \varphi(d_1) \sqrt{T}. \end{aligned} \tag{A7.2}$

The derivation for a put is essentially the same. We just use the fact that $\varphi(x) = \varphi(-x)$ since the normal distribution is symmetric:

$\begin{aligned} \mathcal{V}(P) &= \frac{\partial P}{\partial \sigma} \\ &= K e^{-rT} \varphi(-d_2) \frac{\partial}{\partial \sigma} (-d_2) - S \varphi(-d_1) \frac{\partial}{\partial \sigma} (-d_1) \\ &= -S \varphi(d_1) \frac{\partial}{\partial \sigma} d_2 + S \varphi(-d_1) \frac{\partial}{\partial \sigma} d_1 \\ &= S \varphi(-d_1) \frac{\partial}{\partial \sigma} d_1 - S \varphi(d_1) \frac{\partial}{\partial \sigma} \left[ \frac{\partial}{\partial \sigma} d_1 - \sqrt{T} \right] \\ &= S \varphi(d_1) \sqrt{T}. \end{aligned} \tag{A7.3}$

So $\mathcal{V}(C) = \mathcal{V}(P)$. And we can easily see that we would get the same result using put-call parity.