Estimating ATM Option Prices

A well-known approximation for the Black–Scholes price of an at-the-money (ATM) call or put is

$C = P \approx 0.4 S \sigma \sqrt{T}, \tag{1}$

where $C$ and $P$ are the call and put prices respectively, $S$ is the stock price, $\sigma$ is the stock’s volatility, and $T$ is the time to expiry. To my knowledge, this approximation is from (Brenner & Subrahmanyan, 1988). The goal of this post is to derive Equation $1$ myself.

Consider the Black–Scholes price for a call option:

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

where $K$ is the strike price, $r$ is the interest rate, and $N(x)$ is the cumulative distribution function (CDF) of the standard normal distribution. If the option is ATM, then $S = K e^{-rT}$, and we can write the call as:

$C = S \left[ N (d_1) - N(d_2) \right]. \tag{3}$

Next, if we assume that $r = 0$, then $d_1$ and $d_2$ simplify to

$\begin{aligned} d_1 &= (1/2) \sigma \sqrt{T}, \\ d_2 &= -d_1. \end{aligned} \tag{4}$

This appears to be a simple form, but the CDF is too complicated for quick calculations. So let’s approximate it with a first-order Taylor expansion. Note that the Maclaurin series of $N(x)$ is

$N(x) = N(0) + N^{\prime}(0) x + \frac{N^{\prime\prime}(0) x^2}{2} + \frac{N^{\prime\prime}(0) x^3}{3} + \dots \tag{5}$

So we will linearly approximate $N(x)$ with just the first two terms. In the case of ATM options, this is justified if neither $\sigma$ nor $\sqrt{T}$ is too large, since these terms define $d_1$ and $d_2$. In practice, however, I think $\sigma$ and $\sqrt{T}$ can both be relatively large. For example, when the annualized volatility $\sigma$ is $100\%$ and the time to expiry is $T = 1$ year, the ATM value for $d_1$ is $0.5$ (Equation $4$), and the relative error between $N(d_1)$ and the first-order Taylor approximation of $N(d_1)$ is roughly $1\%$ (Figure $1$). If you want to quickly price an ATM option, this seems like an acceptable error.

Figure 1. The CDF of the normal distribution $N(x)$ (curved blue line), plotted against the first-order Taylor approximation of $N(x)$ (straight red line). The values used are $S=100$, $T=1$, $r=0$, and $\sigma=1$. Then $K$ was chosen such that $S = K e^{-rT}$, which here is just $K=S$. So $d_1 = 0.5$, and the relative error between the true value $N(d_1)$ and the first-order approximation is roughly $1\%$. This is the relative error between red and blue lines at the point $d_1$ (dashed black line).

Now let’s plug the first two terms of the Taylor approximation into Equation $3$. We get

$\begin{aligned} C &= S \left[ \left(N(0) + N^{\prime}(0) d_1\right) - \left( N(0) + N^{\prime}(0) d_2 \right) \right] \\ &= S \left[ N^{\prime}(0) d_1 - N^{\prime}(0) d_2 \right] \\ &= S N^{\prime}(0) 2 d_1. \end{aligned} \tag{6}$

This allows us to approximate the Black–Scholes price of the call as

$C = S N^{\prime}(0) \sigma \sqrt{T}. \tag{7}$

Finally, we just need a good estimate of $N^{\prime}(0)$. This is just the probability density function of the standard normal distribution evaluated at zero:

$N^{\prime}(0) = \frac{1}{\sqrt{2\pi}} \approx 0.4. \tag{8}$

Finally, observe that if $S = K e^{-rT}$, then $P = C$ by put-call parity. Putting this all together, we have

$C = P \approx 0.4 S \sigma \sqrt{T} \tag{9}$

as desired.

Example

At the time of this writing, Apple stock is trading at roughly $$174.5$. An option that expires next Friday (one full trading week) has a time-to-expiry of roughly

$5/260 = 1/52 \approx 0.019. \tag{10}$

Let’s approximate the square root of that as just $0.14$. Finally, imagine we know that the implied volatility of an ATM Apple call is roughly $20\%$. Then the ATM approximation of an Apple call is

$0.4 \times $174.5 \times 0.2 \times 0.14 \approx $1.95. \tag{11}$

And this is pretty close to the last trade price for a call option with strike $$175$ expiring next Friday: $$1.86$. As usual, these data are from Yahoo! Finance.