Proof the Binomial Model Converges to Black–Scholes

This is a companion post to my post on the binomial options-pricing model. Please see that post for an expository treatment of the binomial model. The goal here is to prove that the binomial model converges to Black–Scholes in the limit. This proof is from (Hsia, 1983).

Setup

First, let’s state both models and introduce notation. Let

$\begin{aligned} \text{binomial model:} &\qquad& C &= S B (a; n, \pi_1) - K r^{-t} B(a; n, \pi_2), \\ \text{Black--Scholes:} &\qquad& C &= S N (d_1) - K r^{-t} N(d_2), \end{aligned} \tag{1}$

where

$\begin{aligned} B(x; n, p) &= \sum_{j=x}^n {n \choose j} p^j (1 - p)^{n-1} \\ a &= \frac{\log(K/S) - n \log d}{\log(u/d)} + \zeta, \quad \zeta \in [0, 1), \\ \pi_2 &= (r_0-d)/(u-d), \\ \pi_1 &= (u/r_0) \pi_2, \end{aligned} \tag{2}$

and

$\begin{aligned} N(x) &= \int_{-\infty}^x (1 / \sqrt{2 \pi_2}) e^{-z^2 / 2} dz, \\ d_1 &= \frac{\left[ \log(S/K) + \left[ \log r + (1/2) \sigma^2 \right] t\right]}{\sigma \sqrt{t}}, \\ d_2 &= d_1 - \sigma \sqrt{t}, \end{aligned} \tag{3}$

and where

$\begin{aligned} C &= \text{price of call at time zero} \\ S &= \text{price of stock at time zero} \\ S^{*} &= \text{price of stock at expiry} \\ K &= \text{strike price} \\ t &= \text{time to expiry} \\ n &= \text{number of price changes in time $t$} \\ p &= \text{probability of an up move} \\ \pi_2 &= \text{risk-neutral probability of an up move} \\ u &= \text{up factor} \\ d &= \text{down factor} \\ r_0 &= \text{one plus risk-free interest rate for one period $t/n$} \\ r &= \text{one plus risk-free interest rate for unit time} \\ \mu &= \text{mean of stock price} \\ \sigma^2 &= \text{variance of stock price} \\ a &= \text{smallest number of up moves for the call to end in-the-money.} \end{aligned}$

The binomial model assumes the stock price is a discrete-time process $\{S_n\}$, which is a multiplicative random walk:

$\begin{aligned} S_{n} &= S X_1 X_2 \cdots X_n \\ &= S_{n-1} X_n, \end{aligned}\tag{4}$

where $X_i$ (the up or down factor) is drawn i.i.d. based on a Bernoulli random variable, or

$\mathbb{P}(X_i = u) = p, \qquad \mathbb{P}(X_i = d) = 1-p. \tag{5}$

Black–Scholes assumes the stock price is a continuous-time process $S(t)$, which is geometric Brownian motion. Thus, $S(t)$ is defined as

$S(t) = S e^{Y(t)}, \qquad Y(t) = \sigma B(t) + \mu t, \tag{6}$

where $B(t)$ is Brownian motion, and where $Y(t)$ is Brownian motion with drift:

$Y(t) \sim \mathcal{N}(\mu t, \sigma^2 t), \qquad B(t) \sim \mathcal{N}(0, t). \tag{7}$

$S(t)$ is lognormally distributed with parameters $\mu$ and $\sigma^2$ or equivalently $\log S(t)$ is normally distributed with parameters $\mu$ and $\sigma^2$.

Note that in the main post, the risk-neutral probability $\pi_2$ is denoted $\pi$ and variable here called $\pi_1$ is denoted $\rho$. However, I have adopted slightly different notation here so that variable names “align” in Equation $1$.

Proof

From Equation $1$, it is clear that all we have to do to prove convergence from the binomial model to Black–Scholes is to prove that as $n \rightarrow \infty$,

$\begin{aligned} B (a; n, \pi_1) &\quad\stackrel{d}{\rightarrow}\quad N (d_1), \\ B(a; n, \pi_2) &\quad\stackrel{d}{\rightarrow}\quad N(d_2), \end{aligned} \tag{8}$

where $\stackrel{d}{\rightarrow}$ denotes convergence in distribution.

To do this, we first invoke the DeMoivre-Laplace’s theorem (CLT) to show that each binomial distribution converges to its respective normal distribution. See A1 for a brief review of this central limit theorem. To see this, let $j$ denote the binomial random variable (if $j$ is the outcome of a coin toss, it is heads or an up move with probability $p$). Then by the symmetry of the binomial distribution, we have

$\begin{aligned} B(a; n, q) &= \mathbb{P}(j \geq a) \\ &= \mathbb{P}(j \leq -a) \\ &\stackrel{!}{=} \mathbb{P}\left( \frac{j - \mu_j}{\sigma_j} \leq \frac{-a + \mu_j}{\sigma_j}, \right), \end{aligned} \tag{9}$

where $\mu_j = \mathbb{E}[j]$ and $\sigma_j^2 = \mathbb{V}[j]$. Be careful with the sign in step $!$. Thus, the CLT tells us that as $n \rightarrow \infty$, then

$\begin{aligned} B(a; n, q) \quad\stackrel{d}{\rightarrow}\quad N\!\left( \frac{-a + \mu_j}{\sigma_j} \right). \end{aligned} \tag{10}$

Now let’s give a name to the fraction inside the normal CDF. The fraction looks a lot like the $d_1$ and $d_2$ terms in Black–Scholes, so let’s just call it $d$:

$d = \frac{-a + \mu_j}{\sigma_j}. \tag{11}$

Since $j$ is a binomial random variable, we know $\mu_j = np$ and $\sigma_j = np(1-p)$. And we know that $a$ is

$a = \frac{\log(K/S) - n \log d}{\log(u/d)} + \zeta, \quad \zeta \in [0, 1). \tag{12}$

See A2 for a derivation of $a$. So we can plug these values into Equation $6$ to get

$\begin{aligned} d &= \frac{-a + \mu_j}{\sigma_j} \\ &= -\left( \frac{\log(K/S) - n \log d}{\log(u/d)} + \zeta + np \right) \bigg/ \sqrt{np(1-p)} \\ &= \frac{\log(S/K) + np \log(u/d) + n \log(d)}{\log(u/d) \sqrt{np(1-p)}} + \frac{\zeta}{\sqrt{np(1-p)}}. \end{aligned} \tag{13}$

How can we simplify this? The key insight is to realize that most of the terms in the last line above can be represented as the mean and standard deviation of the log return from $S$ to $S^{*}$. We can see this by computing the mean and variance of the lognormal random variable $\log(S^*/S)$, giving us

$\begin{aligned} \mathbb{E}\left[ \log(S^{*} / S) \right] &= n \left[ p \log(u/d) + \log(d) \right], \\ \mathbb{V}\left[ \log(S^{*} / S) \right] &= np(1-p) \log(u/d)^2. \end{aligned} \tag{14}$

See A3 for a derivation. Finally, observe that the $\zeta$ term will disappear once we will take $n \rightarrow \infty$. So the upper bound in our Gaussian integral can be written as

$d = \frac{\log(S/K) + \mathbb{E}[\log(S^{*} / S)]}{\sqrt{\mathbb{V}_q[\log(S^{*} /S)]}}. \tag{15}$

And we know that the variance of the log return is

$\mathbb{V}[\log(S^{*}/S)] = \mathbb{V}_q[Y(t)] = \sigma^2 t. \tag{16}$

This is a standard property of Brownian and geometric Brownian motion, and it should be clear from Equations $6$ and $7$.

To summarize so far, we have shown that $B(a; n, \pi_1)$ and $B(a; n, \pi_2)$ can both be written as

$B(a; n, \pi) \stackrel{d}{\rightarrow} N(d), \quad d = \frac{\log(S/K) + \mathbb{E}[\log(S^{*} / S)]}{\sigma \sqrt{t}}, \quad \pi \in \{\pi_2,\pi_1\}. \tag{17}$

To finish our proof, we just need to show that $d = d_1$ when the binomial parameter is $\pi_1$ and that $d = d_2$ when the binomial parameter is $\pi_2$. Or in other words,

$\text{as $n \rightarrow \infty$}, \qquad \mathbb{E}[\log(S^{*} / S)] = \begin{cases} \left[ \log r + (1/2) \sigma^2 \right] t & \text{for $\pi=\pi_1$}, \\ \left[ \log r - (1/2) \sigma^2 \right] t & \text{for $\pi=\pi_2$}. \end{cases} \tag{18}$

Let’s first prove this result for $\pi_2$, the risk-neutral probability.

The definition of our multiplicative random walk (Equation $4$) allows us to represent the raw return from zero to $n$ as

$S^* / S = X_1 X_2 \cdots X_n = \prod_{i=1}^n X_i. \tag{19}$

Now the expectation of each multiplicative factor $X_i$ can be written as

$\mathbb{E}[X_i] = \mathbb{E}[S_i / S_{i-1}] = \pi_2 u + (1-\pi_2) d. \tag{20}$

By risk-neutrality, we know the one-period expectation is equal to the one-period interest rate $r_0$ or

$\pi_2 u + (1-\pi_2) d = r_0. \tag{21}$

And since each multiplicative factor is i.i.d., the total expectation is

$\begin{aligned} \mathbb{E} [S^* / S] &= \prod_{i=1}^N (\pi_2 u + (1-\pi_2) d) \\ &= \left[ \pi_2 u + (1-\pi_2) d \right]^n \\ &= r_0^n \\ &= r^t. \end{aligned} \tag{22}$

Finally, we need to use a property of the continuous-time process. If $Z$ is a lognormal random variable with parameters $\nu$ and $s^2$, then its expected value is

$\mathbb{E}[Z] = \exp\!\left( \nu + \frac{1}{2} s^2 \right). \tag{23}$

And note that since $S^{*}$ is lognormally distributed,

$S^* \sim \text{lognormal}\left(\log S + \mu t, \sigma^2 t\right), \tag{24}$

then $S^{*}/S$ is lognormally distributed and $\log(S^*/S)$ is normally distributed,

$\begin{aligned} S^*/S &\sim \text{lognormal}\left(\mu t, \sigma^2 t\right), \\ \log(S^*/S) &\sim \mathcal{N}\left(\mu t, \sigma^2 t\right). \end{aligned} \tag{25}$

This allows us to write expectation and log expectation as

$\begin{aligned} r^t &= \mathbb{E}[S^*/S] = \exp\!\left( \mu t + \frac{1}{2} \sigma^2 t \right), \\ t \log r &= \mu t + \frac{1}{2} \sigma^2 t, \end{aligned}\tag{26}$

which we can rewrite in terms of $\mu t = \mathbb{E}[\log(S^*/S)]$ as

$\mathbb{E}[\log(S^*/S)] = \left[ \log r - \frac{1}{2} \sigma^2 \right] t \tag{27}$

as desired.

The proof of line convergence when $\pi = \pi_1$ is the roughly the same. Let’s repeat our logic from above but for $S/S^*$. First, we can write

$S / S^* = \frac{1}{X_1 X_2 \cdots X_n} \tag{28}$

by Equation $4$. Following the same logic as above, we know the expected value of $S_{i-1}/S_i$ is

$\begin{aligned} \mathbb{E}[S_{i-1}/S_i] &= \pi_1 \left( \frac{1}{u} \right) + (1-\pi_1) \frac{1}{d} \\ &= \frac{\pi_2}{r_0} + \frac{1-\pi_2}{r_0} \\ &= \frac{1}{r_0}. \end{aligned} \tag{29}$

So the expectation of $S/S^*$ can be written as

$\mathbb{E}[S/S^*] = r_0^{-n} = r^{-t}. \tag{30}$

Once again, we can take the log of both sides and use the fact that $S/S^*$ is lognormally distributed:

$\begin{aligned} -t \log r &= \mathbb{E}[S/S^*] \\ &= \mathbb{E}[\log(S/S^*)] + (1/2) \mathbb{V}[\log(S/S^*)] \\ &= \mathbb{E}[-\log(S^*/S)] + (1/2) \mathbb{V}[-\log(S^*/S)] \\ &= -\mathbb{E}[\log(S^*/S)] + (1/2) \sigma^2 t. \end{aligned} \tag{31}$

Putting it all together, we get

$\mathbb{E}[\log(S^*/S)] = \left[ \log r + \frac{1}{2} \sigma^2 \right] t \tag{32}$

as desired.

Conclusion

Perhaps the most notable observation about this proof is that we do not need to specify the risk-neutral probability $\pi_2$. We can choose $\pi_2$ however we would like, and we still get convergence. We will use this fact when fitting the binomial model in the main post. The condition that must hold, however, is the no-arbitrage condition. We used this assumption in Equation $21$.

 

Appendix

A1. De Moivre–Laplace theorem

Let $X_n$ be a binomially distributed random variable with parameters $n$ and $p$, and let $X$ be a normally distributed random variable with parameters $np$ and $np(1-p)$,

$X_n \sim \text{binom}(n, p), \qquad X \sim \mathcal{N}(np, np(1-p)). \tag{A1.1}$

The De Moivre–Laplace (central limit) theorem states that the probability mass function of $X_n$ approximates the probability density function (PDF) of $X$ for large $n$:

$\mathbb{P}(X_n = x) \approx f_X(x) dx. \tag{A1.2}$

We can write this as

$\mathbb{P}\left( X_n = x \right) \approx \frac{1}{\sqrt{2 \pi_2 n p (1-p)}} \exp\left\{ -(x - np)^2 / (2np(1-p)) \right\}. \tag{A1.3}$

If we normalize $X_n$, then the probability approximates the PDF of a standard normal random variable:

$\mathbb{P}\left( \frac{X_n - np}{np(1-p)} = x \right) \approx \frac{1}{\sqrt{2 \pi_2}} \exp\left\{ -x^2 / 2 \right\}. \tag{A1.4}$

Finally, this suggests that we can approximate the binomial CDF as a truncated Gaussian integral:

$\mathbb{P}\left( \frac{X_n - np}{np(1-p)} \leq x \right) \approx \int_{-\infty}^x \frac{1}{\sqrt{2 \pi_2}} \exp\left\{ -z^2 / 2 \right\} dz. \tag{A1.5}$

This fact is used in the main text.

 

A2. Solving for $a$

We have defined $a$ as the smallest integer such as that the option is in-the-money or the smallest integer such that

$u^a d^{n-a} S \gt K. \tag{A2.1}$

To solve for $a$, let’s take the log of both sides:

$\begin{aligned} u^a d^{n-a} S &\gt K \\ a \log u + (n-a) \log d + \log S &\gt \log K \\ a (\log u - \log d) &\gt \log K - \log S - n \log d \\ a &\gt \frac{\log(K/S) - n \log d}{\log(u/d)}. \end{aligned} \tag{A2.2}$

We can remove the inequality sign by introducing a slack variable $\zeta$ such that

$a = \frac{\log(K/S) - n \log d}{\log(u/d)} + \zeta, \quad \zeta \in [0, 1), \tag{A2.3}$

and we’re done.

 

A3. Solving for moments of the log return

The log stock return over the time to expiration is $\log(S^{*} / S)$. We can express this in terms of up and down moves as

$\begin{aligned} \log(S^{*} / S) &= \log\left(\frac{u^j d^{n-j} S}{S} \right) \\ &= j \log u + (n-j) \log d \\ &= j \log(u/d) + n \log d. \end{aligned} \tag{A3.1}$

The mean is

$\begin{aligned} \mathbb{E}\left[ \log(S^{*} / S) \right] &= \mathbb{E}[j] \log(u/d) + n \log d \\ &= np \log(u/d) + n \log d \\ &= n \left[ p \log(u/d) + \log(d) \right]. \end{aligned} \tag{A3.2}$

The variance is

$\begin{aligned} \mathbb{V}\left[ \log(S^{*} / S) \right] &= \mathbb{V}[j] \log(u/d)^2 \\ &= np(1-p) \log(u/d)^2. \end{aligned} \tag{A3.3}$