Applying the Central Limit Theorem to Risk-Neutral Binomial Trees

# 58427


# How to prove that a series of random variables $Z_j = 1$ or $-1$ occurring at risk-neutral probability, converges to normal, using the CLT?












### Context

When pricing options with trees, it is convenient to prove that the asset value at expiry $S_t$ be of log-normal distribution:

$$\log{S_t} = \log{S_0} + \mu T + \sigma \sqrt{\frac{T}{n}} \sum_{j=1}^{n} Z_j$$

where $\mu$ is the mean of the asset value (drift), $\sigma$ its standard deviation (volatility), $T$ is the time to expiry and $n$ the number of steps $\Delta t = T/n$ in the tree.

Last term $\sum_{j=1}^{n} Z_j$ assumes values $Z_j = 1$ or $-1$ with risk-neutral probability $p_{RN}$

$$p_{RN} = \frac{1}{2} \left( 1+ \left( \frac{r - \mu - 0.5 \sigma^2}{\sigma} \right) \sqrt{\Delta t} \right) + \mathcal{O}(\Delta t)$$

Term $\sum_{j=1}^{n} Z_j$ has mean $\nu$

$$\nu = n \left( \frac{r - \mu - 0.5 \sigma^2}{\sigma} \right) \sqrt{\Delta t} + \mathcal{O}(\Delta t)$$

and variance $$1 - \nu^2 = n + \mathcal{O}(1)$$.

Having computed these two quantities, the book I am following hastens to declare that the Central Limit Theorem applies to $\sum_{j=1}^{n} Z_j$, which to me is not immediate to see since its mean and variance are not $0$ and $1$. Then the author proclaims that this allows to prove that $\log{S_T}$ converges to a log-normal distribution:

$$\lim_{n \to \infty} \log{S_T} = \log{S_0} + \left( r - \frac{\sigma^2}{2} \right) T + \sigma \sqrt{T} N(0,1)$$

and thereafter proceeds to derive the Black-Scholes formula.

### Question

How to prove that $$\lim_{n \to \infty} \frac{\sum_{j=1}^{n} Z_j - \nu}{\sqrt{(1-\nu^2)}} = \frac{\sum_{j=1}^{n} Z_j}{\sqrt{n}} = N(0,1)$$ as prescribed by the Central Limit Theorem, holds? (for further clarity, here $N(0,1)$ is the normal distribution, while $n$ is the number of time-steps in the tree)

### EDIT

I had overlooked the fact that mean $\nu$ and variance $1 - \nu^2$ are already defined with respect to the whole sum $\sum_{j=1}^{n} Z_j$, and not just to the single r.v. $Z_j$.

$n$ is already inside the mean and variance definitions, so I have taken it out of terms in the CLT expression (in last equation).

Still, mean $\nu$ directly depends on $n \sqrt{\Delta t} = \sqrt{Tn}$, so for $n \to \infty$, $\nu$ goes to $\infty$, not to $0$ as it would need in order to validate the application of CLT.