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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.

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  • $\begingroup$ Hi: What you wrote at the end after "how to prove that" is the CLT for the sample sum of an RV which is a re-write of the CLT for the sample mean. So, you are already there as far as I can tell. Of course, this assumes that you have the mean and the variance correct and I didn't read that part so disclaimer there. $\endgroup$
    – mark leeds
    Oct 3, 2020 at 15:58
  • $\begingroup$ This looks reasonable as far as an explanation. dartmouth.edu/~chance/teaching_aids/books_articles/… $\endgroup$
    – mark leeds
    Oct 3, 2020 at 16:01
  • $\begingroup$ @markleeds my question should then be rephrased as 'how is it possible to manipulate the first equation, for it to yield the last equation in the 'Context' section of my question, using formulas in between and the CLT ?' What troubles me is the Normal distribution popping up in the last term of the lim of $\log S_T$ $\endgroup$
    – Giogre
    Oct 3, 2020 at 16:12
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    $\begingroup$ Thanks Lingo. That explains the strangeness about the expectation. As far as your question, I would suggest that you email Dr. Joshi but, unfortunately, he died a few years ago unexpectedly. A very kind and generous person based on various things about him that I've seen on the net. As far as travel, I just meant the bus and train ( in NYC ), Not anywhere far !!!!! I'll keep the problem in my bag so I can look at it when the urge arises. All the best. $\endgroup$
    – mark leeds
    Oct 5, 2020 at 23:36
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    $\begingroup$ Ok. No problem. It's been a looooooooooooooooooong time but I remember C&R being really useful when I was trying to understand options for the first time. Maybe, when you get a chance to look at it, it will be helpful. $\endgroup$
    – mark leeds
    Oct 9, 2020 at 3:07

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