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I am trying to prove that the Binomial tree pricing method converges towards the Black and Scholes model, but I am struggling on a specific step.

I don't understand how the limit of p*(1-p) is calculated.

Extract Hull

Things I tried:

I calculated $p(1-p)$:

$\frac{2\cdot\exp\left(\frac{rT}{n}+\sigma\sqrt{\frac{T}{n}}\right)-\exp\left(\frac{2rT}{n}\right)-1}{\exp\left(2\sigma\sqrt{\frac{T}{n}}\right)+\exp\left(-2\sigma\sqrt{\frac{T}{n}}\right)-2}$

I used the following formula to express the exponantial:

$\exp(x) = \sum_{n=0}^{\infty} \frac{x^n}{n!}$

But when replacing the expression of $exp(x)$ in $p(1-p)$, I don't find anything...

I also tried Hospital rule but it looks like an infinite process. In fact I am note sure about the meaning of "expanding the exponential functions in a series"

Thank you for your help!

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I m not sure if this is correct but this is how I would tackle your question.

Given: $ p = \frac{e^{rT/n} - e^{-\sigma \sqrt{T/n}}}{e^{\sigma \sqrt{T/n}} - e^{-\sigma \sqrt{T/n}}}$

In the below taylor expansion, I assume that $x = rT/n$, $\sigma \sqrt{T/n}$, $-\sigma \sqrt{T/n}$

$e^{rT/n}\approx 1 + \frac{rT}{n} +\mathcal{O}((\frac{rT}{n})^2) + \dots$

$e^{\sigma \sqrt{T/n}}\approx 1 + \frac{\sigma \sqrt{T}}{\sqrt{n}} + \frac{1}{2!} (\frac{\sigma \sqrt{T}}{\sqrt{n}})^2 + \dots = 1 + \frac{\sigma \sqrt{T}}{\sqrt{n}} + \frac{1}{2!} \frac{\sigma^2 T}{n} + \mathcal{O}((\frac{\sigma \sqrt{T}}{\sqrt{n}})^3)$ (not sure if correctly used big O notation)

$e^{\sigma \sqrt{T/n}}\approx 1 - \frac{\sigma \sqrt{T}}{\sqrt{n}} + \frac{1}{2!} (\frac{\sigma \sqrt{T}}{\sqrt{n}})^2 + \dots = 1 - \frac{\sigma \sqrt{T}}{\sqrt{n}} + \frac{1}{2!} \frac{\sigma^2 T}{n} + \mathcal{O}((-\frac{\sigma \sqrt{T}}{\sqrt{n}})^3)$

Hence: $p = \frac{1 + \frac{rT}{n} - (1 - \frac{\sigma \sqrt{T}}{\sqrt{n}} + \frac{1}{2!} \frac{\sigma^2 T}{n})}{1 + \frac{\sigma \sqrt{T}}{\sqrt{n}} + \frac{1}{2!} \frac{\sigma^2 T}{n} - (1 - \frac{\sigma \sqrt{T}}{\sqrt{n}} + \frac{1}{2!} \frac{\sigma^2 T}{n})} = \frac{\frac{rT}{n}+ \frac{\sigma \sqrt{T}}{\sqrt{n}} - \frac{1}{2!} \frac{\sigma^2 T}{n}}{2 \frac{\sigma \sqrt{T}}{\sqrt{n}}} = \frac{\frac{T}{n}(r - \frac{1}{2}\sigma^2)+ \frac{\sigma \sqrt{T}}{\sqrt{n}} }{2 \frac{\sigma \sqrt{T}}{\sqrt{n}}} $

$p = \frac{1}{2} + \frac{\sqrt{T} }{2 \sigma\sqrt{n}}(r - \frac{1}{2}\sigma^2) $

$1-p = \frac{1}{2}-\frac{\sqrt{T} }{2 \sigma\sqrt{n}}(r - \frac{1}{2}\sigma^2) $

Using identity ($a^2-b^2) = (a-b)(a+b)$

$p(1-p) = \frac{1}{4} - \frac{T}{4 \sigma^2n}(r - \frac{1}{2}\sigma^2)^2$

From this point, we can see that as $\lim_{n \to \infty}$$\frac{T}{4 \sigma^2n}(r - \frac{1}{2}\sigma^2)^2 \to 0$

Hence, $p(1-p) \to \frac{1}{4}$

I hope this answered your question. Feel free to ask follow up question if there is something you are unsure about. Best of luck!

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  • $\begingroup$ that was great. and interesting also because $p = \frac{1}{2}$ maximizes $p(1-p)$. $\endgroup$
    – mark leeds
    Mar 23, 2023 at 16:48

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