# Binomial tree convergence tree towards BS equation - Struggle with a limit

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.

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"

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

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