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Let $\{Z_k\}_{k=1}^{N}$ be a sequence of i.i.d. random variables with the following distribution $$Z_k = \begin{cases} \alpha &\text{with probability} \ \hat{\pi}\\ -\beta &\text{with probability} \ 1 - \hat{\pi} \end{cases}$$ Then we have $$\ln(S_T) = ln(S_0) + (r - \frac{\sigma^2}{2})T + \sigma\sqrt{T}\frac{1}{\sqrt{N}}\sum_{k=1}^{N}Z_k$$ The only criteria for the convergence of binomial model to Black-Scholes model is that the random variables $Z_k$, $k = 1,\ldots,N$ must satisfy $\hat{\mathbb{E}}[Z_1] = o(\delta)$ and $\hat{\mathbb{E}}[Z_1^2] = 1 + o(1)$ i.e. $$\text{If} \ \hat{\mathbb{E}}[Z_1] = o(\delta), \ \text{and} \ \hat{\mathbb{E}}[Z_1^2] = 1+o(1), \ \text{then}$$ $$\frac{1}{\sqrt{N}}\sum_{k=1}^{N}Z_k \ \text{converges to} \ \mathcal{N}(0,1) \ \text{weakly}$$

Symmetric probability: $$u = \exp(\delta(r - \frac{\sigma^2}{2}) + \sqrt{\delta}\sigma), l = \exp(\delta(r - \frac{\sigma^2}{2}) - \sqrt{\delta}\sigma) , \ \text{and} \ \ R = r\delta$$ Then; $$\hat{\pi}_u = \hat{\pi}_l = \frac{1}{2}$$

Subjective return: $$u = \exp(\delta\nu + \sqrt{\delta}\sigma), l = \exp(\delta\nu - \sqrt{\delta}\sigma), \ \text{and} \ \ R = r\delta$$ Then; $$\hat{\pi}_u = \frac{1}{2}\left(1 + \sqrt{\delta}\frac{r - \nu - \frac{1}{2}\sigma^2}{\sigma}\right) \ \ \text{and} \ \ \hat{\pi}_l = \frac{1}{2}\left(1 - \sqrt{\delta}\frac{r - \nu - \frac{1}{2}\sigma^2}{\sigma}\right)$$

Show $$\hat{\mathbb{E}}[Z_1] = o(\delta), \ \ \text{and} \ \ \hat{\mathbb{E}}[Z_1^2] = 1 + o(1)$$ in the following cases.

a.) symmetric probability

b.) subjective return

Attempted solution a.) $$E[Z_1] = 1\times \frac{1}{2} - 1\times\frac{1}{2} = 0$$ and $$E[Z_1^2] = 1^2\times\frac{1}{2} + (-1)^2\frac{1}{2} = 1$$

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your statement is quite imprecise.

See https://en.wikipedia.org/wiki/Central_limit_theorem

With :

  • $(Z_k)_{k=1\dots n}$ i.i.d with $\mathbb{E}\left[Z_1\right] = \mu$ and $\text{Var}(Z_1)=\mathbb{E}\left[Z_1^2\right] -\mu^2=\sigma^2$

  • and by denoting $\mathcal{N}(m,v)$ a normal variance with mean $m$ and variance $v$

we have : $$ \text{weak}\lim_{N\to\infty}\frac{1}{\sqrt{N}}\sum_{i=1}^n(Z_k-\mu) = \mathcal{N}(0,\sigma^2) $$ or alternatively $$ \text{weak}\lim_{N\to\infty}\frac{1}{\sigma\sqrt{N}}\sum_{i=1}^n(Z_k-\mu) = \mathcal{N}(0,1) $$ you can apply it straightforward to your problem

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  • $\begingroup$ I know how to get the expectation and variance of $N(0,1)$. I guess I am just confused with what $o(\delta)$ means. $\endgroup$ – Wolfy Apr 6 '16 at 16:56
  • $\begingroup$ In your case, if $g(\delta)=o(\delta)$ it means that $\limsup_{\delta\to 0}\frac{g(\delta)}{\delta}=0$ $\endgroup$ – MJ73550 Apr 6 '16 at 17:05
  • $\begingroup$ Ok, I am still unsure how to calculate $E[Z_1]$ could you help me here? $\endgroup$ – Wolfy Apr 6 '16 at 17:11
  • $\begingroup$ $E[Z_1]=\alpha P(Z_1=\alpha)+(-\beta)P(Z_1=-\beta)=\alpha\hat{\pi}-\beta(1-\hat{\pi})$ go to see expected_value on wikipedia $\endgroup$ – MJ73550 Apr 6 '16 at 17:25
  • $\begingroup$ CLT under weak dependence, I am not sure how you think looking there would help $\endgroup$ – Wolfy Apr 6 '16 at 18:02

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