Exercise 3.8. Inis problem presents the convergence or the distribution of stock prices in a sequence of binomial models to the distribution of geometric Brownian motion. In contrast to the analysis of Subsection 3.2.7, here we allow the interest rate to be different from zero.
Let $\sigma>0$ and $r \geq 0$ be given. For each positive integer $n$, we consider a binomial model taking $n$ steps per unit time. In this model, the interest rate per period is $\frac{r}{n}$, the up factor is $u_n=e^{\sigma / \sqrt{n}}$, and the down factor is $d_n=e^{-\sigma / \sqrt{n}}$. The risk-neutral probabilities are then
$$
\tilde{p}_n=\frac{\frac{r}{n}+1-e^{-\sigma / \sqrt{n}}}{e^{\sigma / \sqrt{n}}-e^{-\sigma / \sqrt{n}}}, \quad \tilde{q}_n=\frac{e^{\sigma / \sqrt{n}}-\frac{r}{n}-1}{e^{\sigma / \sqrt{n}}-e^{-\sigma / \sqrt{n}}} .
$$
Let $t$ be an arbitrary positive rational number, and for each positive integer $n$ for which $n t$ is an integer, define
$$
M_{n t, n}=\sum_{k=1}^{n t} X_{k, n}
$$
where $X_{1, n}, \ldots, X_{n, n}$ are independent, identically distributed random variables with
$$
\widetilde{\mathbb{P}}\left\{X_{k, n}=1\right\}=\tilde{p}_n, \quad \widetilde{\mathbb{P}}\left\{X_{k, n}=-1\right\}=\tilde{q}_n, k=1, \ldots, n .
$$
The stock price at time $t$ in this binomial model, which is the result of $n t$ steps from the initial time, is given by (see (3.2.15) for a similar equation)
3.10 Exercises
121
$$
\begin{aligned}
S_n(t) & =S(0) u_n^{\frac{1}{2}\left(n t+M_{n t, n}\right)} d_n^{\frac{1}{2}\left(n t-M_{n t, n}\right)} \\
& =S(0) \exp \left\{\frac{\sigma}{2 \sqrt{n}}\left(n t+M_{n T, n}\right)\right\} \exp \left\{-\frac{\sigma}{2 \sqrt{n}}\left(n t-M_{n t, n}\right)\right\} \\
& =S(0) \exp \left\{\frac{\sigma}{\sqrt{n}} M_{n t, n}\right\} .
\end{aligned}
$$
This problem shows that as $n \rightarrow \infty$, the distribution of the sequence of random variables $\frac{\sigma}{\sqrt{n}} M_{n t, n}$ appearing in the exponent above converges to the normal distribution with mean $\left(r-\frac{1}{2} \sigma^2\right) t$ and variance $\sigma^2 t$. Therefore, the limiting distribution of $S_n(t)$ is the same as the distribution of the geometric Brownian motion $S(0) \exp \left\{\sigma W(t)+\left(r-\frac{1}{2} \sigma\right) t\right\}$ at time $t$.
(i) Show that the moment-generating function $\varphi_n(u)$ of $\frac{1}{\sqrt{n}} M_{n t, n}$ is given by
$$
\varphi_n(u)=\left[e^{\frac{u}{\sqrt{n}}}\left(\frac{\frac{r}{n}+1-e^{-\sigma / \sqrt{n}}}{e^{\sigma / \sqrt{n}}-e^{-\sigma / \sqrt{n}}}\right)-e^{-\frac{u}{\sqrt{n}}}\left(\frac{\frac{r}{n}+1-e^{\sigma / \sqrt{n}}}{e^{\sigma / \sqrt{n}}-e^{-\sigma / \sqrt{n}}}\right)\right]^{n t} .
$$
(ii) We want to compute
$$
\lim _{n \rightarrow \infty} \varphi_n(u)=\lim _{x \downarrow 0} \varphi_{\frac{1}{x^2}}(u),
$$
where we have made the change of variable $x=\frac{1}{\sqrt{n}}$. To do this, we will compute $\log \varphi_{\frac{1}{x^2}}(u)$ and then take the limit as $x \downarrow 0$. Show that