Expected value of function of random walk I am trying to calculate $\lim_{n \to \infty} {E[e^{i \theta \frac{S_n}{n}}]}$. Where $\theta \in \mathbb{R}$, and $S_n$ is simple random walk.
I could simplify it to $\lim_{n \to \infty}E[\cos(\theta \frac{S_n}{n})]$, but I don't know what to do next..
Can you help me?
The hint in the book says that I should use Taylor expansion of $\ln(\cos(x))$ around $x=0$, but I don't see how it can be applied here.
 A: Note that this expected value equals
$$
\sum_{k=0}^n {n \choose k} \frac{e^{i\theta(-1 + \frac{2k}{n})}}{2^n} = \left( \frac{e^{i\theta/n} + e^{-i\theta/n}}{2}\right)^n = \cos(\theta / n)^n
$$
Taking the logarithm results in
$$
n \log(\cos(\theta/n)) = -\frac{\theta^2}{2n}-\frac{\theta^4}{12n^3}-\dotsc
$$
with limit $0$ for $n \rightarrow \infty$.  So the expected value converges to $1$.  It is however not necessary to take the full Taylor expansion.  It suffices to use that $\log(\cos(0)) = 0$ and the derivative of $\log(\cos(\theta x))$ at $0$ is $0$.
A: By the Law of Large Numbers you do have weak convergence $X_n = \displaystyle{\frac{S_n}{n}\Rightarrow 0}$ with $n\to\infty$, so for any measurable and bounded function $f(x)$ it holds that $\mathsf {E}[f(X_n)]\to f(0)$. The function $x\mapsto\mathrm e^{i\theta x}$ is clearly measurable and bounded for all real $\theta$. If you don't see it, you can use
$$
\mathsf E[\mathrm e^{i\theta X_n}] = \mathsf E[\cos(\theta X_n)]+i\mathsf E[\sin(\theta X_n)]
$$
and both functions $\cos(\theta x),\sin(\theta x)$ are bounded and measurable for $\theta\in \mathbb R$.
