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.
Expected value of function of random walk
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0Are you allowed to solve, using [LLN](http://en.wikipedia.org/wiki/Law_of_large_numbers) or [CLT](http://en.wikipedia.org/wiki/Central_limit_theorem)? – 2012-02-20
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0@Ilya Yes, it is even encouraged ) – 2012-02-20
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0nice, I've described such method in my answer. – 2012-02-20
2 Answers
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 $0$.
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0"It suffices to use that log(cos(0))=0 and the derivative of log(cos(θx)) at 0 is 0.". Is it because after second term all other terms contain $\frac{1}{n}$ in them? If so, then in order to know that, we would have to calculate at least the third term as well and then proof that all the rest of the terms will contain $\frac{1}{n}$ in them as well, wouldn't we? Or do you have some other reasoning behind using only first two terms of the expansion? – 2012-02-21
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0No, it is because $\lim_{n->\infty} n f(1/n) = f'(0)$ if $f$ is differentiable and $f(0)=0$. Write it as $\frac{f(1/n)-f(0)}{1/n-0}$ to see this more clearly. – 2012-02-22
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 $\theta\in \mathbb R$.
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0Note that we really don't need anything about weak convergence here. Dominated convergence will work just fine. :) – 2012-02-20
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0@cardinal: I guess you mean that we can use the fact $X_n\to 0$ and since $|\mathrm e^{i\theta X_n}|\leq 1$ then the convergence hold, right? – 2012-02-20
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0This is great method, the reason I didn't select it is because hint asked me to use Taylor series, and also the very next question in the book asks me to verify my limit by substituting $0$ instead of $\frac{S_n}{n}$ :) – 2012-02-20
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0Yes, that's correct. :) – 2012-02-20
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0@Cardinal: I only used the dominated convergence theorem for at least a.s. convergence, so I thought: since I use the weak LLN, why don't apply the definition of weak convergence which perfectly fits this exact problem? :) Please, correct me if I wrong about dominated convergence theorem - or if I misunderstood you idea – 2012-02-20
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0@Ilya: You know that $S_n/n \to 0$ almost surely by the SLLN. At any rate, dominated convergence holds for both convergence in probability (think subsequences) and convergence in distribution (think of coupling or how the result in the first sentence of your answer is proved) in addition to the more standard almost-sure convergence. – 2012-02-20
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0@cardinal: thanks - I see your point – 2012-02-20