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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.

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Are you allowed to solve, using LLN or CLT? –  Ilya Feb 20 '12 at 10:24
    
@Ilya Yes, it is even encouraged ) –  Sunny88 Feb 20 '12 at 10:40
    
nice, I've described such method in my answer. –  Ilya Feb 20 '12 at 11:26
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2 Answers

up vote 4 down vote accepted

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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"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? –  Sunny88 Feb 21 '12 at 12:10
    
No, 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. –  WimC Feb 22 '12 at 20:37
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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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Note that we really don't need anything about weak convergence here. Dominated convergence will work just fine. :) –  cardinal Feb 20 '12 at 11:26
    
@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? –  Ilya Feb 20 '12 at 11:29
    
This 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 $\frac{S_n}{n}$ :) –  Sunny88 Feb 20 '12 at 11:31
    
Yes, that's correct. :) –  cardinal Feb 20 '12 at 11:31
    
@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 –  Ilya Feb 20 '12 at 11:36
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