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This is the last part of a three part problem on characteristic functions, and it's been driving me crazy over the last few days. Any help would be most appreciated.

$X_1,X_2, \ldots, X_n$ are independent, with $P(X_j=j)=P(X_j=-j)=1/2j, P(X_j=0)=1-1/j$ . Show that $S_n/n$ converges to a distribution with characteristic function $ \exp \left( -\int_0^1 \frac{1-\cos(xt)}{x} dx \right) $.

I have made some progress, but I end up with $$\log \psi(t)=\sum_{j=1}^n \frac{\cos(tj/n)-1}{j}.$$ This is close, but still wrong. Any thoughts?

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1 Answer 1

You are almost done. $$\log (\psi_n(t)) = \sum_{j=1}^n \left(\frac{\cos(tj/n)-1}{j} \right) = \sum_{j=1}^n \left(\frac{\cos(tj/n)-1}{j/n} \right) \frac1n $$ This is nothing but a Riemann sum. In the limit as $n \rightarrow \infty$, we get $$\lim_{n \rightarrow \infty} \log (\psi_n(t)) = \lim_{n \rightarrow \infty} \sum_{j=1}^n \left(\frac{\cos(tj/n)-1}{j/n} \right) \frac1n = \int_0^1 \frac{\cos(tx) - 1}{x} dx$$ Hence, $$\log (\psi(t)) = \log \left( \lim_{n \rightarrow \infty} \psi_n(t) \right) = \lim_{n \rightarrow \infty} \log (\psi_n(t)) = \int_0^1 \frac{\cos(tx) - 1}{x} dx$$ For the sake of completeness, a sequence of random variables $X_n$, converges in distribution to a random variable $X$, iff the sequence of characteristic functions of $X_n$, converges to the characteristic function of the random variable $X$ point-wise.

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+1 Nice, complete answer. =) –  Srivatsan Dec 2 '11 at 3:26
    
Thanks a bunch! –  Rasmus Ohman Dec 2 '11 at 3:49
    
@Rasmus If you find the answer helpful, please accept it. [This is considered important feedback in this site.] Thanks, –  Srivatsan Dec 2 '11 at 4:49

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