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Can anyone help me with this problem? The random variable $X_n$ takes the values $\frac{k}{n}$, $k=1,2,\ldots,n$, each with probability $\frac{1}{n}$. Find its characteristic function and the limit as $n\to\infty$. Identify the random variable of the limit characteristic function. Thanks!

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Is this homework? –  Rasmus Feb 13 '11 at 20:08
    
@Rasmus: right! –  kira Feb 13 '11 at 20:16

3 Answers 3

up vote 2 down vote accepted

$$E(e^{itX})=\frac{1}{n}\sum_{k=1}^n {e^{is\frac{k}{n}}}= \frac{1}{n}\frac{e^{\frac{is}{n}}(1-e^{is})}{1-e^{\frac{is}{n}}}\to \frac{e^{is}-1}{is}$$ as $n \to \infty$

which is cf of the uniform distribution on [0,1]

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Hint 1: You should know how to calculate ${\rm E}[f(X)]$, hence, in particular, the characteristic function of $X$.

Hint 2: How can $\int_0^1 {f(x)\,{\rm d}x}$ be approximated by Riemann sum?

Hint 3: ${\rm d}x$ corresponds to which distribution?

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A bit surprised nobody mentioned this yet, but one can use a nice coupling approach here. Let $U$ denote a random variable uniformly distributed on $(0,1)$. For every positive $n$, let $$U_n=n^{-1}\lceil nU\rceil.$$ It should be clear that each $U_n$ is distributed like $X_n$. Furthermore, $|U_n-U|<n^{-1}$ almost surely, hence $U_n\to U$ almost surely, hence $U_n\to U$ in distribution (because almost sure convergence implies convergence in distribution), hence $X_n\to U$ in distribution (because convergence in distribution does not depend on the realization of the distributions by random variables).

This may be called a coupling approach because the random variables $U_n$ are defined all at once on the same probability space (the one used to define $U$). The first chapter of Lectures on the Coupling Method by Torgny Lindvall explains this and some of the many other advantages of the method in different settings, see here for a presentation of this book.

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