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Let $Y_1, Y_1, \dots$ be independent, identically distributed random variables with the uniform[0,1] distribution and let $X_k=k\cdot Y_k,\; S_n=X_1+X_2+ \dots +X_n$. How to prove that $$\frac{S_n}{\frac{n^2}{4}} \,{\buildrel \text{weakly} \over \to_{n \to \infty}}\, 1 $$ and $$\frac{S_n-\frac{n^2}{4}}{\frac16n^{\frac32}} \,{\buildrel \text{weakly} \over \to_{n \to \infty}}\, \mathcal{N}(0,1).$$

Any help would be really appreciated!

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For the first part, you can compute directly that the convergence holds in $L^2$. –  Nate Eldredge May 8 '12 at 22:08

1 Answer 1

up vote 2 down vote accepted

The first result will follow from Chebyshev's inequality, $P(|S_n-n^2/4|>\epsilon n^2/4)\leq \frac{\mbox{Var}(S_n)}{\epsilon^2 n^4/16}$. I calculate the variance below and it is of order $n^3$, so the desired result follows immediately since weak convergence is implied by convergence in probability.

The second result is basically the central limit theorem. You have that $X_i$ have mean $1/2$ and variance $1/12$. Notice then that $Y_k$ has mean $k/2$ and variance $k^2/12$. Let $\sigma_n^2:=\sum_{k=1}^n\frac{k^2}{12}=\frac{1}{12}\frac{n(n+1)(2n+1)}{6}$, so that $\sigma_n\approx \frac{n^3}{36}$. As well, the mean of $S_n$ is $\sum_{k=1}^n\frac{k}{2}=n(n+1)/4\approx n^2/4$.

Since $\sigma_n\rightarrow\infty$, by the generalized CLT,

$\frac{S_n-n^2/4}{\sigma_n}=\frac{S_n-n^2/4}{\frac{1}{6}n^{3/2}}\Rightarrow \mathcal{N}(0,1)$

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