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Suppose $X_2,X_3,\ldots$ are independent random variables.

Assume that $X_k$ has the exponential distribution with parameter $\lambda_k=\dbinom{k}{2} $ for all $k$, which means $ E[X_k] = 1/\lambda_k $ and $\mathrm{Var}(X_k) = 1/(\lambda_k)^2$ for all $k$.

Let $ T_n=\sum_{k=2}^{n}kX_k$.

Prove that

$$\dfrac{T_n}{2\log (n)}\overset {p}{\rightarrow} 1.$$

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Better not include $k=1$ in your sum then, as $\lambda_1 = 0$. – Robert Israel Nov 16 '12 at 6:40
up vote 1 down vote accepted

Hint: estimate the mean and variance, and use Chebyshev's inequality.

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I tried, but it doesn't work... May be I did it in a wrong way. – BigMike Nov 16 '12 at 6:46
What did you get for the mean and variance of $T_n/(2 \log(n))$? – Robert Israel Nov 16 '12 at 6:52
$\dfrac{\sum_{k=2}^{n}1/(k-1)}{log(n)}$ and $\dfrac{\sum_{k=2}^{n}1/(k-1)^2}{log(n)^2}$ – BigMike Nov 16 '12 at 7:03
And what are those sums asymptotically? – Robert Israel Nov 16 '12 at 8:11
1? The thing is I don't know how should I apply the Chebyshev's inequality in this problem. – BigMike Nov 16 '12 at 8:21

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