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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 $\lambda_1 = 0$.2012-11-16

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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.2012-11-16
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    What did you get for the mean and variance of $T_n/(2 \log(n))$?2012-11-16
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    $\dfrac{\sum_{k=2}^{n}1/(k-1)}{log(n)}$ and $\dfrac{\sum_{k=2}^{n}1/(k-1)^2}{log(n)^2}$2012-11-16
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    And what are those sums asymptotically?2012-11-16
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    1? The thing is I don't know how should I apply the Chebyshev's inequality in this problem.2012-11-16
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    $\sum_{k=2}^n 1/(k-1) \approx \log(n)$ while $\sum_{k=2}^\infty 1/(k-1)^2$ converges. For any $\epsilon > 0$ and $M > 0$, show that the interval $(1-\epsilon, 1+\epsilon)$ contains $(\mu_n - M \sigma_n, \mu_n + M \sigma_n)$ if $n$ is large enough, where $\mu_n$ and $\sigma_n$ are the mean and standard deviation of $T_n/(2 \log(n))$.2012-11-16