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Suppose $X_2, X_3, \dots$ are independent random variables. Assume that $X_k$ has the exponential distribution with parameter $\lambda_k = \binom{k}{2}$ for all $k$. Let $$T_n = \sum_{k=2}^n kX_k.$$ Show that $\frac{T_n}{2\log n}$ converges to $1$ in probability.

According to our assumption, we have $\mathrm{E}(X_k)=1/\lambda_k$ and $\mathrm{Var}(X_k)=1/\lambda_k^2$. I'm trying to apply the weak law of large numbers to $\{kX_k\}_{k=1}^\infty$ but where does the $\log n $ come from? Any help is appreciated!

Jay
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  • A small hint: Use the fact that $kX_{k} \sim Exp (\lambda_{k}/k)$. So for each component of sum you'll have $E(kX_{k})= \frac{k-1}{2}$. And $Var(kX_{k})=\frac{(k-1)^2}{4}$. You can use for example Chebyshev's inequility. – kolobokish Nov 13 '19 at 18:43
  • Could you show some details? I didn't see how we could use Chebyshev's inequality. – Jay Nov 13 '19 at 19:10

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Let $Y_k=\lambda_k X_k$. The $Y_k$ are identically distributed random variables satisfying the exponential distribution with parameter $\lambda=1$; since $P(X_k\le x)=1-e^{-\lambda_kx}$ we can deduce that $P(Y_k\le y)=1-e^{-y}$.

Then $T_n=\sum_{k=2}^n\frac{kY_k}{\lambda_k}=\sum_{k=2}^{n}\frac{2Y_k}{k-1}$.

We can calculate that $\mathbb{E}[T_n]=\sum_{k=2}^n\mathbb{E}\left[\frac{2Y_k}{k-1}\right]=\sum_{k=2}^n \frac{2}{k-1}$ and $\text{var}(T_n)=\sum_{k=2}^n \text{var}\left(\frac{2Y_k}{k-1}\right)=\sum_{k=2}^{n} \frac{4}{(k-1)^2}$

It is proved here that $\frac{\sum_{k=2}^n\frac{1}{k-1}}{\log n}\rightarrow 1$ as $n\rightarrow \infty$. So $\mathbb{E}\left[\frac{T_n}{2\log n}\right]$ converges to $1$. It can also be shown that $\frac{\sum_{k=2}^n\frac{1}{(k-1)^2}}{\log n}\rightarrow 0$ as $n\rightarrow \infty$ and hence $\text{var}\left[\frac{T_n}{2\log n}\right]$ converges to zero.

Angela Pretorius
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