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Here's the problem: Let $X_{1},\dots,X_{n}$ be a sequence of independent random variables with $X_{i}\sim P_{i}$. Let $E\left[X_{i}\right]=\mu_{i}$ and $Var\left[X_{i}\right]=\sigma^{2}$. Suppose that

$\frac{1}{n}\sum_{1\leq i\leq n}\sigma_{i}^{2}\to0$

We want to show that $\bar{X}_{n}-\bar{\mu}_{n}\overset{P}{\to}0$, where

$\bar{\mu}_{n}=\frac{1}{n}\sum_{1\leq i\leq n}\mu_{i}$

I'm absolutely sure that we have to use Markov's inequality for this and link it up with the facts about the variance in the set up

$P\left\{ \left|\bar{X}_{n}-\bar{\mu}_{n}\right|>\varepsilon\right\} \leq\frac{E\left[\left(\bar{X}_{n}-\bar{\mu}_{n}\right)^{2}\right]}{\varepsilon^{2}}$

But I'm not sure where to proceed from here. I'm also asked to show that if $\bar{\mu}_{n}\to\mu$ then $\bar{X}_{n}\overset{P}{\to}\mu$, but I'm also confused about how to show this.

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  • $\begingroup$ Can you elaborate on the $X_i \sim P_i$ note? $\endgroup$ – Aaron Montgomery Oct 4 '17 at 2:58
  • $\begingroup$ I says that each $X_i$ is drawn from a different probability distribution $P_i$. In this case the $X_1, \dots, X_n$ are not iid. $\endgroup$ – kittykat Oct 4 '17 at 3:05
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  1. Using independence $$ \mathsf{E}(\bar{X}_n-\bar{\mu}_n)^2=\frac{1}{n^2}\sum_{i=1}^n\sigma_i^2. $$
  2. $$ |\bar{X}_n-\mu|\le |\bar{X}_n-\bar{\mu}_n|+|\bar{\mu}_n-\mu|. $$
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  • $\begingroup$ Thank you very much. This was very helpful. $\endgroup$ – kittykat Oct 4 '17 at 3:34

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