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I have this problem at hand

$X_1,X_2,\cdots $ are iid random variables with finite second moments.Define$$Y_n={2\over n(n+1)}\sum_{i=1}^n iX_i$$ Show that $Y_n\xrightarrow{P}E(X_1)$.

I know that $\sum\limits_{i=1}^n {(X_i-\mu)\over \sum\limits_{i=1}^n \sigma_i^2}\xrightarrow{P}0$ as $n\to\infty$. And I can replace $\sum\limits_{i=1}^n \sigma_i^2$ with $n$ and things will still hold provided that ${\sum\limits_{i=1}^n \sigma_i^2\over n^2} \to 0 $ as $n\to \infty$

I tried (since $\sigma_i$'s are finite) that say $\sigma_i \lt k$ ($k$ finite). $${\sum\sigma_i^2\over n^2}\le {k^2\over n}\to 0 \text{ as }n\to \infty$$ How ever I got stuck when I realised that Var$(X)=\sigma^2\implies $Var$(iX)=i^2\sigma^2$ implying$${\sum\limits_{i=1}^ni^2 \sigma_i^2\over n^2}\le k^2{n(n+1)(2n+1)\over 6n^2}\not\to 0 \text{ as }n\to \infty$$

So please help me proceed.

Qwerty
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  • Starting from "I know that", what you are doing is a mystery. Why not simply compute/estimate $E(Y_n)$ and $\mathrm{var}(Y_n)$? Note that $$\left(\frac2{n(n+1)}\right)^2\sum_{i=1}^ni^2=\Theta\left(\frac1n\right).$$ – Did Jun 13 '16 at 14:25
  • I got the point. Thanks.. Actually I am new to convergence, so I tend to mess things up... – Qwerty Jun 13 '16 at 14:36

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