Let $\{X_i\}_{i\geq 1}$ be IID with finite second moment, and
$$
Y_n = \frac{2}{n(n+1)}\sum_{i=1}^n \,i\cdot X_i \, , \qquad n\geq 1 \, .
$$
Could you please tell me how can I show that $Y_n$ converges in probability to $\mathrm{E}[X_1]$?
I'm thinking Kolmogorov Convergence criterion. But seems like I cannot prove it using that. Any suggestion?
Best Answer
Actually, we can even show that $\mathbb E|Y_n-\mathbb E[X_1]|^2\to 0$. Indeed, since $\sum_{j=1}^nj=n(n+1)/2$ and $\mathbb E[X_j]=\mathbb E[X_1]$ for all $j$, $$Y_n-\mathbb E[X_1]=\frac 2{n(n+1)}\sum_{j=1}^nj(X_j-\mathbb E[X_j]),$$ hence $$\tag{1}\mathbb E|Y_n-\mathbb E[X_1]|^2=\frac 4{n^2(n+1)^2}\sum_{i,j=1}^n ij\mathbb E\left[(X_i-\mathbb E[X_i])(X_j-\mathbb E[X_j])\right].$$ If $i\neq j$, then by independence $\mathbb E\left[(X_i-\mathbb E[X_i])(X_j-\mathbb E[X_j])\right]=0$ and plugging it in (1), $$\tag{2}\mathbb E|Y_n-\mathbb E[X_1]|^2=\frac 4{n^2(n+1)^2}\sum_{j=1}^n j^2\mathbb E\left[(X_j-\mathbb E[X_j])^2\right].$$ Using now the fact that $X_j$ has the same distribution as $X_1$ and bounding $\sum_{j=1}^nj^2$ by $n^2(n+1)$, equality (2) becomes $$\mathbb E|Y_n-\mathbb E[X_1]|^2\leqslant\frac 4{n+1}\mathbb E\left[(X_0-\mathbb E[X_0])\right]^2$$ and we are done.