# Solved – Convergence in probability, \$X_i\$ IID with finite second moment

Let \${X_i}_{igeq 1}\$ be IID with finite second moment, and
\$\$
Y_n = frac{2}{n(n+1)}sum_{i=1}^n ,icdot X_i , , qquad ngeq 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?

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Actually, we can even show that \$mathbb E|Y_n-mathbb E[X_1]|^2to 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 ijmathbb Eleft[(X_i-mathbb E[X_i])(X_j-mathbb E[X_j])right].\$\$ If \$ineq j\$, then by independence \$mathbb Eleft[(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^2mathbb Eleft[(X_j-mathbb E[X_j])^2right].\$\$ 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]|^2leqslantfrac 4{n+1}mathbb Eleft[(X_0-mathbb E[X_0])right]^2\$\$ and we are done.

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