Show that \(V_n=Z_n - W_n \rightarrow^p \mu^2\)

What I think:

\(Z_n-W_n=\frac{1}{n}\sum_{i=1}^nX_i^2-(X_i-\bar{X})^2\), so \((X_i-\bar{X})^2=X_i^2-2X_i\bar{X}+\bar{X}^2\)

now

\(Z_n-W_n=\frac{1}{n}\sum_{i=1}^n2X_i\bar{X}-\bar{X}^2\) doing \(S_n=\sum_{i=1}^n2X_i\bar{X}-\bar{X}^2 \rightarrow Z_n-W_n=\frac{S_n}{n}\)

\(E[S_n]=E[\sum_{i=1}^n2X_i\bar{X}-\bar{X}^2]\) assuming \(X_i\) and \(\bar{X}\) are independent.

\(E[S_n]=2n*E[X_1]E[\bar{X}]-n*E[\bar{X}]^2=2n*\mu*\mu-n\mu^2=n\mu^2\) now applying chebyshev's

\(P(|Z_n-W_n)-\mu^2|\geq\epsilon)=P(|\frac{S_n-E[S_n]}{n}|\geq\epsilon)=P(S_n-E[S_n]\geq n\epsilon)\leq\frac{Var(S_n)}{n^2\epsilon^2}\rightarrow0\)

Finally

\(V_n=Z_n - W_n \rightarrow^p \mu^2\)