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Given $E[f^2(X)] < \infty$ and $X_i \sim_{iid} X$, need to show $$ E\left[ \frac{1}{n} \left( \sum_{i=1}^{n} (f(X_i) - E[f(X)] \right)^2 \right] \leq E[f^2(X)]. $$

My try:

$$E\left[ \frac{1}{n} \left( \sum_{i=1}^{n} (f(X_i) - E[f(X)] \right)^2 \right] = \frac{1}{n}E\left[ \left( \sum_{i=1}^{n} f(X_i) \right)^2 - n E[f(X)] \sum_{i=1}^{n} f(X_i) \right].$$

Now applying Cauchy-Schwarz inequality to the first term we get, $$ E\left[ \frac{1}{n} \left( \sum_{i=1}^{n} (f(X_i) - E[f(X)] \right)^2 \right] \leq n \cdot Var[f(X)] . $$

I am not able to get rid of $n$ on RHS. Any help to fix this is appreciated!

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The trick is that your random variables $X_i$ are independent, so that $$E[(f(X_i)-m)(f(X_j)-m)]=E[f(X_i)-m]E[f(X_j)-m]=0,$$ where $m=E[X]$.

This shows that $$E\left[\frac1n\left(\sum_if(X_i)-E[f(X)]\right)^2\right]=\frac1n\sum_iE\left[\left(f(X_i)-E[f(X)]\right)^2\right]=\mathrm{Var}(f(X)).$$

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