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$X_1, \cdots ,X_n$ are a random sample from a population. $f(x)$ is a function such that $\mathrm{Var}$ $f(X_1)$ exists. I want to show that $$ \mathrm{Var} (\sum^n_{i=0} f(X_i) ) = n (\mathrm{Var} f(X_1)).$$

This is my attempt.

$$\begin{eqnarray*} \mathrm{Var} (\sum^n_{i=0} f(X_i) ) & = & \mathbb{E}\left[\sum_{i=0}^{n}f(X_{i})-\mathbb{E}\left(\sum_{i=0}^{n}f(X_{i})\right)\right]^{2}\\ & = & \mathbb{E}\left[\sum_{i=0}^{n}f(X_{i})-\left(\sum_{i=0}^{n}\mathbb{E}\, f\left(X_{i}\right)\right)\right]^{2}\\ & = & \mathbb{E}\left[\sum_{i=0}^{n}\left(f\left(X_{i}\right)-\mathbb{E}\, f\left(X_{i}\right)\right)\right]^{2}\\ \end{eqnarray*}$$

This is where I got stuck. The last expression seems to involve the covariance, but I can't bring it out. I'd appreciate all the help I can get. Thanks.

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up vote 1 down vote accepted

Make your life easier, just call $y_i = f(x_i)$

Then demonstrate the result for two vars: $Var(A+B) = Var(A) + Var(B) = 2 Var(A)$ for any $A$ $B$ iid.

Then generalize.

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Thanks for making my life easier :) – Godwin Jun 1 '11 at 3:57

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