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I have random variables $X_1, \ldots, X_N$ and two functions: a bounded function $|f|\leq M$ and another function $0 \le g \le 1$. I would like to upper bound the variance $$ \mathbb{V}\left[\sum_{i=1}^N f(X_i)g(X_i) \right] $$ by a bound that does not depend on $f$ (happy for it to depend on $g$). Is it possible?

Attempt

I tried by using the usual variance formula, but I am unsure if I can do this type of bounds $$ \begin{align} \mathbb{V}\left[\sum_{i=1}^N f(X_i)g(X_i) \right] &= \mathbb{E}\left[\left(\sum_{i=1}^N f(X_i)g(X_i)\right)^2\right] - \mathbb{E}\left[\sum_{i=1}^N f(X_i)g(X_i)\right]^2 \\ &\leq \mathbb{E}\left[\left(\sum_{i=1}^N M g(X_i)\right)^2\right]- \mathbb{E}\left[\sum_{i=1}^N f(X_i)g(X_i)\right]^2 \\ &= M^2 \mathbb{E}\left[\left(\sum_{i=1}^N g(X_i)\right)^2\right] - \mathbb{E}\left[\sum_{i=1}^N f(X_i)g(X_i)\right]^2 \\ &\leq M^2 \mathbb{E}\left[\left(\sum_{i=1}^N g(X_i)\right)^2\right] \end{align} $$ but I was wondering if I can do anything better. One idea I had was to work directly on the formula for the variance of a sum $$ \mathbb{V}\left[\sum_{i=1}^N f(X_i)g(X_i)\right] = \sum_{i=1}^N \mathbb{V}[f(X_i)g(X_i)] + 2 \sum_{i=1}^N \sum_{j=i+1}^N \text{Cov}\left(f(X_i)g(X_i), f(X_j)g(X_j)\right) $$

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  • $\begingroup$ That inequality isn't valid as written (what if $\sum g(X_i) = 0$?); you'd need $\sum M|g(X_i)|$ instead. $\endgroup$ Mar 21 at 15:58
  • $\begingroup$ @GregMartin Mmmm I think I forgot to write a condition. Here I know that $g$ is non-negative as well $\endgroup$ Mar 21 at 16:00
  • $\begingroup$ $(X_n)_{n=1,..,N}$ are i.i.d? $\endgroup$
    – NN2
    Mar 21 at 16:34
  • $\begingroup$ @NN2 Yes, they are! $X_1, \ldots X_n \overset{\text{iid}}{\sim} \mu$ $\endgroup$ Mar 21 at 16:56

1 Answer 1

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From the assumption that $(X_i)_{i=1,..,N}$ are i.i.d, we deduce that $(f(X_i)g(X_i))_{i=1,..,N}$ are also i.i.d. As a consequence $$\begin{align} L:=\mathbb{V}\left(\sum_{i=1}^N f(X_i)g(X_i) \right) &= n\cdot \mathbb{V}\left( f(X_1)g(X_1) \right) \\ &= n\cdot \mathbb{E}\left(f^2(X_1)g^2(X_1) \right)-n\cdot \mathbb{E}^2\left(f(X_1)g(X_1) \right)\\ \end{align}$$ Without any further information about $f,g$ and $X$, a best upper bound is $$L\stackrel{(1)}{\le}n\cdot \mathbb{E}\left(f^2(X_1)g^2(X_1) \right)\stackrel{(2)}{\le}\color{red}{nM^2\cdot \mathbb{E}\left(g^2(X_1) \right)}$$ The equality of $(1)$ occurs if for example $f$ is an odd function, $g$ is an even function and $X$ follows a symmetric distribution.

The equality of $(2)$ occurs if and only if $|f(x)| = M$ for all $x$.

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