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I am trying to prove the following statement: $$ E[g(X)] E[X^2g(X)]\ge E[Xg(X)] E[Xg(X)] $$ where $X$ is a random variable, $E[\cdot]$ denotes the expectation operator with respect to $X$, and $g(\cdot)$ is a well-behaved,positive-valued, bounded function.

For $X$ being discrete, I have both the intuition and the proof. For simplicity, assume $X$ takes only two values $x_1$ and $x_2$, with respective probabilities $p_1$ and $p_2$. Further, let $g(x_1)=g_1$ and $g(x_2)=g_2$. Then, the left hand side of the above inequality is given as \begin{align} E[g(X)] E[X^2g(X)]&=[g_1p_1+g_2p_2]\times[{x_1}^2g_1p_1+{x_2}^2g_2p_2] \\ &=x_1^2g_1^2p_1^2+x_2^2g_1g_2p_1p_2+x_1^2g_1g_2p_1p_2+x_2^2g_2^2p_2^2. \end{align} Similarly, the right hand side is given as \begin{align} E[Xg(X)] E[Xg(X)]&=[x_1g_1p_1+x_2g_2p_2]\times[x_1g_1p_1+x_2g_2p_2] \\ &=x_1^2g_1^2p_1^2+2x_1x_2g_1g_2p_1p_2+x_2^2g_2^2p_2^2. \end{align} So the difference between these two expressions is $$ E[g(X)] E[X^2g(X)] - E[Xg(X)] E[Xg(X)] = (x_1-x_2)^2g_1g_2p_1p_2\ge0, $$ which implies that the given inequality is correct. This procedure can be applied to the case where $X$ admits $n$ discrete data points instead of just two.

My question is how to prove the same relationship for continuous random variables, i.e., show that $$ \int g(x)f(x)dx \int x^2 g(x) f(x) dx \ge \int x g(x) f(x) dx \int x g(x) f(x) dx $$ where, $f(\cdot)$ is the probability density function of $X$ and $g(\cdot)\ge0$.

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  • $\begingroup$ In your "simple case" you seem to be assuming that $g$ takes positive values (because you want $g_1g_2\geq 0$ $\endgroup$
    – Yemon Choi
    Commented Feb 4, 2015 at 3:26
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    $\begingroup$ If you assume that $g\geq 0$ in the general problem then the result follows immediately from the Cauchy-Schwarz inequality: $Xg(X)= Xg(X)^{1/2}\cdot g(X)^{1/2}$ $\endgroup$
    – Yemon Choi
    Commented Feb 4, 2015 at 3:27
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    $\begingroup$ And if you don't require $g \ge 0$ then this is false, even in the discrete case. Let $X = \pm 1$ with probability 1/2 each (the Rademacher distribution) and let $g(x) = x$ on $[-1,1]$ (you can extend it to a smooth bounded function on $\mathbb{R}$ if you like). Then your equation reads $E[X] E[X^3] \ge E[X^2] E[X^2]$ which is false; the left side is 0 and the right side is 1. For an example with a continuous random variable, let $X$ be uniformly distributed on $[-1,1]$. $\endgroup$
    – Nate Eldredge
    Commented Feb 4, 2015 at 4:19
  • $\begingroup$ Yes, I should have clearly stated that $g(\cdot) \ge 0$. Otherwise, based on the counterexamples you suggested the assertion is not necessarily true. Yemon, can you elaborate more about using the Cauchy-Schwarz inequality? My understanding is that one can show $E[X^2g(X)]\ge(E[Xg(X)^{1/2}])^2$. However, this is not exactly what I need. $\endgroup$
    – Aliveli
    Commented Feb 4, 2015 at 14:23

3 Answers 3

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Converting Yemon Choi's comment to an answer:

If $E[X g(X)] \le 0$, we are done because the left side of your proposed inequality is nonnegative. So assume $E[X g(X)] \ge 0$.

First, by the triangle inequality we have $$E[X g(X)] \le E [|X| g(X)].\tag{1}$$

The Cauchy-Schwarz inequality says that for any random variables $Y,Z$ we have $$E[|YZ|] \le \sqrt{E[Y^2]\, E[Z^2]}.$$ Applying this with $Y = \sqrt{g(X)}$, $Z = X \sqrt{g(X)}$ we obtain $$E[|X| g(X)] \le \sqrt{E[g(X)] \,E[X^2 g(X)]}. \tag{2}$$ Now combine (1) with (2) and square both sides.

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  • $\begingroup$ Thanks Nate - I was in a rush earlier so you have saved me the trouble of writing it out $\endgroup$
    – Yemon Choi
    Commented Feb 4, 2015 at 16:52
  • $\begingroup$ Thanks a lot! This is indeed a very nice answer! In my case, X only takes positive values so the proof is even easier. Thanks! $\endgroup$
    – Aliveli
    Commented Feb 6, 2015 at 3:13
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If $(\Omega,\mathcal F, P)$ is the probability space you are using, then $$ \mu(A) = \frac{E[A g(X)]}{E[g(X)]} $$ for $A \in \mathcal F$ is another probability measure. Your inequality follows from $$ \|X\|_2 \ge \|X\|_1 $$ using norms of the spaces $L^2(\Omega,\mathcal F, \mu)$ and $L^1(\Omega,\mathcal F, \mu)$. I leave proof of this to you...

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You can prove this working backwards from your last expression:

$(y - x)^2 g(x) g(y) f(x) f(y) > 0$, now if you expand this term you obtain

$g(x) f(x) y^2 g(y) f(y) + x^2 g(x) f(x) g(y) f(y) \geq 2 x g(x) f(x) y g(y) f(y)$.

If you now integrate this inequality with respect to $ x$ and $y$ you get the result you want.

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