We would like to prove the following fact:<\/p>\n
For any non-negative random variable \\(X\\) having finite first and second moments, \\(\\mathbb P(X>0) \\ge (\\mathbb EX)^2\/\\mathbb EX^2\\).<\/p>\n
The proof isn’t difficult. Here are three different ones. The statement would also work for non-positive \\(X\\), of course; and an analogous statement can be made for arbitrary \\(X\\) comparing \\(\\mathbb P(X\\ne 0)\\) with some combination of moments for the positive and negative parts of \\(X\\).<\/p>\n Proof 2.<\/strong> Apparently this problem can also be proved by an application of the Cauchy-Schwarz inequality. Assume the probability space \\((\\Omega, \\mathcal F, \\mathbb P)\\). The space of finite second-moment real-valued random variables \\(L_2(\\Omega)=\\{X(\\omega):\\Omega \\to R\\}\\) with the inner product \\(\\langle X,Y\\rangle_{L_2(\\Omega)}=\\mathbb E XY\\) and induced norm \\(\\Vert X\\Vert_{L_2(\\Omega)}=\\sqrt{\\mathbb EX^2}\\) is a Hilbert space (modulo \\(L_2\\) equivalence). Given this, let us apply Cauchy-Schwarz on the two random variables \\(X\\) and \\(\\mathbf 1_{X>0}\\):<\/p>\n \\(\\langle X, \\mathbf 1_{X>0} \\rangle^2 \\le \\Vert X \\Vert^2 \\Vert \\mathbf 1_{X>0} \\Vert^2\\), by Cauchy-Schwarz
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\nProof 1.<\/strong> We already know from Jensen’s inequality that \\(\\mathbb E f(X) \\ge f(\\mathbb E X)\\) if \\(f\\) is convex. This gives \\((\\mathbb EX)^2\/\\mathbb EX^2 \\le 1\\) for any \\(X\\). The trick to make it be \\(\\le \\mathbb P(X>0)\\) is to note that the density at \\(X=0\\) contributes nothing to any moment. In particular, if \\(F_X(t)\\) is the distribution function of \\(X\\), then define a random variable \\(Y\\) that is \\(X\\) without the probability at zero, that is, according to the distribution function \\(F_Y(t)=(F_X(t)-F_X(0))\/(1-F_X(0))\\). Then Jensen’s gives \\((\\mathbb EY)^2\/\\mathbb EY^2 \\le 1\\). However, \\(\\mathbb EY = \\mathbb EX \/ (1-F_X(0)) = \\mathbb EX \/ \\mathbb P(X>0)\\), and \\(\\mathbb EY^2 = \\mathbb EX^2 \/ (1-F_X(0)) = \\mathbb EX^2 \/ \\mathbb P(X>0)\\), so \\((\\mathbb EX)^2\/\\mathbb EX^2 = [(\\mathbb EY)^2 \\mathbb P(X>0)^2] \/ [\\mathbb EY^2 \\mathbb P(X>0)] \\le \\mathbb P(X>0)\\). \\(\\blacksquare\\)<\/p>\n
\n\\((\\mathbb E X \\mathbf 1_{X>0})^2 \\le \\mathbb EX^2 \\mathbb E\\mathbf 1_{X>0}\\), specializing to \\(L_2(\\Omega)\\)
\n\\((\\mathbb E X)^2 \\le \\mathbb EX^2 \\mathbb P(X>0)\\), by noting that \\(X = X \\mathbf 1_{X>0}\\). \\(\\blacksquare\\)<\/p>\n