### a problem of moments

We would like to prove the following fact:

For any non-negative random variable $$X$$ having finite first and second moments, $$\mathbb P(X>0) \ge (\mathbb EX)^2/\mathbb EX^2$$.

The proof isn’t difficult. Here are three different ones.

Proof 1. 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 $$(\mathbb EX)^2/\mathbb EX^2 = [(\mathbb EY)^2 \mathbb P(X>0)^2] / [\mathbb EY^2 \mathbb P(X>0)] \le \mathbb P(X>0)$$. $$\blacksquare$$

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 $$X$$.

Proof 2. 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 $$\mathbf 1_{X>0}$$:

$$\langle X, \mathbf 1_{X>0} \rangle^2 \le \Vert X \Vert^2 \Vert \mathbf 1_{X>0} \Vert^2$$, by Cauchy-Schwarz
$$(\mathbb E X \mathbf 1_{X>0})^2 \le \mathbb EX^2 \mathbb E\mathbf 1_{X>0}$$, specializing to $$L_2(\Omega)$$
$$(\mathbb E X)^2 \le \mathbb EX^2 \mathbb P(X>0)$$, by noting that $$X = X \mathbf 1_{X>0}$$. $$\blacksquare$$

This is a special case of something called the Paley-Zygmund inequality. I didn’t know such a thing existed.

Proof 3. This one only proves the discrete case. It is well known that for positive discrete random variables $$X$$, $$\mathbb EX = \sum_{k=0}^\infty \mathbb P(X>k) = \mathbb P(X>0)+\mathbb P(X>1)+\cdots$$. Basically $$\mathbb P(X=1)$$ is counted once, $$\mathbb P(X=2)$$ is counted twice, and so on. The analogous thing can be derived for $$\mathbb EX^2$$, except now we need to count in squares. Happily we also know that squares accumulate by odd integers, i.e. $$n^2=1+3+5+\cdots+(2n-1)$$, so $$\mathbb EX^2 = \sum_{k=0}^\infty \mathbb (2k+1) \mathbb P(X>k) = \mathbb P(X>0)+3\mathbb P(X>1)+5\mathbb P(X>2)+\cdots$$.

Let’s simplify the notation a bit. Put $$q_k=\mathbb P(X>k)$$, then $$q_0\ge q_1\ge q_2 \ge \cdots$$. We just need to prove that $$q_0\ge (q_0+q_1+q_2+\cdots)^2 / (q_0+3q_1+5q_2+\cdots)$$, which is to say, $$(q_0+q_1+q_2+\cdots)^2 \le q_0(q_0+3q_1+5q_2+\cdots)$$. The two sides both have limits, so this just requires some accounting. On the left hand side, $$(q_0+q_1+q_2+\cdots)^2$$ expands to $$q_0^2+(q_1^2+2q_0q_1)+(q_2^2+2q_0q_2+2q_1q_2)+\cdots = Q_0+Q_1+Q_2+\cdots$$, where $$Q_k \triangleq q_k^2 + 2 \sum_{i=0}^{k-1} q_iq_k \le (2k+1)q_0q_k \triangleq R_k$$. But $$R_0+R_1+R_2+\cdots$$ is exactly the right hand side. So the left hand sum is dominated by the right hand sum. $$\blacksquare$$

With some real analysis, this proof could be made to work for random variables that are not discrete, but it might also turn into a special case of Proof 1. In any case, it’s interesting in its own right.