# How to derive Paley-Zigmund Inequality proof

The Paley-Zygmund inequality is given by

$$$$\operatorname{P}( Z > \theta\operatorname{E}[Z] ) \ge (1-\theta)^2 \frac{\operatorname{E}[Z]^2}{\operatorname{E}[Z^2]}$$$$

I want to prove it. I found this Wikipedia article on the Paley-Zygmund inequality which states that I can decompose the random variable $$X$$ like so

$$$$E[Z] = E[Z\mathbf{1}_{\{Z\leq\theta E[Z]\}}]+E[Z\mathbf{1}_{\{Z>\theta E[Z]\}}]$$$$

the article then states that \begin{align} E[Z\mathbf{1}_{\{Z\leq\theta E[Z]\}}] \leq & \; \theta E[Z] \;\;\; \text{and}\\ E[Z\mathbf{1}_{\{Z>\theta E[Z]\}}] \leq & \; \sqrt{E[Z^2]P(Z>\theta E[Z])} \end{align}

Plugging this into the first equation yields

\begin{align} E[Z] \leq & \; \theta E[Z] + \sqrt{E[Z^2]P(Z>\theta E[Z])} \\ \iff (1-\theta)^2 \frac{E[Z]^2}{E[Z^2]} \leq & \; P(Z>\theta E[Z]). \end{align}

I have two questions: 1) The decomposition makes intuitive sense. Either way, exactly one of the conditions is true, so the equality holds. However, is there a more formal way / analytical approach to support this statement? 2) How are the upper bounds for the two expected values of the indicator variable derived? I would appreciate an answer which is kept quite simple as I am not a (pure) mathematician.

/edit: Possibly should have posted this to math.stackexchange. Maybe a mod can move the question.

• Check this math.stackexchange.com/questions/1501927/… might be helpful. Commented Apr 13, 2021 at 17:41
• Thanks, I saw this question before. The answer states my procedure exactly, but without explanation on my two questions. Commented Apr 13, 2021 at 17:45
• I hope that the following answer will help. Commented Apr 13, 2021 at 18:13

For the first part of your question we have that: We have that $$Z\geq 0$$ thus the domain can be decomposed as $$Z\geq 0= (Z\leq \theta \mathbb{E}[Z]) \cup (Z> \theta \mathbb{E}[Z])$$

$$\mathbb{E}[Z] = \int_{Z\geq 0} z p(z)dz = \int_{Z\leq \theta \mathbb{E}[Z]} zp(z)dz + \int_{Z> \theta \mathbb{E}[Z]}zp(z) = \\ \mathbb{E}[Z\times 1_{Z\leq \theta \mathbb{E}[Z]}] + \mathbb{E}[Z\times 1_{Z>\theta \mathbb{E}[Z]}]$$

For the second part:

$$\mathbb{E}[Z\times 1_{Z\leq \theta \mathbb{E}[Z]}] = \int_{Z\leq \theta \mathbb{E}[Z]} zp(z)dz \leq \int_{Z\leq \theta \mathbb{E}[Z]} max(z)p(z)dz \\ = \theta \mathbb{E}[Z]\int p(z)dz = \theta \mathbb{E}[Z]$$

Also, the $$Z$$ has finite variance thus $$\mathbb{E}[Z^{2}]<\infty,$$ $$Z$$ is square integrable, so we can use the Cauchy–Schwarz inequality, https://en.wikipedia.org/wiki/Cauchy%E2%80%93Schwarz_inequality for $$L^{2}$$,

$$\mathbb{E}[Z\times 1_{Z>\theta \mathbb{E}[Z]}] = \int z1_{Z> \theta\mathbb{E}[Z]}p(z) \leq (\int z^{2}p(z)dz)^{1/2} (\int 1_{Z> \theta \mathbb{E}[Z]}^{2}p(z)dz)^{2}\\ =\mathbb{E}[Z^{2}]^{1/2}\mathbb{P}(Z> \theta \mathbb{E}[Z])^{1/2}$$

• Thank you, this is very clear! Commented Apr 13, 2021 at 19:44
• @MarcelB In the interval $Z\leq \theta \mathbb{E}[Z]$ the largest possible value that $Z$ can take is $\theta \mathbb{E}[Z]$. You mean why the integral on $q(z)$ disappears? Commented Apr 13, 2021 at 20:16
• You are quick, I wanted to find an answer myself, that's why I deleted the question. Thanks for the response! And no nevermind that, I misread! Commented Apr 13, 2021 at 20:20
• But yeah, actually I wonder why the integral disappears. It must equal 1, which would make sense for a pdf on the full domain but I'm not operating on the full domain.. why is it 1? Commented Apr 13, 2021 at 20:23
• Another way to view this is to consider that you take the max of $(z1_{Z\leq \theta \mathbb{E}[Z]})$ and you let the integral lie on the whole domain i.e. $Z\geq 0$. Commented Apr 13, 2021 at 20:28

It is possible to understand the argument without writing any integrals.

Why the decomposition is true? Because it is a pointwise inequality. Let $$a = \theta E[Z]$$. The following holds for any random variable $$Z$$ and any $$a \in \mathbb R$$: $$Z = Z \cdot 1_{Z \le a} + Z \cdot 1_{\{Z > a\}}. \quad (*)$$ To see this recall that random variables are just real-valued functions on the sample space. The inequality reads $$Z(\omega) = Z(\omega) \cdot 1_{\{Z(\omega) \le a\}} + Z(\omega) \cdot 1_{\{Z(\omega) > a\}}.$$ A bit more simply, note that $$1 = 1_{\{Z \le a\}} + 1_{\{Z > a\}}$$ and multiply both sides by $$Z$$.

Next, we take the expectation of both sides of (*) to get the desired result.

Why the two bounds work? The first one is again a pointwise inequality: $$Z \cdot 1_{\{Z \le a\}} \le a.$$ The LHS is either 0 or when nonzero it is $$\le a$$. We then take the expectation of both sides which is valid since expectation is a "monotone" operator (i.e., preserves inequalities).

The second bound is the Cauchy-Schwarz inequality which for random variables reads $$E|XY| \le \sqrt{E X^2} \sqrt{E Y^2}.$$ Apply this with $$X = Z$$ and $$Y = 1_{\{Z > a\}}$$ and note that $$E Y^2 = E Y = P(Z > a)$$. The first equality is since $$Y^2 = Y$$ (why?) and the second equality is since the expectation of an indicator variable is the probability of the underlying event, basically by the definition of expectation.

• Thanks for the response! Commented Aug 12, 2022 at 0:45
• You are welcome. Commented Aug 12, 2022 at 1:24