The Paley-Zygmund inequality is given by

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

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

\begin{equation} E[Z] = E[Z\mathbf{1}_{\{Z\leq\theta E[Z]\}}]+E[Z\mathbf{1}_{\{Z>\theta E[Z]\}}] \end{equation}

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.

  • $\begingroup$ Check this math.stackexchange.com/questions/1501927/… might be helpful. $\endgroup$
    – Fiodor1234
    Commented Apr 13, 2021 at 17:41
  • $\begingroup$ Thanks, I saw this question before. The answer states my procedure exactly, but without explanation on my two questions. $\endgroup$ Commented Apr 13, 2021 at 17:45
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    $\begingroup$ I hope that the following answer will help. $\endgroup$
    – Fiodor1234
    Commented Apr 13, 2021 at 18:13

2 Answers 2


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 $L^{2}$,

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

  • $\begingroup$ Thank you, this is very clear! $\endgroup$ Commented Apr 13, 2021 at 19:44
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    $\begingroup$ @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? $\endgroup$
    – Fiodor1234
    Commented Apr 13, 2021 at 20:16
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    $\begingroup$ 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! $\endgroup$ Commented Apr 13, 2021 at 20:20
  • $\begingroup$ 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? $\endgroup$ Commented Apr 13, 2021 at 20:23
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    $\begingroup$ 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. $Z\geq 0$. $\endgroup$
    – Fiodor1234
    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.

  • $\begingroup$ Thanks for the response! $\endgroup$ Commented Aug 12, 2022 at 0:45
  • $\begingroup$ You are welcome. $\endgroup$
    – passerby51
    Commented Aug 12, 2022 at 1:24

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