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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.

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  • $\begingroup$ Check this math.stackexchange.com/questions/1501927/… might be helpful. $\endgroup$
    – Fiodor1234
    Apr 13 at 17:41
  • $\begingroup$ Thanks, I saw this question before. The answer states my procedure exactly, but without explanation on my two questions. $\endgroup$
    – MarcelB
    Apr 13 at 17:45
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    $\begingroup$ I hope that the following answer will help. $\endgroup$
    – Fiodor1234
    Apr 13 at 18:13
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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 $L^{2}$,

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

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  • $\begingroup$ Thank you, this is very clear! $\endgroup$
    – MarcelB
    Apr 13 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
    Apr 13 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! $\endgroup$
    – MarcelB
    Apr 13 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? $\endgroup$
    – MarcelB
    Apr 13 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
    Apr 13 at 20:28

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