How to derive Paley-Zigmund Inequality proof 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.
 A: 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 $L^{2}$,
$$\mathbb{E}[Z\times 1_{Z>\theta \mathbb{E}[Z]}] = \int z1_{Z> \theta\mathbb{E}[Z]}p(z) \leq (\int z^{2}p(z)dz)^{1/2} (\int 1_{Z> \theta \mathbb{E}[Z]}^{2}p(z)dz)^{2}\\
=\mathbb{E}[Z^{2}]^{1/2}\mathbb{P}(Z> \theta \mathbb{E}[Z])^{1/2} $$
A: 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.
