Deriving a variant of Paley-Zygmund inequality for Gaussian variables

Question

Recently I have encountered the following inequalities:

Let $Z \sim \mathcal{N}(0,\sigma^2)$, then \begin{equation} \forall t \in \mathbb{R},\text{ } P(|t+Z| \geqslant \sigma) \geqslant P(|Z| \geqslant \sigma) \geqslant 3/10,\quad\quad\quad(1) \end{equation} which is called Palye-Zygmund(PZ) lower bound by the author.

While the first inequality may come from properties of Gaussian variables (I have not figured out which exact property it is), I suppose that the second inequality is derived from the standard PZ inequality. However, I encountered some difficulties in deriving it.

Considering the standard PZ inequality: For a non-negative random variable $X$ and any $0 \leqslant \theta < 1$, it holds $$ P(X > \theta E(X)) \geqslant (1-\theta)^2 \frac{E^2(X)}{E(X^2)}. \quad\quad\quad(2) $$

To derive the second inequality in $(1)$, what I tried was to plug $X=Z^2$ into $(2)$, which then leads to $$ P(|Z|> \sqrt{\theta} \sigma) \geqslant (1-\theta)^2 \frac{1}{3}. $$ I did not manage to derive $(1)$ from here. Any suggestions about the derivation are appreciated.

Answer 1

In this answer I will derive $P(|Z|\geq \sigma)\geq\frac{3}{10}$ using first principles. Because the normal random variable $Z$ is centered (i.e. mean zero), we have $$P(|Z|\geq \sigma)=P(Z\leq -\sigma)+P(Z\geq \sigma)=2\cdot P(Z\geq \sigma).$$ So it suffices to prove that $P(Z\geq \sigma)\geq \frac{3}{20}$. We have that $Z/\sigma$ is a standard normal random variable. So if $X\sim\mathcal{N}(0,1)$, it further suffices to prove $P(X\geq 1)\geq 3/20$, which reduces the problem to showing

$$P(0\leq X\leq 1)=\int_0^1 \frac{1}{\sqrt{2\pi}}e^{-x^2/2}dx\leq \frac{1}{2}-\frac{3}{20}=\frac{7}{20}.$$ From the inequality $\exp({-x^2/2})\leq 1-x^2/2+x^4/8$ for all $x\in [0,1]$ (in fact for all $x\in\mathbb{R}$), we get that

$$\int_0^1\frac{1}{\sqrt{2\pi}}e^{-x^2/2}dx\leq \int_0^1\frac{1}{\sqrt{2\pi}}(1-\frac{x^2}{2}+\frac{x^4}{8})dx=\frac{103}{120\sqrt{2\pi}}\leq \frac{7}{20},$$ finishing the derivation.

Edit: To prove $\exp({-x^2/2})\leq 1-x^2/2+x^4/8$ (by making the substitution $t=x^2/2$) it suffices to prove that $$\exp(-t)\leq 1-t+\frac{t^2}{2}.$$ for all $t\geq 0$. Write $f(t):=\exp(-t)$ and $g(t) :=1-t+t^2/2$. It is enough to show that (a) $f(0)\leq g(0)$, and (b) $f'(t)\leq g'(t)$ for all $t\geq 0$. It is easy to check (a), and checking (b) reduces to showing that $1-t\leq e^{-t}$ for $t\geq 0$. There are many ways to show this, but I will do a probabilistic argument (for $t>0$, because the inequality is trivial when $t=0$). Let $U$ have a Poisson distribution with rate parameter $t>0$. From Markov's inequality, $$P(U\geq 1)\leq E[U]\implies 1-e^{-t}\leq t,$$ giving us the desired inequality.