Can we apply $P(|X|\geq t)=2 P(X\geq t)$ for a discrete centered and symmetric random variable?

Question

I am reading here that for a centered variable $X$ we have

$$P(|X|\geq t)=P(X\leq -t)+P(X\geq t)=2\cdot P(X\geq t)\tag{1}$$ I suppose $X$ has also to be symmetric in order to have (1). For a Rademacher r.v. $X$, which is centered and symmetric, we have that

$$P(X\geq t) \leq \sqrt{(1+t)^{-1-t}(1-t)^{-1+t}}\tag{2}$$ Is it correct to use (1) in order to write for a Rademacher r.v. $X$ that

$$P(|X|\geq t) = 2\cdot P(X\geq t) \leq 2 \sqrt{(1+t)^{-1-t}(1-t)^{-1+t}} \tag{3}$$ with $t < 1$.

In order to study the correctness of (3), I have tried to compute the upper bound of $P(|X|\geq t)$ using Cramer-Chernoff method. We have the logarithm of the MGF of |X|

$$\psi_{|X|} (\lambda) = \ln E[e^{\lambda |X|}] = \ln \left(\frac{1}{2}\left( e^{\lambda \cdot |1|} + e^{\lambda \cdot |-1|}\right)\right) = \lambda \tag{4}$$

Then we have the Chernoff's inequality

$$\operatorname P (|X| \geq t) \leq \exp (-\psi_{|X|}^*(t)) \tag{5}$$

where

$$\psi_{|X|}^* (t) = \sup_{\lambda \geq 0 } (\lambda t - \psi_{|X|} (\lambda)) = 0\tag{6}$$

with $-1 < t < 1 \stackrel{\lambda \geq 0}{\Leftrightarrow} - \lambda \leq t \lambda - \psi_{|X|} (\lambda) \leq 0$, which implies $\psi_{|X|}^* (t)=0$, which in turn implies $\operatorname P (|X| \geq t) \leq 0$ which combined with the fact that for every probability $\operatorname P (|X| \geq t) \geq 0$, we conclude that $\operatorname P (|X| \geq t) = 0$.

Am I doing something wrong? This means also that (1) can not hold for a Rademacher r.v.?

Could you please someone cast some light

Answer 1

I assume $t>0$ throughout my answer.

The first equality, i.e., $P(|X|\geq t)=P(X\leq -t)+P(X\geq t)$, is always true by definition.

The second equality, i.e., $P(X\leq -t)+P(X\geq t) = 2 \cdot P(X\geq t)$, is true if and only if $P(X\leq -t) = P(X\geq t)$. Centered (i.e., zero-mean) random variables do not necessarily have this property, as you are aware of. You are correct that the random variable being symmetric is a sufficient condition for this second equality to hold (and in fact necessary if you want it for all $t$, by definition).