I have been thinking about the following problems:
If two random variable have the relationship $|X_1|<|X_2|$, we have that their tail probabilities satisfy the inequality that: $$P(|X_1|>t)\leq P(|X_2|>t)$$
This is trivially true since the event $\{|X_1|>t\}$ will imply the event $\{|X_2|>t\}$.
What if we add some independent unimodal random variable $Y$ whose pdf is symmetric around the origin, e.g., $Y\sim N(0,1)$.
Then for the random variable $X_1+Y$ and $X_2+Y$
Do we have the following relationship
$$P(|X_1+Y|>t)\leq P(|X_2+Y|>t)?$$
Moreover, if we weaken the condition "$|X_1|<|X_2|$ almost surely" to "$P(|X_1|>t)\leq P(|X_2|>t)$ for any $t>0$", with all else being the same, does the result $P(|X_1+Y|>t)\leq P(|X_2+Y|>t)$ still hold?
This problem interests me in that I wonder to what extent, the tail probablity comparison is invariant to an outer-source noise term. I wonder if anyone have seen some previous research on this topic.
Thank you.
Suppose $Y$ has PDF $\varphi$ satisfying $\varphi(y) = \varphi(-y)$, and $\varphi(y)$ is a decreasing function of $y > 0$.
Let \begin{align} f(x) = P(|x+Y|>t) = 1 - P(-x-t\le Y\le-x+t) .\end{align} Note that $f(x)$ is an even function. Also note $$ f'(x) = \varphi(x-t) - \varphi(x+t) $$ is positive for $x > 0$.
So if $0 < |x| < |y| $, then $f(x,t) \le f(y,t) $.
So if $|X_1(\omega)| \le |X_2(\omega)|$, then $$ P(|X_1(\omega)+Y|>t) \le P(|X_2(\omega)+Y|>t) .$$ Integrate both sides over $\omega$, and you get $$ P(|X_1+Y|>t) \le P(|X_2+Y|>t) .$$