Given that $X \sim N(0,t)$ and that $Z$ and $X$ are independent, show that for any $a>0$
$$P[|Z-X|>a] \ge P[|X|>a]$$
My approach was to prove this by first for when $Z $ is a constant function, extend the proof to when $Z $ is a simple function and then treat the general $Z $ as a limit of simple functions. I can do the first two steps but not the last.
This is what I have done: since the density of $X$ is given by $\frac{1}{\sqrt{2 \pi t}}e^{-\frac{x^2}{2t}} $ which is symmetric about $0$ and has it's maximum at $0$, we see that $x \mapsto P[|X-x|>a]$ is minimized at $x=0$. Thus the inequality holds when $Z$ is a constant random variable.
Now let $Z$ be such that it may take only two different values $a$ and $b$. Then
\begin{equation*} \begin{split} P[|Z-X|>a] &= \\ &= P[\{Z=a \} \cap \{|Z-X|>a\}]+P[\{Z=b \} \cap \{|Z-X|>a\}] \\ &= P[\{Z=a \} \cap \{|a-X|>a\}]+P[\{Z=b \} \cap \{|b-X|>a\}] \\ &= P[\{Z=a \}] P[ \{|a-X|>a\}]+P[\{Z=b \}] P[\{|b-X|>a\}] \\ &\ge P[\{Z=a \}] P[ \{|X|>a\}]+P[\{Z=b \}] P[\{|X|>a\}] \\ &= P[\{|X|>a\}] \end{split} \end{equation*} Where the last equality follows since $P[\{Z=a \}] + P[\{Z=b \}]=1$
Exactly the same proof works if $Z$ is a simple function taking $n$ different values.
May I in some way by taking $Z $ to be a limit of simple functions extend this proof to general $Z $?
Note: $Z $ is of course assumed to be a random variable.
Much grateful for any help provided!
Isn't $P[|Z-X|>a]= E[ P(|Z-X|>a|Z) ]=E[g(Z)]$ where $g(z)=P[|z-X|>a]$, by the law of iterated expectations, a.k.a. Fubini's theorem? As you point out, $g(z)\ge g(0)$, so the result follows immediately.