Let $X$ be an exponential random variable (say with mean 1) and $\Bbb{P}(X \in A) < \Bbb{P}(X \in B)$ for two events $A, B \subset\Bbb{R}$. Is $\Bbb{P}(aX \in A) < \Bbb{P}(aX \in B)$ for $a > 0$, or for $a > 1$?
One strategy I've tried to figure this out is to look at the integrals corresponding to this probability, and rephrasing this way suggests that I seek to show that $$\int \exp(-x)\Bbb{I}_A(x)dx<\int \exp(-x)\Bbb{I}_B(x)dx \implies \int\left( \exp(-x)\Bbb{I}_A(x)\right)^{1/a}dx<\int\left( \exp(-x)\Bbb{I}_B(x)\right)^{1/a}dx$$ but this doesn't seem all that helpful.
This is a question I came up with to help figure out another problem so it may be ill formed.
Not necessarily.
Take for instance $A=(x,+\infty)$ and $B=(0,y)$ for any $x,y>0$ such that $1-\exp(-x)>\exp(-y)$, so that $\mathbb P(X\in A)<\mathbb P(Y\in B)$. For instance, $x=\ln(3)$ and $y=\ln(2)$.
Then $\mathbb P(aX\in A)=\mathbb P(X>\frac xa)$ tends to $1$ as $a$ goes to $+\infty$, and $\mathbb P(aX\in B)=\mathbb P(X<\frac ya)$ vanishes as $a$ goes to $+\infty$. Therefore $\mathbb P(aX\in A)>\mathbb P(aX\in B)$ for $a$ large enough.