Is $\mathbb{P}(\mathbb{X}\leq a-\epsilon) \leq \mathbb{P}(\mathbb{X}\leq a-\mathbb{Y}| |\mathbb{Y}| \leq \epsilon)?$

Question

This is perhaps a trivial question concerning implication of events. Suppose $\mathbb{X},\mathbb{Y}$ are random variables on a common space equipped with the measure $\mathbb{P}$. Let $a,\epsilon$ be positive constants s.t. $\epsilon < a$. Then, is it true that $\mathbb{P}(\mathbb{X}\leq a-\epsilon) \leq \mathbb{P}(\mathbb{X}\leq a-\mathbb{Y}| |\mathbb{Y}| \leq \epsilon)?$. This seems to be trivially true since the event $\{\mathbb{X}\leq a-\epsilon\}$ implies the event $\{\mathbb{X}\leq a-\mathbb{Y}| |\mathbb{Y}| \leq \epsilon\}$. Is this correct or am I missing some subtlety that appears due to the conditional operator? In particular, can we draw implications between such events, where one is a conditional event, and the other is not?

Answer 1

Define $X$ and $Y$ as follows. Flip a fair coin. If it comes up heads, set $X=0$ and $Y=1$. If it comes up tails, set $X=2$ and $Y=0$. Let $a=1$ and $\epsilon=1/2$. The probability that $X\leq a-\epsilon=\frac12$ (meaning the coin came up heads) is $\frac12$. But the conditional probability that $X\leq a-Y$ given $Y\leq\epsilon$ is zero. (Details about that zero: The only time $Y$ is less than $\epsilon$ is when the coin flip came up tails, and then $X=2$ while $a-Y=1$ so $X$ is not less than $a-Y$.)

Answer 2

You cannot draw such conclusion. The probability $\mathbb P(\mathbb X\le a-\epsilon)$ does not involve $\mathbb Y$ at all, whereas in $\mathbb P(\mathbb X\le a-\mathbb Y| | \mathbb Y | <\epsilon)$, it involves the dependence between $\mathbb X $ and $\mathbb Y$. For instance, if we take $\mathbb X = a +\epsilon- 2 \epsilon \mathbf 1 _{\{|\mathbb Y |\ge \epsilon\}}$, then $\mathbb P(\mathbb X\le a-\mathbb Y| | \mathbb Y | <\epsilon)=0$ but generally $\mathbb P(\mathbb X\le a-\epsilon) = \mathbb P( |\mathbb Y |\ge \epsilon) >0$ if $\mathbb P( |\mathbb Y |\ge \epsilon) >0$.

On the other hand, the inequality is correct if $\mathbb X$ and $\mathbb Y$ are independent.