I don't know why I am having such trouble with this. I've tried using De Morgan but haven't made much progress yet.
Prove if a>0, b>0, P(X\cdot Y\le b)\ge P(X\le b/a)-P(Y>a), where $X,Y$ random variables?
0
$\begingroup$
probability
-
0Nevermind, did it via De Morgan. You are welcome to post another technique though. – 2011-11-16
2 Answers
2
This is actually false without additional assumptions on the supports of $X$ and $Y$.
Try $a=b=1$, with $X=Y=-2$ (deterministic) for a counter-example, \begin{align*} \mathbb P(XY\leq a) &= 0\\ \mathbb P(X\leq b/a) &= 1\\ \mathbb P(Y> a) &= 0 \end{align*}
However, you may either suppose additionally that:
- $X>0$ almost surely;
- $Y>0$ almost surely;
- or simply replace the formula with $P(X\cdot Y\le b)\ge P(\lvert X\rvert\le b/a)-P(\lvert Y\rvert>a)$
1
$ \begin{array}{rcl} P(XY ≤ b) & \ge & P (X ≤ b/a, Y \le a) \quad \quad \leftarrow \mbox{caution, see correction below}\\ P( Y > a) & \ge & P (X ≤ b/a, Y > a)\\ P(XY ≤ b)+P( Y > a) & \ge & P (X ≤ b/a, Y \le a) + P (X ≤ b/a, Y > a)=P (X ≤ b/a) \end{array} $
-
0$P(XY \le b) \ge P (X \le b/a, Y \le a)$ is true when everything is non-negative but FelixCQ's counter example applies here too. – 2011-11-16