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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.

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    Nevermind, did it via De Morgan. You are welcome to post another technique though.2011-11-16

2 Answers 2

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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 $P(X\cdot Y\le b)\ge P(\lvert X\rvert\le b/a)-P(\lvert Y\rvert>a)$
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$ \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} $

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    $P(XY \le b) \ge P (X \le b/a, Y \le a)$ is true when everything is non-negative but FelixCQ's counter example applies here too.2011-11-16