Sorry for the probably very easy question;

From a mathematical perspective I would have said yes, it holds because

$$\mathbb P (\max(X,Y) \leq t \vert X \geq Y)= \mathbb P(X \leq t \vert X \geq Y)$$

and now the condition about $Y$ becomes unimportant since it is not part of the probability anymore and can be omitted - however if I know that some random variable is bigger than another (or in a bigger example bigger than a bunch of random variables $Y_1,...,Y_n$), then the probability that $X$ is very big should also be higher or am I wrong?


Counterexample. Even if we imagine that $X$ and $Y$ are independent, suppose that $Y$ always equals $1$, $X$ equals $0$, $2$, or $4$ with probability $1/3$ each, and $t = 3$. Then

$$ P(\max(X, Y) \leq t \mid X \geq Y) = \frac{P(\max(X, Y) \leq t, X \geq Y)}{P(X \geq Y)} = \frac{1/3}{2/3} = \frac12 $$


$$ P(X \leq t) = \frac23 $$

  • 1
    $\begingroup$ Thank you, so my intuition was correct $\endgroup$ – user299124 Feb 15 at 1:51

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