# How to show that $P(Y \geq y) \leq P(Y^2 \geq y^2)$?

I am wondering why it is the case that $P(Y \geq y) \leq P(Y^2 \geq y^2)$. In general, I am unclear what are the rules for transformations within the probability sign itself. Could anyone give me a hint? thanks.

It is not true in general.

If e.g. $Y$ only takes values in $(-1,1)$ then $P(Y\geq-1)=1$ and $P(Y^2\geq (-1)^2)=0$.

If $y\geq0$ then $Y\geq y\implies Y^2\geq y^2$ or equivalently $\{Y\geq y\}\subseteq\{Y^2\geq y^2\}$.

So in that case $P(Y\geq y)\leq P(Y^2\geq y^2)$.

Check that $$P(Y^2 \ge y^2) = P(Y\ge |y|) + P(Y\le -|y|)$$

Hint if $y>0$

$Y^2 \geq y^2$ if and only if $Y\geq y \;\vee Y\leq -y$

• Yes, thanks for the correction – user26977 Jan 29 '16 at 14:11