Find $P(Y>13)$ if $Y = X^2$

Question

Let $F_X$ be a CDF of a random variable $X$ and let $Y = X^2$. Find $P(Y > 13)$.

We know that $P(Y>13)= 1 - P(Y \leq 13)$. Let's examine the second part.

$$\begin{align} P(Y \leq 13) =& P(X^2 \leq 13) \\ =& P(|X| \leq \sqrt{13}) \\ =& P(X \leq \sqrt{13}) + P(X \geq - \sqrt{13}) \\ =& P(X \leq \sqrt{13}) + 1 - P(X < - \sqrt{13}) \\ =& F_X(\sqrt{13}) + 1 - F_X(-\sqrt{13}) \end{align}$$

Thus I get

$$P(Y > 13) = F_X(-\sqrt{13}) - F_X(\sqrt{13})$$

Is this solution correct?

Answer 1

Your solution is incorrect but you were almost there. You used at the third equal sign that the events are disjoint, which is not true. You should have done the following $$ P(|X|\leq \sqrt{13}) = 1 - P(|X|> \sqrt{13}) = 1 -( P(X > \sqrt{13} ) + P(X < -\sqrt{13})) = 1 - (1 - P(X \leq \sqrt{13}) +P(X < -\sqrt{13})) = F_X(\sqrt{13}) - F_X(-\sqrt{13})$$

I hope this helps!

Answer 2

Note that $$\Pr\{|X| \le a\} = \Pr\{-a \le X \le a\} = F_{X}(a)-F_{X}(-a).$$

Answer 3

No, but easy to fix. Notice $$ P\left(|X| \leq \sqrt{13}\right) = P(-\sqrt{13} \leq X \leq \sqrt{13}) = F(\sqrt{13}) - F(-\sqrt{13}) $$