# Continous random variable short proof

I am given the problem:

If X is a continuous random variable with cumulative distribution function F and density function f, show that the random variable Y = X^2 is also continuous and express its cumulative distribution function in terms of F and f.

Isn't a continuous random variable multiplied by itself also a continuous random variable? If so, can't we just say Y = F^2?

$$F_Y(y)=P(Y\leq y)=P(X^2\leq y)=P(\sqrt{y}\geq X\geq -\sqrt{y})=F_X(\sqrt{y})-F_X(-\sqrt{y})$$ when $$y\geq 0$$. If you want pdf of Y, just take the derivative of this.
• You need to subtract $P(X < a)$ from $P(X < b)$ to get $P(a < X < b)$. (Ignoring equalities because of continuity) – gunes Jan 18 at 6:16
• Ah, thank you for the reminder. As for the pdf I believe it would be $\frac {1}{2}fx(\frac {1}{2\sqrt{y}}) - \frac {1}{2}fx(\frac {-1}{2\sqrt{y}})$ – Crystal McMillian Jan 18 at 6:26
• @CrystalMcMillian derivative of $\sqrt{y}$ goes out of $f$. Review the chain rule. – gunes Jan 18 at 6:40