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?


It’s again continuous since its support is all non-negative numbers. But, CDF is wrong. We deal with transformations as follows in general:

$$F_Y(y)=P(Y\leq y)=P(X^2\leq y)=P(\sqrt{y}\geq X\geq -\sqrt{y})=F_X(\sqrt{y})-F_X(-\sqrt{y})$$ when $y\geq 0$. If you want pdf of Y, just take the derivative of this.

| cite | improve this answer | |
  • $\begingroup$ Shouldn't the last part of the expression be -Fx(sqrt(y)), not -Fx(-sqrt(y))? $\endgroup$ – Crystal McMillian Jan 18 '19 at 5:36
  • $\begingroup$ Also apologies for the syntax, I am new here and I don't know how to use the mathematical syntax in my posts in the comments. $\endgroup$ – Crystal McMillian Jan 18 '19 at 5:36
  • $\begingroup$ You need to subtract $P(X < a)$ from $P(X < b) $ to get $P(a < X < b)$. (Ignoring equalities because of continuity) $\endgroup$ – gunes Jan 18 '19 at 6:16
  • $\begingroup$ Ah, thank you for the reminder. As for the pdf I believe it would be $\frac {1}{2}fx(\frac {1}{2\sqrt{y}}) - \frac {1}{2}fx(\frac {-1}{2\sqrt{y}})$ $\endgroup$ – Crystal McMillian Jan 18 '19 at 6:26
  • $\begingroup$ @CrystalMcMillian derivative of $\sqrt{y}$ goes out of $f$. Review the chain rule. $\endgroup$ – gunes Jan 18 '19 at 6:40

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.