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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?

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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.

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  • $\begingroup$ 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

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