Let random variable X have the probability density function

f(x)= $\frac{x}{2}$ for $0<x<2$

0 otherwise

Find the pdf of $Y=X^3$

Now I’m quite new to this, I’m trying to find $P(Y<y)$ and then finding the CDF from it integrating which I can just differentiate to get the pdf of Y but I’m not sure about the change of variable.



$$P(Y\leq y)=P(X^3\leq y)=P(X\leq y^{1/3})$$

  • $\begingroup$ is this right: $g^{-1}(Y)=y^{1/3}$ $\frac{dg^{-1}(y)}{dy} = \frac{1}{3}y^{-\frac{2}{3}}$ $F_{y}Y = \frac{y^{-\frac{2}{3}}}{6}$ $\endgroup$ – user655883 Apr 25 '20 at 17:51
  • $\begingroup$ If you want to use the Jacobian method:$f_Y(y)=|J|f_X(y^{1/3})$ where $J=\frac{1}{3} y^{-2/3}$ $\endgroup$ – Masoud Apr 25 '20 at 17:56

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