# Finding Probability density function of a random variable with transformation

Let random variable X have the probability density function

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

0 otherwise

Find the pdf of $$Y=X^3$$

Now I’m quite new to this, I’m trying to find $$P(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.

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