This is from MITx's Intro to Probability and Statistics course, the problem is on this page.

Suppose $X \sim \textrm{Uniform}(0,1)$ and $Y=X^3$. Find the pdf for $Y$.

Since it's a uniform distribution, I believe the pdf for X is $f(x) = \frac{1}{b-a}$ so in this case just 1. The inverse function of Y is, I think, $g^{-1}(y)=\sqrt[3]{y}$ . The formula I have for finding the pdf for y is $f_y(y)=f_x(g^{-1}(y)) \left|\frac{d}{dy}g^{-1}(y)\right|$. So essentially it seems to be the pdf for y should be the derivative of the inverse function $g^{-1}(y)=\sqrt[3]{y}$.

I've run this many times and I keep getting the answer to be $\frac{1}{3y^{2/3}}$, but the course webpage says the answer is $\frac{1}{3}y^{-2}$. Could anyone explain to me why this is so?

  • $\begingroup$ Please add the [self-study] tag & read its wiki. $\endgroup$
    – Silverfish
    Jul 16, 2016 at 15:12

1 Answer 1


One way to check whether you are right, or the webpage is right is to see if the pdfs integrate to 1. $X$ is defined between 0 and 1, so $Y = X^3$ is also defined between 0 and 1.

$$\int_0^1 \dfrac{1}{3} \dfrac{1}{y^2}dy = \dfrac{1}{3}\left[-\dfrac{1}{y} \right]^1_0 = \text{does not converge}. $$

So clearly the webpage is wrong.


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