Question: Suppose X~ Uniform(0,1) and $Y=x^3$.What is the pdf for Y?

my attempt: so, X follows a uniform distribution with minimum of 0 and max 1,so $f_X(x)=1.$

Using the definition that the pdf of X is related to $F_X$ by $f_X(x)=F'_X(x)$

let $y=x^3 \Rightarrow dy=3x^2dx \Rightarrow dx=\dfrac{dy}{3x^2}=\dfrac{dy}{3y^{2/3}} $


giving $f_Y(y)=\dfrac{dy}{3y^{2/3}} $, however the answer is given as $\dfrac{1}{3y^2},$ any help as to where i may have went wrong?

sources: http://ocw.mit.edu/courses/mathematics/18-05-introduction-to-probability-and-statistics-spring-2014/readings/MIT18_05S14_Reading5d.pdf


  • 1
    $\begingroup$ Your solution is right! The "answer" you gave does not even have a converging integral in the $[0,1]$ interval. Probably some kind of typo. $\endgroup$ – MBW Jun 9 '15 at 1:37

We have that for $y\in(0,1)$ that $$ P(Y \leq y) = P(X^3 \leq y)=P(X \leq y^{1/3}) =\int_0^{y^{1/3}} f_X(x)dx =y^{1/3} $$ and therefore $$ f_Y(y)=\frac{d}{dy}P(Y \leq y)= \frac{1}{3y^{2/3}} $$ so nothing is wrong.


No, you were correct and the mark-book has a typographical error.

Given $f_X(x) = \mathbf 1_{x\in[0;1]}$ and $Y = X^3$

The map from the support of $X$ onto $Y$ is one-to-one, so the change of variables transformation is a straightforward: $$f_Y(y) = f_X(x)\,\Big\lvert \dfrac{\mathrm d x}{\mathrm d y}\Big\rvert$$

Where $\;x = y^{1/3}\;$, $\;\dfrac{\mathrm d x}{\mathrm d y} = \tfrac 1 3 y^{-2/3}\;$, and $\;\mathbf 1_{x\in[0;1]} = \mathbf 1_{y\in[0;1]}\;$.

So, therefore we do have: $$f_Y(y) = \frac{1}{3\,y^{2/3}}\;\mathbf 1_{y\in[0;1]}$$

As you had.

We can also confirm that $\;\int_0^1 \frac{1}{3\,y^{2/3}} \operatorname d x = 1\;$ , as is required for a probability density function.

  • $\begingroup$ Thanks, shows the importance of thinking from first principles... I just assumed i messed up somewhere as I usually do. Didn't actually think to assess if their answer made sense. $\endgroup$ – Ryan Joseph Jun 9 '15 at 2:13

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.