# Manipulating continous random variables

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

$f_X(x)dx=dx=\dfrac{dy}{3y^{2/3}}=f_Y(y)dy$

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?

http://ocw.mit.edu/courses/mathematics/18-05-introduction-to-probability-and-statistics-spring-2014/readings/reading-questions-5d/

• 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. – MBW Jun 9 '15 at 1:37

## 2 Answers

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.

• 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. – Ryan Joseph Jun 9 '15 at 2:13