# Probability Density Function - help with domain

Let X be a continuous random variable with pdf

$f(x) = \begin{cases} 1/4, & \text{if$x\in$(-2,2)} \\ 0, & \text{otherwise} \end{cases}$

Find the probability density functiions of i) Y=$X^3$ and ii) Z= $X^4$

So far I have done this for i)

$F_Y(Y) = P(Y \le y) = P(X \le y^{1/3}) = F_X(y^{1/3}) = \int1/4.dt$

Carrying this through, I get an answer that is 1/2 of what I should be getting.

For the integral (the last part I was up to) I am using $y^{1/3}$ and $-y^{1/3}$ as my upper and lowers but I am pretty sure I should be using $y^{1/3}$ and 0. Can someone tell me why?

• First, sketch a graph of $Y$ versus $X$ and $Z$ versus $X$ to determine the range of values that $Y$ and $Z$ take on. Then work the problem again. – Dilip Sarwate Jun 13 '17 at 11:55

## 2 Answers

$f_X(x) = \tfrac 14\mathbf 1_{x\in[-2;2]}$

The CDF of $Y:=X^3$ will be :

$$F_{X^3}(y) ~{= \mathsf P(X^3\leq y) \\= \mathsf P(X\leq \sqrt{y}) \\= F_X(\sqrt{y})}$$

Now, you could evaluate this by taking the integral of the pdf for $X$, but as we only seek the pdf for $Y$, we can save quite a bit of work by immediately applying the chain rule to the unsigned derivative.

$$f_{X^3}(y) ~{= \begin{vmatrix}\dfrac{\mathrm d~F_{X}(\sqrt{y})}{\mathrm d ~y~\qquad}\end{vmatrix} \\= \begin{vmatrix}\tfrac 1{3y^{2/3}}\end{vmatrix}f_X(\sqrt{y})}$$

Likewise the pdf for $Z:=X^4$ can be found by:

\begin{align}F_{X^4}(z)~&{=~ \mathsf P(X^4\leq z) \\ =~ \mathsf P(-\sqrt{z}\leq X\leq \sqrt{z})\\ ~~\vdots}\end{align}

Do you know the transformation of pdf theorem? That is, for $Y=g(X)$, $g:\mathbb{R}\to\mathbb{R}$ is one-to-one, $$f_Y(y)=f_X(g^{-1}(y))\left|\frac{d}{dy}g^{-1}(y)\right|.$$ For $Z$ you have to apply another approach. Probably you can figure it out by yourself.