Uniform distribtion: clarification of $f_X(x)$

Question

I have $Y=2(X-1)^2 -1$ where $X$ is uniform distributed on $[0,2]$

I want to find the pdf of $Y$ and expected value of $Y$.

My question is just: Does $X$ have pdf $f_X(x)= \frac{1}{2}$?

Answer 1

A uniformly distributed R.V. on $[0,2]$ has a p.d.f. of $$ f_X(x) = \begin{cases} \frac{1}{2} & x \in [0,2] \\ 0 & \text{ otherwise} \end{cases} $$ One easy way to ensure you're using the correct with the p.d.f. for some uniform R.V. Z on some set $S$ is by noting that by definition of a uniform random variable we have $$ f_Z(z) = \begin{cases} c & z \in S \\ 0 & z \not\in S \end{cases} $$ where $c \in \mathbb{R}$. Now in order to find $c$ we note that $$ \int_{\text{All Space}} f_Z(s) \,\mathrm{d}z = 1 \implies \int_S c \, \mathrm{d}z = 1 \implies c = \frac{1}{|S|} $$ where $|S| = \int_S \,\mathrm{d}z$ denotes the size of the set (if you're familiar with measure theory and you're using some measure $\mu$ then it's $\mu(S)$).

Now one method to find the derived p.d.f. of the random variable $Y = 2 ( X - 1 )^2 - 1$ is to find the c.d.f. of $Y$ and then differentiate to find the p.d.f.. In order to find the c.d.f. we look at the following: $$ F_Y(y) = P(Y \le y) = P(2 ( X - 1 )^2 - 1 \le y) = P\left(-\sqrt{\frac{y+1}{2}} + 1 \le X \le \sqrt{\frac{y+1}{2}} + 1\right) $$ Now we can integrate the p.d.f. of X over this set above in order to find the c.d.f. $Y$ and finally differentiate with respect to $y$ in order to find the p.d.f. of $Y$. I'm leaving the rest to do for the OP since from here on out it's plug and chug. @Katie, if you need more help comment on this and I'll try to help more!

EDIT: Let's take a closer look at the c.d.f. and the p.d.f. of $Y$. Since I can see @Katie understands the plug and chug process I don't feel wrong writing it all out here.

Note first that we calculate the c.d.f. as follows: $$ F_Y(y) = \int_{-\sqrt{\frac{y+1}{2}} + 1}^{\sqrt{\frac{y+1}{2}} + 1} f_X(x) \, \mathrm{d} x $$ Here we need to be careful with evaluating this integral. We need to find when (for what $y$) we will be integrating over $[0,2]$ and when (for what $y$) we will not. We have a couple cases to consider here:

• $y < -1$: When we took the square root we implicitly restricted the range of $Y$ and thus the domain of the c.d.f. (and p.d.f.) so that $y$ cannot be $< -1$.
• $-1 \le y \le 1$: For any $y$ in this interval we will be integrating over some subset of $[0,2]$.
• $y > 1$: Here we will be integrating over some superset of $[0,2]$, that is some set which contains $[0,2]$

Taking all of this into account we can define our c.d.f. as follows:

$$ F_Y : [-1, \infty) \to \mathbb{R} \\ F_Y(y) = \begin{cases} \int_{-\sqrt{\frac{y+1}{2}} + 1}^{\sqrt{\frac{y+1}{2}} + 1} \frac{1}{2} \, \mathrm{d} x & y \in [-1,1] \\ 1 & \text{otherwise} \end{cases} $$

which can be simplified, and extended to the entire real line as follows: $$ F_Y(y) = \begin{cases} 0 & y < -1 \\ \sqrt{\frac{y+1}{2}} & y \in [-1,1] \\ 1 & \text{otherwise} \end{cases} $$

You can easily check that this is in fact continuous and and differentiable on the entire real line so we can define the p.d.f. as follows: $$ f_Y(y) = \frac{\mathrm{d}F_Y(y)}{\mathrm{d} y} = \frac{1}{2 \sqrt{\frac{y+1}{2}}} \cdot \frac{1}{2} = \frac{1}{4 \sqrt{\frac{y+1}{2}}} \text{ on } [-1,1] $$ and since the derivative is zero all elsewhere our final p.d.f. is: $$ f_Y(y) = \begin{cases} \frac{1}{4 \sqrt{\frac{y+1}{2}}} & y \in [-1,1] \\ 0 & \text{otherwise} \end{cases} $$