Let $$f_X(x)=\begin{cases}c \cdot x&\text{for }0 \leq x \leq 1\\
0&\text{otherwise }\end{cases}$$
with $c > 0 $ be the probability density function of the random variable $X$. Find the probability density function of $Y:= X^2.$
First we find $c:$
$$\int_{-\infty}^\infty {f_X(x) \text{ }dx = 1} = \int_{0}^1 cx \text{ } dx \Rightarrow c = 2. $$
Since $f_Y(y)$ is the derivative of the cumulative distribution function $F_Y(y)$, we first need to calculate $F_Y(y):$
$$F_Y(y) = P(Y \leq y) = P(X^2 \leq y) = P(X \leq \sqrt{y}) = F_X(\sqrt{y}).$$
Now to find $F_X(t):$
$$F_X(t) = \int_{0}^t 2x \text{ } dx = t^2 \Rightarrow F_X(t)=\begin{cases}0&\text{for }t \in ]-∞, 0[ \\
t^2&\text{for } t \in [0, 1] \\
1 &\text{for } t \in ]1, ∞[\end{cases}$$
$\Rightarrow F_Y(y) = P(X \leq \sqrt{y}) = y$ if $0 \leq \sqrt{y} \leq 1$ and $1$ if $\sqrt{y} > 1$.
$\Rightarrow f_Y(y) = F_Y(y)' = 1 $ if $0 \leq y \leq 1$ and $0$ if $y > 1$.
Can you please check my work? I'm not sure about the last part since $F_Y(y)$ is not defined for all $y \in R$ (it's undefined for $y < 0$). Thank you.
Best Answer
That's pretty ok. You found that
$$f_Y(y)=\mathbb{1}_{[0;1]}(y)$$
In other words
$$Y\sim U[0;1]$$
Yes it is. $F_Y(y)=0$ $\forall y<0$
$$ F_Y(y) = \begin{cases} 0, & \text{if $y<0$ } \\ y, & \text{if $0\leq y<1$ } \\ 1, & \text{if $y \geq 1$ } \end{cases}$$
In this kind of exercise it is easier to directly find $f_Y(y)$ without passing by its CDF:
$$ \bbox[5px,border:2px solid black] { f_Y(y)=f_X[g^{-1}(y)]\bigg|\frac{d}{dy}g^{-1}(y)\Bigg| \ } $$
Substituting you get immediately
$$f_Y(y)=2\sqrt{y}\frac{1}{2\sqrt{y}}=1$$
$y \in[0;1]$ (and $0$ elsewhere)