Question: Let $X\sim U(0,1)$.Determine the probability density function of the following variable $Y=X^2$.
Defining the p.d.f for $X$ we have $f_X(x)=\begin{cases}0\:\:if\:x\notin[0,1]\\1\:\:if\:x\in[0,1]\end{cases}$
I know that $E(Y)=\int_\limits{R}{}g(x)f_x(x)dx=\int_\limits{0}^{1} x^2\times 1=\frac{1}{2}$, for $Y=g(X)=X^2$
$Var(Y)=E(Y^2)-(E(Y))^2=\frac{1}{5}-\frac{1}{9}=\frac{4}{45}$
According to my resolution the p.d.f should remain the same $f_X(x)=\begin{cases}0\:\:if\:x\notin[0,1]\\1\:\:if\:x\in[0,1]\end{cases}$.
Question: What is the point of askin the p.d.f for $Y$? Or is my answer wrong? If so, why?
Thanks in advance!
Best Answer
No idea why do you compute expectation of $Y$ but
$$\int_0^1 x^2 \, dx = \frac13 $$
The density must have changed.
A possible way to find pdf is to first find the CDF then differentiate it to find pdf:
Let $y \in (0,1)$, \begin{align} Pr(Y \le y) &= Pr(X^2 \le y) \\ &= Pr( X \le \sqrt{y}) \end{align}
Can you complete the task?
Edit:
$$F_Y(y) = \begin{cases} 0 & ,y < 0\\ \sqrt{y} &, y \in [0,1) \\ 1 &, y \ge 1 \end{cases}$$