Density tranformation theoren, n=1 - exercise solution

Question

Let X be uniform on $[0,1]$, and let $Y=\sqrt{x}$. Find $E(Y)$ by

a) finding density of $Y$ and then finding the expectation, and

b) by using definition $\mathbb{E}[g(x)]=\int g(x)f(x)dx$.

attempted SOLUTION:

$$f^{-1}(x): x=y^2$$

$$ \frac {d}{dy} f^{-1}(y) = 2y $$

According to the denity transfornation theorem:

$$ f(y) = f_x(f^{-1}(y))|\frac{d}{dy}f^{-1}(y)|= y \space 2y$$

Provided SOLUTION:

$$f(y)= 2y$$

I have not finished the exercise because i need the correct expression for $f(y)$ in order to calculate the $\mathbb{E}[Y]$. Where is my mistake?

Answer 1

Density of uniform distribution is $1$ on $(0,1)$. So $f_x(f^{-1} (y))=1$. Now you should be able to use either of the two methods to get $EY=\frac 2 3 $

Answer 2

For $y\in(0,1)$ we have \begin{align} F_Y(y) :&= \mathbb P(Y\leqslant y)\\ &= \mathbb P(\sqrt X\leqslant y)\\ &= \mathbb P(X\leqslant y^2)\\ &= y^2, \end{align} and the density of $Y$ is obtained by differentiating: $$ f_Y(y) = \frac{\mathsf d}{\mathsf dy} F_Y(y) = 2y,\ 0<y<1. $$ It follows that $$ \mathbb E[Y] = \int_0^1 2y\ \mathsf dy = \frac23, $$ or by using the law of the unconscious statistician, $$ \mathbb E[Y] = \mathbb E[\sqrt X] = \int_0^1 \sqrt x\ \mathsf dx = \frac23. $$