Random variable $X$ has uniform distribution on section $[0,2]$. What's the expected value of variable $Y=\frac{X^{4}}{2}$

Question

Random variable $X$ has uniform distribution on section $[0,2]$. What's the expected value of variable $Y=\frac{X^{4}}{2}$

I don't know how to start this task. I know formula for density of this distribution:

$f(x))\left\{\begin{matrix} \frac{1}{b-a} &,x\in [a,b] \\ 0 & ,x\notin [a,b]\ \end{matrix}\right.$

If I put values from section I will get:

$f(x))\left\{\begin{matrix} \frac{1}{2} &,x\in [0,2] \\ 0 & ,x\notin [0,2]\ \end{matrix}\right.$

Can I use this to solve this task or should I take a go at it with different method?

Answer 1

There's probably a shortcut somewhere, so you can find $\mathbf{E}Y$ without the pdf, but I don't know it. So first, it's the easiest to find CDF of $Y$: $$ P(Y<y) = P(X^4<2y) = P(0<X<(2y)^{\frac{1}{4}}) $$ Since $X$ is positive, and $X^4$ is a strictly increasing funciton, it's easy enough to find it. From that, take a derivative to get pdf. Now, follow @BrianTung's suggestion by using the definition of $\mathbf{E}X$

Answer 2

The expected value of a probability distribution, $g(x)$, is defined as $\int_{\mathbb{R}}x g(x) dx$. In your case, $g(x)$ is the distribution of $\frac{X^4}{2}$, where $X \in U(0, 2)$. Using the substitution $x = \frac{u^4}{2}$, which is equivalent to the law of the unconscious statistician, the integral becomes $$\int_{0}^{2} \frac{u^4}{2} \cdot\frac{1}{2-0} \ \mathrm d u$$

Can you finish from here?