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

Mathematics Asked by fakeraker p on December 5, 2020

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:

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

If I put values from section I will get:

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

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

2 Answers

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 $$int_{0}^{2} frac{u^4}{2} cdotfrac{1}{2-0} mathrm d u$$

Can you finish from here?

Answered by Varun Vejalla on December 5, 2020

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 $mathbf{E}X$

Answered by Alex on December 5, 2020

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