Question - Evaluate the mean and variance of $Y$ if $X$=$U^3$ where $U$ is a uniform random variable in the unit interval [-1,1] and $a$$=$$1/2$ .
Doubt - if Y= g(X) , then i find myself perfectly able to solve the problem;
I would start with $E[Y]$= $\int_{-\infty}^\infty g(X)\,f(x) dx$ and proceed.
However I'm a little bit unsure about the substitution of the required expression here, especially the limits that will vary.
If anyone could solve this, please help out. cheers.
Let $h(x)=x^3$ and $f$ be the PDF of $U$. Then we aim at finding the mean and variance of $Y=g(h(U))$. For the mean we have $$ {\rm E}[Y]=\int_{-\infty}^\infty g(h(x))f(x)\,\mathrm dx=\int_{-1}^1g(x^3)f(x)\,\mathrm dx $$ since $f(x)=0$ outside of $[-1,1]$. Now, note that $x\mapsto g(x^3)$ is an odd function (since it's a composition of two odd functions) and $f$ is an even function. Therefore $x\mapsto g(x^3)f(x)$ is an odd function and hence ${\rm E}[Y]=0$.
To compute the variance we note that $\mathrm{Var}(Y)={\rm E}[Y^2]$ since ${\rm E}[Y]=0$. For this we have $$ {\rm E}[Y^2]=\int_{-1}^1 g(x^3)^2 f(x)\,\mathrm dx. $$ Now we just need to find an expression for the function $x\mapsto g(x^3)^2$. First we have that $$ g(x^3)= \begin{cases} -a,\quad&\text{if }\; x\leq -b,\\ x^3,&\text{if }\; -b< x<b,\\ a,&\text{if }\; x\geq b, \end{cases} $$ with $b=a^{1/3}$ and hence $$ g(x^3)^2= \begin{cases} x^6,\quad &\text{if }\; -b<x<b,\\ a^2,&\text{otherwise}. \end{cases} $$
This enables us to split up the integral accordingly: $$ \int_{-1}^1 g(x^3)^2f(x)\,\mathrm dx=\int_{-1}^{-b} g(x^3)^2 f(x)\,\mathrm dx+\int_{-b}^b g(x^3)^2f(x)\,\mathrm dx+\int_b^1 g(x^3)^2f(x)\,\mathrm dx. $$