The variance of $Z=Y^2 + Y + 2017$

Question

Find the variance of $Z=Y^2 + Y + 2017$ where X is uniformly distributed over [-1,3] and $Y$ is the distance from $X$ to the nearest endpoint of $[-1, 3]$

I know that $Var(Z)$ = $Var(Y^2 + Y + 2017)$ = $Var(Y^2+Y)$. I defined

$Y=3-X$ when $X<1$ and

$Y=X+1$ when $x>1$.

So $Y$ can be written as $Y= 2 - |1-X|$

I also calculated $f(x) = 1/4$ on $[-1,3]$ and $0$ otherwise. I tried to plug $P=Y^2+Y$ into $Var(P) = E[P^2] - (E[P])^2$ definition but couldn't solve it. Any help or hint is appreciated.

Answer 1

Hint: Repeat the below for $k = 1, \dots, 4$: $$\mathbb{E}[Y^k]=\int_{-1}^{3}[2-|1-x|)^kf(x)\text{ d}x=\int_{-1}^{1}[2-(1-x)]^k\cdot \dfrac{1}{4}\text{ d}x+\int_{1}^{3}[2-(x-1)]^k\cdot \dfrac{1}{4}\text{ d}x$$ Notice $$\begin{align} \text{Var}(Y^2+Y)&=\mathbb{E}[(Y^2+Y)^2]-\left(\mathbb{E}[Y^2+Y] \right)^2 \\&= \mathbb{E}[Y^4]+2\mathbb{E}[Y^3]+\mathbb{E}[Y^2]-\left(\mathbb{E}[Y^2]+\mathbb{E}[Y] \right)^2 \end{align}$$