$Y=AX + BX^3$. Compute Var(Y|X=x) $X$ has distribution $N(1,1)$ and $A$ & $B$ both have distribution $U(-1,1)$.

Question

$Y=AX + BX^3$, $X$ has distribution $N(1,1)$ and $A$ & $B$ both have distribution $U(-1,1)$. $X,A, B$ are all independent. May assume without proof that X and Y are jointly continuous and that $B$ and $X^k$ are independent for all $k \in \Bbb N$. How do I compute $Var(Y|X=x)$? (In previous parts of the question I was asked to show $E[Y|X] = 0$, compute $E[Y]$, which I showed was $0$, and compute $corr(X,Y)$ which I also showed was $0$). Please point me in the right direction with an appropriate hint. (Could I also check my answer once computed to check my understanding?) Thanks.

Answer 1

Note that $Var(A)=Var(B)=\frac{(1-(-1))^2}{12}=\frac{1}{3}$.

Now, intuitively,

$\begin{align}Var(Y|X=x) &= & Var(AX+BX^3|X=x)\\&=&Var(Ax+Bx^3)\\&=&x^2Var(A)+(x^3)^2Var(B)\\&=&\frac{x^2+x^6}{3}\end{align}$.

Can you formalize the idea?