Let $X,Y$ and $Z$ be three indenependent real valued random variables. Al with finite second momennt and all with mean $0$ and variance $1$. Define $$ W= \frac{X+YZ}{\sqrt{1+Z^2}} $$
Show that $V(W\mid Z)=1$, a.e.
First we note that because on independence we have that $$ V(W\mid Z)=V(W) $$ Then writing the variance in expectation form \begin{align*} V(W) & = E(W^2)-(EW)^2 \end{align*} First I find $EW$. But because of lineariy of the expectation operator and the independence of the variables, we can write this as $$ EW= \left(EX+EYEZ\right)E\frac{1}{\sqrt{1+Z^2}} $$ but as $EX=EY=EZ=0$ then $EW=0$.
Now lets find $E(W^2)$ $$ E(W^2)=E\frac{X^2+Y^2Z^2+\overbrace{2XYZ}^0}{1+Z^2}=(1+1)E\frac{1}{1+Z^2} $$
I am stuck here. I am not sure how to find the expectation of $1/(1+Z^2)$.
First, $$ V(W|Z)=\frac{V(X+YZ|Z)}{1+Z^2}. $$ Second, $$ V(X+YZ|Z)=V(X|Z)+V(YZ|Z)=V(X)+Z^2V(Y)=1+Z^2. $$ Q.E.D.