Past exam of the mathematics course of the University of Cambridge / 2006 / ii / Paper 3 / 24J / c / Solution

Condition on $X$. The normal characteristic function gives $\mathbb{E}(e^{iuXY}\mid X)=e^{-u^2{X}^2/2}$, whose Gaussian integral is $(1+u^2{)}^{-1/2}$. A second Gaussian integral gives $\mathbb{E}(e^{-iu{Z}^2})=(1+2iu{)}^{-1/2}$, taking the square-root branch continuous from one at $u=0$. Independence of $(X,Y)$ and $Z$ therefore gives