I'm trying to understand the solution to this question:
Suppose that $(X, Y)$ is a normally distributed random vector with $X \sim N(0, 1)$, $Y \sim N(0, 1)$, $\text{Cov}(X, Y) = 0$.
Determine the characteristic function $\phi_{XY}(u) = E(e^{iuXY})$ of the product $XY$ for all $u \in \mathbb{R}$.
Hint: Use conditional expectations.
Given solution:
$\phi (u) = \mathbb{E}(\mathbb{E}(e^{iuXY}|Y)) = \mathbb{E}(e^{-\frac{u^2Y^2}{2}}) = \large\int^\infty_{-\infty}e^{-\frac{u^2x^2}{2}}\frac{1}{\sqrt{2\pi}}e^{-\frac{x^2}{2}}dx$
I don't understand how $\mathbb{E}(e^{-\frac{u^2Y^2}{2}})$ is obtained, using the tower property, is this correct:
$\mathbb{E}(\mathbb{E}(e^{iuXY}|Y)) = \mathbb{E}(\mathbb{E}(e^{iuX}|Y)\cdot Y)$
When $X$ is standard normal, then it has a characteristic function of: $$\begin{align}\phi_X(t)&=\mathsf E(\mathrm e^{itX})\\&={\mathrm e^{-t^{2}/2}}\end{align}$$
When also, $X,Y$ are independent, $\mathsf E(\mathrm e^{iuXY}\mid Y)=\phi_X(uY).$
So, the application of the tower property gives us:
$$\begin{align}\mathsf E(\mathrm e^{iuXY}) &=\mathsf E(\mathsf E(\mathrm e^{iuXY}\mid Y))\\[1ex]&=\mathsf E(\phi_X(uY))\\[1ex]&=\mathsf E(\mathrm e^{-u^2Y^2/2})\end{align}$$