characteristic function of $\sqrt{Y}Z$, $Y$ exponential, $Z$ normal How can I compute the characteristic function of $\sqrt{Y}Z$ when $Y$ is an exponential distribution with parameter $\lambda$ and $Z$ is a standard normal distribution? 
Edit : we assume that $Y$ and $Z$ are independant.
 A: The conditional characteristic function of $Z$ given that $Y$ has some particular value $y$ is
\begin{align}
\chi_{\sqrt Y \,Z\,\mid\,Y=y} (t) = {} & \operatorname E\left(e^{it\sqrt Y\, Z} \mid Y=y \right) \\[8pt]
= {} & \chi_Z\left(t\sqrt y\right) = e^{-(t\sqrt y)^2/2} \\
& \quad\quad\quad \uparrow \text{This is of course lower-case $y$.} \\[8pt]
= {} & e^{-t^2 y/2}.
\end{align}
Therefore,
\begin{align}
\chi_{\sqrt Y\,Z} (t) = {} & \operatorname E\left( e^{-t^2 Y/2} \right) \\
& \quad\quad\quad\uparrow\ldots\text{and this is of course capital } Y. \\[10pt]
= {} & \int_0^\infty e^{-t^2y/2} e^{-\lambda y} (\lambda\,dy) \\
& \quad\quad\quad \uparrow \text{and of course lower case again.} \\[10pt]
= {} & \int_0^\infty e^{-\left( \lambda + t^2/2 \right) y} \lambda\, dy \\[8pt]
= {} & \frac \lambda {\lambda + t^2/2} = \underbrace{ \frac 1 {1 + \left(\frac t {\sqrt {2\lambda}} \right)^2}.}_\text{This form may be useful.}
\end{align}
A: While Michael's approach is rather elegant, I wanted to show a more direct approach. Both answers of course amount to the same calculations. Let $f_Y(y)$ and $f_Z(z)$ denote the densities of $Y$ and $Z$, respectively, while $\varphi_Z(t) = e^{-t^2/2}$ denotes the characteristic function of $Z$. Then
\begin{align*}
\varphi_{\sqrt Y Z}(t) &= E[e^{it\sqrt{Y}Z}] \\
&= \int_{-\infty}^\infty \int_{-\infty}^\infty e^{it\sqrt yz} f_Z(z)f_Y(y) \,dz\,dy \\
&= \int_{-\infty}^\infty \left(\int_{-\infty}^\infty e^{i(t\sqrt y)z} f_Z(z)\,dz\right)f_Y(y)\,dy \\
&= \int_{-\infty}^\infty \varphi_Z(t\sqrt y) f_Y(y)\,dy \\
&= \int_0^\infty e^{-i(t\sqrt y)^2/2} \lambda e^{-\lambda y} \,dy \\
&= \lambda \int_0^\infty e^{-(\lambda + t^2/2)y} \,dy \\
&= \frac{\lambda}{\lambda + t^2/2}.
\end{align*}
(I kept all calculations in probabilistic terms as I assume you don't know measure theory)
