characteristic function of $WX$ where $W$ and $X$ are independent standard normal random variables

Question

Let $W$ and $X$ be independent random variables, both standard normal distributed. I have to show that for the characteristic function of $WX$ it holds that

$\phi_{WX}(u) = \frac {1}{\sqrt{1+u^2}}$.

I tried to calculate

$\phi_{WX}(u) = E[e^{iwxu}] = \int_{\mathbb R} e^{iyu} dF_{WX}(y) = \int_{\mathbb R} e^{iyu} f_{WX}(y) dy$,

where $f_{WX}(y)$ is the density of $WX$ for which I used the density formula:

Let $X$ and $Y$ be independent with densities $f_X$ and $f_Y$. Then for the density $f_{XY}$ of the product $XY$ it holds

$f_{XY} = \int_{\mathbb R} \frac{1}{|t|} f_X (t) f_Y(\frac{z}{t}) dt$.

I used $f_X=f_Y= \frac{1}{\sqrt{2 \pi}} e^{-\frac{x^2}{2}}$ but the calculation was too hard and lead to nothing.

I would be very thankful for some hints or a nice trick.

Answer 1

Since $W$ and $X$ are independent, we have

$$\phi(\xi) := \mathbb{E}e^{i \xi (W \cdot X)} = \mathbb{E} \left[ \mathbb{E}(e^{i \xi W \cdot x}) \big|_{x=X} \right]$$

for any $\xi \in \mathbb{R}$. As $W \sim N(0,1)$ this gives

$$\phi(\xi) = \mathbb{E} \exp \left( - \frac{\xi^2 X^2}{2} \right),$$

and so

$$\begin{align*} \phi(\xi) &= \frac{1}{\sqrt{2\pi}} \int_{\mathbb{R}} \exp \left(- \frac{\xi^2 y^2}{2} \right) \exp \left(- \frac{y^2}{2} \right) \, dy \\ &= \frac{1}{\sqrt{2\pi}} \int_{\mathbb{R}} \exp \left( - \frac{y^2}{2} (\xi^2+1) \right) \, dy. \tag{1}\end{align*}$$

Note that $$p(y) := \sqrt{\frac{\xi^2+1}{2\pi}} \exp \left(- \frac{y^2}{2} (\xi^2+1) \right)$$ is the density of the Gaussian distribution with mean $0$ and variance $\sigma^2 = 1/(\xi^2+1)$; in particular $\int_{\mathbb{R}} p(y) \, dy = 1$, i.e.

$$\int_{\mathbb{R}} \exp \left( - \frac{y^2}{2} (\xi^2+1) \right) \, dy = \sqrt{\frac{2\pi}{\xi^2+1}}.$$

Plugging this into $(1)$ gives $\phi(\xi) = 1/\sqrt{\xi^2+1}$.