Find the distribution of sum and product of standard normal random variables

Question

Let $X,Y$ and $Z$ be three independent real valued random variables. All with finite second moment and all with mean $0$ and variance $1$. Define $$ W= \frac{X+YZ}{\sqrt{1+Z^2}} $$

Find the distribution of $W$ under the additional assumption that $X,Y$ and $Z$ all have a marginal $\mathcal{N}(0,1)$-distribution

Hint: Compute $P(W \le w )$ by Tonellie, integrating out of the joint distribution of $(X,Y)$ first.

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I believe that $P(W \le w )$ can be written as

\begin{align} P(W \le w )&=P\left(\frac{X+YZ}{\sqrt{1+Z^2}}\le w\right)\\ & =P(X+YZ \le w\sqrt{1+Z^2})\\ & =P(X \le w\sqrt{1+Z^2})+P(Y\le w\sqrt{1+Z^2})P(Z \le w\sqrt{1+Z^2}) \end{align} But I am not totally sure. Further I do not know how to proceed from here.

Answer 1

Note: For independent $T_1\sim\mathcal{N}(\mu_{T_1} , \sigma_{T_1}^2)$ and $T_2\sim\mathcal{N}(\mu_{T_2} , \sigma_{T_2}^2)\,,$ we have

$\Phi\left(\dfrac{\mu_{T_2}-\mu_{T_1}}{\sqrt{\sigma_{T_1}^2{}+{}\sigma_{T_2}^2}}\right){}={}P\left(T_1<T_2\right){}={}\displaystyle\dfrac{1}{\sigma_{T_2}\sqrt{2\pi}}\int\limits_{{-}\infty}^{\infty}{}\Phi\left(\dfrac{t-\mu_{T_1}}{\sigma_{T_1}}\right)e^{-\frac{1}{2}\left(t-\mu_{T_2}\right)^2/\sigma^2_{T_2}},,\mathrm dt,,$

where $\Phi$ is the CDF for the standard normal distribution.

So, after using the definition of conditional probability, the independence of $X$ and $\left\{Y, Z\right\}$ and Tonelli's theorem (to justify the iterated integrals and "associativity" thereof), we can use the observation above to solve the integral as follows:

$$ \begin{eqnarray*} P\left(W<w\right)&{}={}&P\left(\dfrac{X+YZ}{\sqrt{1+Z^2}}<w\right){}={}P\left(X+YZ<w\sqrt{1+Z^2}\right)\newline &{}={}&P\left(X<w\sqrt{1+Z^2}-YZ\right)\newline &{}={}&\dfrac{1}{\left(\sqrt{2\pi}\right)^2}\int\limits^{\infty}_{{-}\infty}\int\limits^{\infty}_{{-}\infty} P\left(X<w\sqrt{1+z^2}-yz\,\bigg|\,\,Y=y,Z=z\right)e^{-\frac{1}{2}\left(y^2+z^2\right)}\mathrm dy\,\mathrm dz\newline &{}={}&\dfrac{1}{\left(\sqrt{2\pi}\right)^2}\int\limits^{\infty}_{{-}\infty}\int\limits^{\infty}_{{-}\infty} P\left(X<w\sqrt{1+z^2}-yz\right)e^{-\frac{1}{2}\left(y^2+z^2\right)}\mathrm dy\,\mathrm dz\newline &{}={}&\dfrac{1}{\left(\sqrt{2\pi}\right)^2}\int\limits^{\infty}_{{-}\infty}\left(\,\,\,\int\limits^{\infty}_{{-}\infty} P\left(X<w\sqrt{1+z^2}-yz\right)e^{-\frac{1}{2}y^2}\mathrm dy\right)e^{-\frac{1}{2}z^2}\,\mathrm dz\newline &{}={}&\dfrac{1}{\left(\sqrt{2\pi}\right)^2}\int\limits^{\infty}_{{-}\infty}\left(\,\,\,\int\limits^{\infty}_{{-}\infty} \Phi\left(w\sqrt{1+z^2}-yz\right)e^{-\frac{1}{2}y^2}\mathrm dy\right)e^{-\frac{1}{2}z^2}\,\mathrm dz\newline &{}={}&\dfrac{1}{\sqrt{2\pi}}\int\limits^{\infty}_{{-}\infty}\dfrac{1}{z\sqrt{2\pi}}\left(\,\,\,\int\limits^{\infty}_{{-}\infty} \Phi\left(u\right)e^{-\frac{1}{2}\left(u-w\sqrt{1+z^2}\right)^2/z^2}\mathrm dy\right)e^{-\frac{1}{2}z^2}\,\mathrm dz\newline &{}={}&\dfrac{1}{\sqrt{2\pi}}\int\limits^{\infty}_{{-}\infty} \Phi\left(w\right)e^{-\frac{1}{2}z^2}\,\mathrm dz\newline &{}={}&\Phi\left(w\right)\,. \end{eqnarray*} $$

Therefore, $W\sim\mathcal{N}\left(0,1\right)$.