If $W = \frac{X + YZ}{\sqrt{1 + Z^2}}$ where all variables involved are standard-normally distributed and independent, what is the distribution of $W$?
The solution I am reading begins with $$P(W \le w) = \int \int \int 1_{(-\infty, w)} ( (x+yz)/(1+z^2)) \frac{1}{2\pi} \exp^{- (x^2 + y^2)/2} \ dx dy \frac{1}{\sqrt{2\pi}} e^{-z/2} dz $$
but I don't find that particularly helpful. What happened?
This step is mostly notation. To understand what happened, you can see that $W\leq w$ if and only if the vector $(X,Y,Z)$ takes some value in $\mathbb{R}^3$ such that $\frac{X+YZ}{\sqrt{1+Z^2}}\leq w$. So to measure the event $(-\infty,w]\subseteq\mathbb{R}$ with the distribution of $W$ you can measure the corresponding event in $\mathbb{R}^3$.
To measure events $A$ that are subsets of $\mathbb{R}$ you do $\int_A f(x) dx=\int1_A(x)f(x)dx$ where $f$ is the density of your distribution.
To measure events $A\subseteq \mathbb{R}^3$ in $\mathbb{R}^3$ you do $\int_{A}g(x,y,z)d(x,y,z)$.
For a random vector of 3 i.i.d. random variables your density can be expressed as a product $g(x,y,z)=f(x)f(y)f(z)$ and the integral can then be expressed as $\int\int\int 1_A f(x)f(y) dxdy f(z) dz$ which is how you get to that formula, where $f$ is the density of the normal distribution.
