# Why $\frac{X+YZ}{\sqrt{1+Z^2}} \sim \mathcal{N}(0,1)$ if $X,Y,Z$ are i.i.d. $\mathcal{N}(0,1)$?

The original question was: Suppose $X,Y,Z$ are i.i.d. $\mathcal{N}(0,1)$, find a nonnegative continuous function $g$ such that $\frac{X+YZ}{g(Z)} \sim \mathcal{N}(0,1)$.

The solution says, since $E(X+YZ\mid Z=z)=0$ and $Var(X+YZ\mid Z=z)=1+z^2$, for all $z \in \mathcal{R}$. So $g=\sqrt{1+Z^2}$.

I see the calculation of the expectation and the variance, but not sure why $\frac{X+YZ}{g(Z)}$ follows a normal distribution. The solution says, since $X+YZ\mid Z=z$ follows a normal distribution for all $z$, thus $X+YZ$ is normal.

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First we find the conditional distribution given $Z$. That means treating $Z$ as constant. So we have $1\cdot X+c\cdot Y$ where $c$ is constant. That is normally distributed with standard deviation $\sqrt{1^2+c^2}$. In this case, that is $\sqrt{1+Z^2}$. That implies that the conditional distribution of $$T=\frac{X+YZ}{\sqrt{1+Z^2}}%\tag{1}$$ given $Z$ is $\mathcal{N}(0,1)$.
Finally, observe that the conditional distribution of $T$ given $Z$ does not depend on $Z$. From that, two things follow: (1) It is the same as the marginal (or "unconditional") distribution of $T$, and (2) $T$ is actually probabilistically independent of $Z$.