Finding distribution of $\frac{X_1+X_2 X_3}{\sqrt{1+X_3^2}}$

Solution 1:

If $(X_1,X_2)$ is i.i.d. standard normal then, for every $(a,b)$, $aX_1+bX_2$ is centered normal with variance $a^2+b^2$. Thus, by independence of $X_3$ and $(X_1,X_2)$, $$Z=\frac{X_1+X_3X_2}{\sqrt{1+X_3^2}}=\frac{1}{\sqrt{1+X_3^2}}X_1+\frac{X_3}{\sqrt{1+X_3^2}}X_2$$ is centered normal with variance $$\left(\frac1{\sqrt{1+X_3^2}}\right)^2+\left(\frac{X_3}{\sqrt{1+X_3^2}}\right)^2=1$$ that is, $Z$ is standard normal.

Solution 2:

@user587389 I had the same exact concerns! How can we interpret an entire random variable as a constant?

Below is an argument to make Did's idea a bit more formal. Did's idea of "freezing" $X_3$ is equivalent to conditioning on an outcome of $X_3$. Using Baye's Theorem, we can write $$f_{Z}(z)=\int_{-\infty}^{\infty}f_{Z|X_{3}=x}(z|x)f_{X_3}(x)dx$$ Now $Z|X_3=x \sim N(0,1)$ from Did's remarks, so that $$f_{Z|X_3=x}(z|x)=\frac{1}{\sqrt{2\pi}}e^{-z^2/2}$$ Finally, $$f_{Z}(z)=\frac{1}{\sqrt{2\pi}}e^{-z^2/2}\int_{-\infty}^{\infty}f_{X_3}(x)dx=\frac{1}{\sqrt{2\pi}}e^{-z^2/2}$$ which proves that $Z\sim N(0,1)$.