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

I am trying to find distribution of random variable:

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

where $X_i \sim \mathcal{N}(0,1)$ and independent.

My thinking so far is to use random variable algebra and to find distributions step by step. However, I don't know what to do with the variable: $\sqrt{1+X_3^2}$ (it seems like some variant of $\chi$-distribution).

Moreover, I'm worried that my thinking is to complicated and that the solution should be more elegant.

• Commented Jun 3, 2020 at 18:50

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.
• @expiTTp1z0 Mentally "freezing" $X_3$ led to the first sentence of the answer, I guess.
• But isn't $X_3$ a random variable? How can we use it as a constant? Commented Jun 3, 2020 at 23:02
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 $$Z\sim N(0,1)$$.