# 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.

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. – Did Jun 26 '17 at 12:47