Distribution of $Z=\frac{UX+VY}{\sqrt{U^2+2\rho UV+V^2}}$ when $(X,Y)$ is bivariate normal and $U,V$ independent of $X$ and $Y$ $X$ and $Y$ have bivariate normal distribution with zero means, unit variances, and correlation $\rho$. Let $U$ and $V$ be independent of $X$ and $Y$. 
How can we find the distribution of $Z = \displaystyle\frac{UX + VY}{\sqrt{U^2 + 2\rho UV + V^2}}$
I was thinking of using the characteristic function by considering the characteristic function of $Z$, but I don't really know how to proceed.
 A: From the fact that distribution of linear combination of jointly normal variables is univariate normal, one can show that $uX+vY$ is normally distributed, where $u,v$ are real constants. You can verify this using moment generating functions. The MGF of $uX+vY$ is given by 
\begin{align}
M(t)&=E(\exp(tuX+tvY))
\\&=E[\exp(tuX)E(\exp(tvY)\mid X)] 
\\&=E[\exp(tuX)\exp(\rho tv X+(1-\rho^2)t^2v^2/2)]
\\&=\exp(1-\rho^2)t^2v^2/2)E(\exp((tu+tv\rho)X))
\\&=\exp((1-\rho^2)t^2v^2/2)\exp((tu+tv\rho)^2/2)
\\&=\exp(t^2(v^2+u^2+2uv\rho)/2)
\end{align}
, which is the MGF of a $\mathcal{N}(0,u^2+v^2+2uv\rho)$ distribution. 
Known facts used above are that $Y\mid X\sim\mathcal N(\rho X,1-\rho^2)$ and the MGF of a univariate normal variable. Also note that $X$ is standard normal.
By uniqueness of MGF, we conclude that $$uX+vY\sim\mathcal{N}(0,u^2+v^2+2uv\rho)$$
That is, $$Z\mid (U=u,V=v)=\frac{uX+vY}{\sqrt{u^2+2\rho uv+v^2}}\sim\mathcal{N}(0,1)$$
So the distribution of $Z$ conditioned on $U=u$ and $V=v$ is independent of $u$ and $v$. This means the unconditional distribution would be the same as the conditional distribution.
Hence, $$Z=\frac{UX+VY}{\sqrt{U^2+2\rho UV+V^2}}\sim\mathcal{N}(0,1)$$
A direct solution is also possible from the MGF or the characteristic function of $Z$, conditioned on $U$ and $V$.
