Let $X$ and $Y$ be independent standard normal random variables and consider the following linear transformations, $$U = X$$ and $$V = \rho X + \sqrt{1 – \rho^2}Y,$$ where $\rho \in (0, 1)$. Show that the joint density of $U$ and $V$ is symmetric in $u$ and $v$.
I have found the joint density to be $$
f_{U, V}(u, v) = \frac 1 {2\pi\sqrt{1 – \rho^2}} \exp\left\{-\frac {u^2 – 2\rho uv + v^2} {2(1 – \rho^2)}\right\}.$$
However, what does it mean to show that the joint density of $U$ and $V$ is symmetric in $u$ and $v$ and what is the simplest way to go about showing this? Can I just "observe"? Any intuitive explanations will be greatly appreciated 🙂
Best Answer
It is enough to observe that $f(u,v) = f(v,u)$. A function being symmetric in $k$ variables means that it is unchanged for any permutation of the variables.