Testing the independence of two jointly normal variables

Question

Variables $u$ and $v$ are jointly normal, correlated with zero mean. $X$ is a linear combination of $u$ and $v$:

\begin{align*} X := \frac{u}{\sqrt{E(u^2)}}-\rho\frac{v}{\sqrt{E(v^2)}} \end{align*}

where:

\begin{align*} \rho =\frac{E(uv)}{\sqrt{E(u^2)}\sqrt{E(v^2)}}. \end{align*}

I need to find if $X$ and $v$ are independent. If they are then $Cov(X,v)=0$

How to prove this? My guess would be:

\begin{align*} Cov(X,v)=E(Xv)-E(X)E(v)=E(X)E(v)-E(X)E(v)=0. \end{align*}

Answer 1

Recall two useful facts about (generalized) normal random variables:

1. If two random variables $u,v$ have a joint normal distribution and $u^{\prime},v^{\prime}$ are random variables obtained by applying a $2\times 2$ matrix to the random vector $(u,v)^T$, then $u^{\prime}$ and $v^{\prime}$ also have a joint normal distribution.

2. If two random variables $X,Y$ have a joint normal distribution, then they are independent if and only if they are uncorrelated.

For your problem, write $\sigma_u^2=\mathbb{E}[u^2]$ and $\sigma_v^2=\mathbb{E}[v^2]$. Then $v$ and $X=\frac{u}{\sigma_u}-\rho\frac{v}{\sigma_v}$ have a joint normal distribution by the first fact above, and moreover $$ \mathrm{cov}(X,v)=\mathbb{E}[Xv]=\frac{\mathbb{E}[uv]}{\sigma_u}-\frac{\rho\mathbb{E}[v^2]}{\sigma_v}=\frac{\rho\sigma_u\sigma_v}{\sigma_u}-\frac{\rho\sigma_v^2}{\sigma_v}=0 $$

Therefore $X$ and $v$ are independent by the second fact above.