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I'm taking a probability class and my prof used the following theorem IIRC.

Let $g\sim\mathcal{N}(\mu,\Sigma)$ where $\Sigma$ is diagonal( I don't know if this condition is necessary) and $\langle u,v\rangle=0$, then $\langle g,u\rangle$ and $\langle g,v\rangle$ are independent.

Is this correct? If so, how to prove this? I believe the following is a special case of the theorem: Are the random variables $X + Y$ and $X - Y$ independent if $X, Y$ are distributed normal? I tried to use the same technique to prove the theorem but got stuck.

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  • $\begingroup$ If $\color{blue}{ u^T \Sigma v = 0}$, then you can show that $u^T g$ and $v^T g$ are independent. It won't generally hold if just $u^T v = 0$. Hint for showing this: write $z := (u^Tg, v^Tg)^T$ in the form $Ag$ for some matrix $A$, and use that the covariance matrix of $Ag$ is $A\Sigma A^T$. Also recall facts about linear transformations of normal random vectors and when such a vector has independent components based on its covariance matrix). $\endgroup$ – Minus One-Twelfth Feb 17 at 2:54
  • $\begingroup$ So actually $\Sigma$ doesn't need to be diagonal? $\endgroup$ – Yihan Zhou Feb 17 at 5:57
  • $\begingroup$ Correct! All that matters is that $u^T \Sigma v = 0$. $\endgroup$ – Minus One-Twelfth Feb 17 at 5:59
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Elaborating on Minus One-Twelfth's comment:

The pair $(g^\top u, g^\top v)$ is jointly normal. (Why?)

Thus independence is equivalent to $\text{Cov}(g^\top u, g^\top v) = 0$. The covariance is $$\begin{align}\text{Cov}(g^\top u, g^\top v) &= E[(g^\top u - E[g^\top u])(g^\top v - E[g^\top v])] \\ &= E[((g - \mu)^\top u)((g - \mu)^\top v)] \\ &= E[u^\top (g - \mu)(g-\mu)^\top v] \\ &= u^\top E[(g-\mu)(g-\mu)^\top] v \\ &= u^\top \Sigma v. \end{align}$$


If $\Sigma$ is a multiple of the identity matrix and if $\mu = 0$, then independence is equivalent to $u^\top v = 0$. However, in general you have to use the above expression $u^\top \Sigma v$.

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  • $\begingroup$ Why is it jointly normal? Is this a property of the multivariate Gaussian variable? I know that $g^Tu$ and $g^Tv$ are both normal. $\endgroup$ – Yihan Zhou Feb 17 at 3:36
  • $\begingroup$ @YihanZhou One way to check that it is jointly normal is to check that any linear combination $a(g^\top u) + b (g^\top v)$ is [univariate] normal. Indeed, any such linear combination is itself a linear combination of the components of $g$. $\endgroup$ – angryavian Feb 17 at 3:40
  • $\begingroup$ Why $E[(u^Tg)(g^Tv)]=u^T\Sigma v$? Sorry I'm not stat major and don't know much about multivariate Gaussian. $\endgroup$ – Yihan Zhou Feb 17 at 4:04
  • $\begingroup$ @YihanZhou See my edit; I made a mistake earlier. $\endgroup$ – angryavian Feb 17 at 4:13

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