Gaussian random variable in $\mathbb{R}^n$ question Let $X=(X_1,...,X_n)$ is a Gaussian random variable in $\mathbb{R}^n$ with mean $\mu$ and covariance matrix $V$. 
I want to show that we can write $X_2$ in the form $X_2 = aX_1 + Z$, where $Z$ is independent of $X_1$, and I want to find the distribution of $Z$.
Any help with this would be really appreciated.
Thanks!
 A: The claim is not valid unless a restriction is placed on the factor $a$. Specifically, if $Z$ has to be independent of $X_1$, we must have $a<\sigma_2/\sigma_1$, where $\sigma_1$ and $\sigma_2$ are the standard deviations of $X_1$ and $X_2$.
We know that:


*

*Sums and differences of normal rv's are normal.

*If two rv's are independent, their covariance is zero. 

*If two normal rv's have zero covariance, they are independent. (This is not true in general for non-normal rv's...)

*If the rv $X_1$ has mean and variance $\mu_1$ and $\sigma_1^2$, the mean and variance of $aX_1$ are given by $a\mu_1$ and $a^2\sigma_1^2$, resp.

*The rv $Z$, which satisfies $X_2=aX_1+Z$, is supposed to be independent of $X_1$.


Since $X_1$ and $X_2$ are elements of a multivariate normal rv, their individual distributions are (univariate) normal. 
Since $Z=X_2-aX_1$, the difference of two normal rv's, the distribution of $Z$ must also be normal. Since normal rv's are completely described by the mean and variance, we need to find $\mu_Z$ and $\sigma_Z^2$.
Let the means of $X_1$ and $X_2$ be denoted by $\mu_1$ and $\mu_2$, resp. Since $Z=X_2-aX_1$, we must have $$\mu_Z=\mu_2-a\mu_1$$ in order for the mean of $aX_1+Z$ to equal $\mu_2$.
Since $X_1$ and $Z$ are supposed to be independent, their covariance is zero. Hence,  $\sigma_2^2=Var(X_2)=Var(aX_1+Z)=a^2\sigma_1^2+\sigma_Z^2$. Put differently, we must have $$\sigma_Z^2=\sigma_2^2-a^2\sigma_1^2.$$  
But $\sigma_Z^2>0$ -- which is required for $Z$ to be normal -- can only be true if $\sigma_2^2>a^2\sigma_1^2$, i.e., if $$a<\frac{\sigma_2}{\sigma_1}.$$
If $Z$ and $X_1$ are allowed to be dependent, the restriction on $a$ can be lifted. 
A: Assuming that $X_2 = aX_1 + Z$ holds (with $Z$ and $X_1$ independent), you can find $a$ in terms of particular entries of the covariance matrix. Once you have $a$, you know that $Z=X_2 - aX_1$, so you can find the distribution of $Z$, and check that it is independent of $X_1$.
A: Let $\bar{X}$ be the normalization of $X$ (so that its components have zero mean and unit variance). Then note that it suffices to show $$\bar{X}_2=a\bar{X}_1+Z$$ for some $Z$ independent of $\bar{X}_1$. So we can assume that the components of $X$ have unit variance and zero mean. Now write 
$$
\mathbb{E}[X_2X_1]=a \implies a=\sigma_{12}.
$$
So you want to show that $\sigma_Z=\mathrm{var}(Z)$ can be determined such that the  equality 
$$
X_2=\sigma_{12}X_1+Z
$$
holds (a.e.). This is equivalent to showing that
$$
\mathbb{E}[(X_2-\sigma_{12}X_1-Z)^2]=0.
$$ 
Expand both sides and solve for $\sigma_Z$.
