I understand the arithmetic derivation of the PDF of a conditional distribution of a multivariate Gaussian, as explained here, for example. Does anyone know of a more conceptual (perhaps, co-ordinate free) proof of the same result, perhaps one that uses characterising properties of the Gaussian?


2 Answers 2


A multivariate Gaussian (or Normal) random variable $X=(X_1,X_2,\ldots,X_n)$ can be defined as an affine transformation of a tuple of independent standard Normal variates $Z=(Z_1,Z_2,\ldots, Z_m)$. This easily implies the desired result, because when we condition $X$, we impose linear constraints among the $Z_j$. (If this is not obvious, please read on through the details.) This merely reduces the number of "free" $Z_j$ contributing to the variation among the $X_i$--but those $X_i$ nevertheless remain affine combinations of independent standard Normals, QED.

We can obtain this result in three steps of increasing generality. First, the distribution of $X$ conditional on its first component is Normal. Second, this implies the distribution of $X$ conditional on some linear constraint $C^\prime X = d$ is Normal. Finally, that implies the distribution of $X$ conditional on any finite set of $r$ such linear constraints is Normal.


By definition,

$$X = \mathbb{A} Z + B$$

for some $n\times m$ matrix $\mathbb{A} = (a_{ij})$ and $n$-vector $B = (b_1, b_2, \ldots, b_n)$. Because one affine followed by another is still an affine transformation, notice that any affine transformation of $X$ is therefore also Normal. This fact will be used repeatedly.

Fix a number $x_1$ in order to consider the distribution of $X$ conditional on $X_1=x_1$. Replacing $X_1$ by its definition produces

$$x_1 = X_1 = b_1 + a_{11}Z_1 + a_{12}Z_2 + \cdots + a_{1m}Z_m.$$

When all the $a_{1j}=0$, the two cases $x_1=b_1$ and $x_1\ne b_1$ are easy to dispose of, so let's move on to the alternative where, for at least one index $k$, $a_{1k}\ne 0$. Solving for $Z_k$ exhibits it as an affine combination of the remaining $Z_j,\, j\ne k$:

$$Z_k = \frac{1}{a_{1k}}\left(x_1 - b_1 - (a_{11}Z_1 + \cdots + a_{1,k-1} + a_{1,k+1} + \cdots + a_{1m}Z_m)\right).$$

Plugging this in to $\mathbb{A}Z + B$ produces an affine combination of the remaining $Z_j$, explicitly exhibiting the conditional distribution of $X$ as an affine combination of $m-1$ independent standard normal variates, whence the conditional distribution is Normal.

Now consider any vector $C=(c_1, c_2, \ldots, c_n)$ and another constant $d$. To obtain the conditional distribution of $X$ given $C^\prime X = d$, construct the $n+1$-vector

$$Y = (Y_1,Y_2,\ldots, Y_{n+1})=(C^\prime X, X_1, X_2, \ldots, X_n) + (d, b_1, b_2, \ldots, b_n).$$

It is an affine combination of the same $Z_j$: the matrix $\mathbb{A}$ is row-augmented (at the top) by $C^\prime \mathbb{A}$ (an $n+1\times m$ matrix) and the vector of means $B$ is augmented at the beginning by the constant $d$. Therefore, by definition, $Y$ is multivariate Normal. Applying the preceding result to $Y$ and $d$ immediately shows that $Y$, conditional on $Y_1 = d$, is multivariate Normal. Upon ignoring the first component of $Y$ (which is an affine transformation!), that is precisely the distribution of $X$ conditional on $C^\prime X = d$.

The distribution of $X$ conditional on $\mathbb{C}X = D$ for an $r\times n$ matrix $\mathbb{C}$ and an $r$-vector $D$ is obtained inductively by applying the preceding construction one term at a time (working row-by-row through $\mathbb{C}$ and component-by-component through $D$). The conditionals are Normal at every step, whence the final conditional distribution is Normal, too.


The device used in the answer that you cite will also get you the conditional distribution. Here is a self-contained derivation with a slight change in notation.

Partition the column vector $X:=(X_1, X_2,\ldots, X_n)^T$ into subvectors $X_a$ and $X_b$: $$ X = \left(\begin{matrix}X_a\\X_b\end{matrix}\right) $$ and correspondingly partition the mean vector $\mu$ and covariance matrix $\Sigma$ of $X$: $$ \mu = \left(\begin{matrix}\mu_a\\ \mu_b\end{matrix}\right);\qquad \Sigma=\left(\begin{matrix}\Sigma_{a,a}&\Sigma_{a,b}\\\Sigma_{b,a}&\Sigma_{b,b}\end{matrix}\right)$$ The key is to find a matrix $C$ of constants such that $$Z:=X_a- C X_b\tag1$$ is uncorrelated with $X_b$; and since $Z$ and $X_b$ are both Gaussian, they are also independent. For $Z$ and $X_b$ to be uncorrelated we demand $$ 0= \operatorname{cov} (Z, X_b)=\operatorname{cov} (X_a - CX_b, X_b)=\Sigma_{a,b}-C\Sigma_{b,b}.\tag2 $$ Such a $C$ can always be found: If $\Sigma_{b,b}$ is invertible, then $$ C:=\Sigma_{a,b}\Sigma_{b,b}^{-1}\tag3$$ will do; otherwise you can take $\Sigma_{b,b}^{-1}$ to be the Moore-Penrose pseudoinverse of $\Sigma_{b,b}$.

Now the conditional distribution of $X_a$ given $X_b=x_b$ is easily obtained: $$P(X_a\in A\mid X_b=x_b)=P(Z+CX_b\in A\mid X_b=x_b) \stackrel{(*)}=P(Z+Cx_b\in A),\tag4 $$ where in (*) we use the fact that $Z$ and $X_b$ are independent. But $Z+Cx_b$ clearly has a Gaussian distribution, since it's an affine transformation of the original vector $X$... and we're done!

This same device gets you the conditional mean: $$\begin{align} E(X_a\mid X_b=x_b)&=E(Z + C X_b\mid X_b=x_b)\\ &=E(Z\mid X_b=x_b) + Cx_b\\ &\stackrel{(*)}=E(Z) + Cx_b\\ &= E(X_a)- CE(X_b) + C(x_b)\\ &= \mu_a + C(x_b - \mu_b) \end{align} $$ and the conditional variance: $$\begin{align} \operatorname{var}(X_a\mid X_b=x_b)&=\operatorname{var}(Z + C X_b\mid X_b=x_b)\\ &=\operatorname{var}(Z\mid X_b=x_b)\\ &\stackrel{(*)}=\operatorname{var}(Z)\\ &= \operatorname{cov}(Z, X_a-CX_b)\\ &=\operatorname{cov}(Z, X_a) - \underbrace{\operatorname{cov}(Z, X_b)}_0 C^T\\ &=\operatorname{cov}(X_a-CX_b, X_a)\\ &=\Sigma_{a,a}-C\Sigma_{b,a} \end{align} $$


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