Deriving the conditional distributions of a multivariate normal distribution

Question

We have a multivariate normal vector ${\boldsymbol Y} \sim \mathcal{N}(\boldsymbol\mu, \Sigma)$. Consider partitioning $\boldsymbol\mu$ and ${\boldsymbol Y}$ into $$\boldsymbol\mu = \begin{bmatrix} \boldsymbol\mu_1 \\ \boldsymbol\mu_2 \end{bmatrix} $$ $${\boldsymbol Y}=\begin{bmatrix}{\boldsymbol y}_1 \\ {\boldsymbol y}_2 \end{bmatrix}$$

with a similar partition of $\Sigma$ into $$ \begin{bmatrix} \Sigma_{11} & \Sigma_{12}\\ \Sigma_{21} & \Sigma_{22} \end{bmatrix} $$ Then, $({\boldsymbol y}_1|{\boldsymbol y}_2={\boldsymbol a})$, the conditional distribution of the first partition given the second, is $\mathcal{N}(\overline{\boldsymbol\mu},\overline{\Sigma})$, with mean
$$ \overline{\boldsymbol\mu}=\boldsymbol\mu_1+\Sigma_{12}{\Sigma_{22}}^{-1}({\boldsymbol a}-\boldsymbol\mu_2) $$ and covariance matrix $$ \overline{\Sigma}=\Sigma_{11}-\Sigma_{12}{\Sigma_{22}}^{-1}\Sigma_{21}$$

Actually these results are provided in Wikipedia too, but I have no idea how the $\overline{\boldsymbol\mu}$ and $\overline{\Sigma}$ is derived. These results are crucial, since they are important statistical formula for deriving Kalman filters. Would anyone provide me a derivation steps of deriving $\overline{\boldsymbol\mu}$ and $\overline{\Sigma}$ ?

Answer 1

You can prove it by explicitly calculating the conditional density by brute force, as in Procrastinator's link (+1) in the comments. But, there's also a theorem that says all conditional distributions of a multivariate normal distribution are normal. Therefore, all that's left is to calculate the mean vector and covariance matrix. I remember we derived this in a time series class in college by cleverly defining a third variable and using its properties to derive the result more simply than the brute force solution in the link (as long as you're comfortable with matrix algebra). I'm going from memory but it was something like this:

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It is worth pointing out that the proof below only assumes that $\Sigma_{22}$ is nonsingular, $\Sigma_{11}$ and $\Sigma$ may well be singular.

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Let ${\bf x}_{1}$ be the first partition and ${\bf x}_2$ the second. Now define ${\bf z} = {\bf x}_1 + {\bf A} {\bf x}_2 $ where ${\bf A} = -\Sigma_{12} \Sigma^{-1}_{22}$. Now we can write

\begin{align*} {\rm cov}({\bf z}, {\bf x}_2) &= {\rm cov}( {\bf x}_{1}, {\bf x}_2 ) + {\rm cov}({\bf A}{\bf x}_2, {\bf x}_2) \\ &= \Sigma_{12} + {\bf A} {\rm var}({\bf x}_2) \\ &= \Sigma_{12} - \Sigma_{12} \Sigma^{-1}_{22} \Sigma_{22} \\ &= 0 \end{align*}

Therefore ${\bf z}$ and ${\bf x}_2$ are uncorrelated and, since they are jointly normal, they are independent. Now, clearly $E({\bf z}) = {\boldsymbol \mu}_1 + {\bf A} {\boldsymbol \mu}_2$, therefore it follows that

\begin{align*} E({\bf x}_1 | {\bf x}_2) &= E( {\bf z} - {\bf A} {\bf x}_2 | {\bf x}_2) \\ & = E({\bf z}|{\bf x}_2) - E({\bf A}{\bf x}_2|{\bf x}_2) \\ & = E({\bf z}) - {\bf A}{\bf x}_2 \\ & = {\boldsymbol \mu}_1 + {\bf A} ({\boldsymbol \mu}_2 - {\bf x}_2) \\ & = {\boldsymbol \mu}_1 + \Sigma_{12} \Sigma^{-1}_{22} ({\bf x}_2- {\boldsymbol \mu}_2) \end{align*}

which proves the first part. For the covariance matrix, note that

\begin{align*} {\rm var}({\bf x}_1|{\bf x}_2) &= {\rm var}({\bf z} - {\bf A} {\bf x}_2 | {\bf x}_2) \\ &= {\rm var}({\bf z}|{\bf x}_2) + {\rm var}({\bf A} {\bf x}_2 | {\bf x}_2) - {\bf A}{\rm cov}({\bf z}, -{\bf x}_2) - {\rm cov}({\bf z}, -{\bf x}_2) {\bf A}' \\ &= {\rm var}({\bf z}|{\bf x}_2) \\ &= {\rm var}({\bf z}) \end{align*}

Now we're almost done:

\begin{align*} {\rm var}({\bf x}_1|{\bf x}_2) = {\rm var}( {\bf z} ) &= {\rm var}( {\bf x}_1 + {\bf A} {\bf x}_2 ) \\ &= {\rm var}( {\bf x}_1 ) + {\bf A} {\rm var}( {\bf x}_2 ) {\bf A}' + {\bf A} {\rm cov}({\bf x}_1,{\bf x}_2) + {\rm cov}({\bf x}_2,{\bf x}_1) {\bf A}' \\ &= \Sigma_{11} +\Sigma_{12} \Sigma^{-1}_{22} \Sigma_{22}\Sigma^{-1}_{22}\Sigma_{21} - 2 \Sigma_{12} \Sigma_{22}^{-1} \Sigma_{21} \\ &= \Sigma_{11} +\Sigma_{12} \Sigma^{-1}_{22}\Sigma_{21} - 2 \Sigma_{12} \Sigma_{22}^{-1} \Sigma_{21} \\ &= \Sigma_{11} -\Sigma_{12} \Sigma^{-1}_{22}\Sigma_{21} \end{align*}

which proves the second part.

Note: For those not very familiar with the matrix algebra used here, this is an excellent resource.

Edit: One property used here this is not in the matrix cookbook (good catch @FlyingPig) is property 6 on the wikipedia page about covariance matrices: which is that for two random vectors $\bf x, y$, $${\rm var}({\bf x}+{\bf y}) = {\rm var}({\bf x})+{\rm var}({\bf y}) + {\rm cov}({\bf x},{\bf y}) + {\rm cov}({\bf y},{\bf x})$$ For scalars, of course, ${\rm cov}(X,Y)={\rm cov}(Y,X)$ but for vectors they are different insofar as the matrices are arranged differently.

Answer 2

The answer by Macro is great, but here is an even simpler way that does not require you to use any outside theorem asserting the conditional distribution. It involves writing the Mahalanobis distance in a form that separates the argument variable for the conditioning statement, and then factorising the normal density accordingly.

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Rewriting the Mahalanobis distance for a conditional vector: This derivation uses a matrix inversion formula that uses the Schur complement $\boldsymbol{\Sigma}_* \equiv \boldsymbol{\Sigma}_{11} - \boldsymbol{\Sigma}_{12} \boldsymbol{\Sigma}_{22}^{-1} \boldsymbol{\Sigma}_{21}$. We first use the blockwise inversion formula to write the inverse-variance matrix as:

$$\begin{equation} \begin{aligned} \boldsymbol{\Sigma}^{-1} = \begin{bmatrix} \boldsymbol{\Sigma}_{11} & \boldsymbol{\Sigma}_{12} \\ \boldsymbol{\Sigma}_{21} & \boldsymbol{\Sigma}_{22} \\ \end{bmatrix}^{-1} = \begin{bmatrix} \boldsymbol{\Sigma}_{11}^* & \boldsymbol{\Sigma}_{12}^* \\ \boldsymbol{\Sigma}_{21}^* & \boldsymbol{\Sigma}_{22}^* \\ \end{bmatrix}, \end{aligned} \end{equation}$$

where:

$$\begin{equation} \begin{aligned} \begin{matrix} \boldsymbol{\Sigma}_{11}^* = \boldsymbol{\Sigma}_*^{-1} \text{ } \quad \quad \quad \quad & & & & & \boldsymbol{\Sigma}_{12}^* = -\boldsymbol{\Sigma}_*^{-1} \boldsymbol{\Sigma}_{12} \boldsymbol{\Sigma}_{22}^{-1}, \quad \quad \quad \\[6pt] \boldsymbol{\Sigma}_{21}^* = - \boldsymbol{\Sigma}_{22}^{-1} \boldsymbol{\Sigma}_{21} \boldsymbol{\Sigma}_*^{-1} & & & & & \boldsymbol{\Sigma}_{22}^* = \boldsymbol{\Sigma}_{22}^{-1} + \boldsymbol{\Sigma}_{22}^{-1} \boldsymbol{\Sigma}_{21} \boldsymbol{\Sigma}_*^{-1} \boldsymbol{\Sigma}_{12} \boldsymbol{\Sigma}_{22}^{-1}. \text{ } \\[6pt] \end{matrix} \end{aligned} \end{equation}$$

Using this formula we can now write the Mahalanobis distance as:

$$\begin{equation} \begin{aligned} (\boldsymbol{y} &- \boldsymbol{\mu})^\text{T} \boldsymbol{\Sigma}^{-1} (\boldsymbol{y} - \boldsymbol{\mu}) \\[6pt] &= \begin{bmatrix} \boldsymbol{y}_1 - \boldsymbol{\mu}_1 \\ \boldsymbol{y}_2 - \boldsymbol{\mu}_2 \end{bmatrix}^\text{T} \begin{bmatrix} \boldsymbol{\Sigma}_{11}^* & \boldsymbol{\Sigma}_{12}^* \\ \boldsymbol{\Sigma}_{21}^* & \boldsymbol{\Sigma}_{22}^* \\ \end{bmatrix} \begin{bmatrix} \boldsymbol{y}_1 - \boldsymbol{\mu}_1 \\ \boldsymbol{y}_2 - \boldsymbol{\mu}_2 \end{bmatrix} \\[6pt] &= \quad (\boldsymbol{y}_1 - \boldsymbol{\mu}_1)^\text{T} \boldsymbol{\Sigma}_{11}^* (\boldsymbol{y}_1 - \boldsymbol{\mu}_1) + (\boldsymbol{y}_1 - \boldsymbol{\mu}_1)^\text{T} \boldsymbol{\Sigma}_{12}^* (\boldsymbol{y}_2 - \boldsymbol{\mu}_2) \\[6pt] &\quad + (\boldsymbol{y}_2 - \boldsymbol{\mu}_2)^\text{T} \boldsymbol{\Sigma}_{21}^* (\boldsymbol{y}_1 - \boldsymbol{\mu}_1) + (\boldsymbol{y}_2 - \boldsymbol{\mu}_2)^\text{T} \boldsymbol{\Sigma}_{22}^* (\boldsymbol{y}_2 - \boldsymbol{\mu}_2) \\[6pt] &= \quad (\boldsymbol{y}_1 - \boldsymbol{\mu}_1)^\text{T} \boldsymbol{\Sigma}_*^{-1} (\boldsymbol{y}_1 - \boldsymbol{\mu}_1) - (\boldsymbol{y}_1 - \boldsymbol{\mu}_1)^\text{T} \boldsymbol{\Sigma}_*^{-1} \boldsymbol{\Sigma}_{12} \boldsymbol{\Sigma}_{22}^{-1} (\boldsymbol{y}_2 - \boldsymbol{\mu}_2) \\[6pt] &\quad - (\boldsymbol{y}_2 - \boldsymbol{\mu}_2)^\text{T} \boldsymbol{\Sigma}_{22}^{-1} \boldsymbol{\Sigma}_{21} \boldsymbol{\Sigma}_*^{-1} (\boldsymbol{y}_1 - \boldsymbol{\mu}_1) + (\boldsymbol{y}_2 - \boldsymbol{\mu}_2)^\text{T} \boldsymbol{\Sigma}_{22}^{-1} (\boldsymbol{y}_2 - \boldsymbol{\mu}_2) \\[6pt] &\quad + (\boldsymbol{y}_2 - \boldsymbol{\mu}_2)^\text{T} \boldsymbol{\Sigma}_{22}^{-1} \boldsymbol{\Sigma}_{21} \boldsymbol{\Sigma}_*^{-1} \boldsymbol{\Sigma}_{12} \boldsymbol{\Sigma}_{22}^{-1} (\boldsymbol{y}_2 - \boldsymbol{\mu}_2) \\[6pt] &= (\boldsymbol{y}_1 - (\boldsymbol{\mu}_1 + \boldsymbol{\Sigma}_{12} \boldsymbol{\Sigma}_{22}^{-1} (\boldsymbol{y}_2 - \boldsymbol{\mu}_2)))^\text{T} \boldsymbol{\Sigma}_*^{-1} (\boldsymbol{y}_1 - (\boldsymbol{\mu}_1 + \boldsymbol{\Sigma}_{12} \boldsymbol{\Sigma}_{22}^{-1} (\boldsymbol{y}_2 - \boldsymbol{\mu}_2))) \\[6pt] &\quad + (\boldsymbol{y}_2 - \boldsymbol{\mu}_2)^\text{T} \boldsymbol{\Sigma}_{22}^{-1} (\boldsymbol{y}_2 - \boldsymbol{\mu}_2) \\[6pt] &= (\boldsymbol{y}_1 - \boldsymbol{\mu}_*)^\text{T} \boldsymbol{\Sigma}_*^{-1} (\boldsymbol{y}_1 - \boldsymbol{\mu}_*) + (\boldsymbol{y}_2 - \boldsymbol{\mu}_2)^\text{T} \boldsymbol{\Sigma}_{22}^{-1} (\boldsymbol{y}_2 - \boldsymbol{\mu}_2) , \\[6pt] \end{aligned} \end{equation}$$

where $\boldsymbol{\mu}_* \equiv \boldsymbol{\mu}_1 + \boldsymbol{\Sigma}_{12} \boldsymbol{\Sigma}_{22}^{-1} (\boldsymbol{y}_2 - \boldsymbol{\mu}_2)$ (which will be the conditional mean vector in the normal case). Note that this result is a general result that does not assume normality of the random vectors involved in the decomposition. It gives a useful way of decomposing the Mahalanobis distance so that it consists of a sum of quadratic forms on the marginal and conditional parts. In the conditional part the conditioning vector $\boldsymbol{y}_2$ is absorbed into the mean vector and variance matrix. To clarify the form, we repeat the equation with labelling of terms:

$$(\boldsymbol{y} - \boldsymbol{\mu})^\text{T} \boldsymbol{\Sigma}^{-1} (\boldsymbol{y} - \boldsymbol{\mu}) = \underbrace{(\boldsymbol{y}_1 - \boldsymbol{\mu}_*)^\text{T} \boldsymbol{\Sigma}_*^{-1} (\boldsymbol{y}_1 - \boldsymbol{\mu}_*)}_\text{Conditional Part} + \underbrace{(\boldsymbol{y}_2 - \boldsymbol{\mu}_2)^\text{T} \boldsymbol{\Sigma}_{22}^{-1} (\boldsymbol{y}_2 - \boldsymbol{\mu}_2)}_\text{Marginal Part}.$$

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Deriving the conditional distribution: Now that we have the above form for the Mahalanobis distance, the rest is easy. We have:

$$\begin{equation} \begin{aligned} p(\boldsymbol{y}_1 | \boldsymbol{y}_2, \boldsymbol{\mu}, \boldsymbol{\Sigma}) &\overset{\boldsymbol{y}_1}{\propto} p(\boldsymbol{y}_1 , \boldsymbol{y}_2 | \boldsymbol{\mu}, \boldsymbol{\Sigma}) \\[12pt] &= \text{N}(\boldsymbol{y} | \boldsymbol{\mu}, \boldsymbol{\Sigma}) \\[10pt] &\overset{\boldsymbol{y}_1}{\propto} \exp \Big( - \frac{1}{2} (\boldsymbol{y} - \boldsymbol{\mu})^\text{T} \boldsymbol{\Sigma}^{-1} (\boldsymbol{y} - \boldsymbol{\mu}) \Big) \\[6pt] &\overset{\boldsymbol{y}_1}{\propto} \exp \Big( - \frac{1}{2} (\boldsymbol{y}_1 - \boldsymbol{\mu}_*)^\text{T} \boldsymbol{\Sigma}_*^{-1} (\boldsymbol{y}_1 - \boldsymbol{\mu}_*) \Big) \\[6pt] &\overset{\boldsymbol{y}_1}{\propto}\text{N}(\boldsymbol{y}_1 | \boldsymbol{\mu}_*, \boldsymbol{\Sigma}_*). \\[6pt] \end{aligned} \end{equation}$$

This establishes that the conditional distribution is also multivariate normal, with the specified conditional mean vector and conditional variance matrix.

Answer 3

Since I regularly come back to this, here is my attempt to reduce the gore of Ben's answer (wether or not you think this is true will depend on your tastes).

With this approach it turns out to be a bit easier to calculate $Y_2\mid Y_1$ so weare going to do that instead. Notationwise we define $$ \bar{y} := y-\mu $$ So that we can essentially rid ourselves of the mean without loss of generality.

Cholesky based proof

The main insight for this proof is the use of the cholesky decomposition of the covariance matrix, i.e. $\Sigma = LL^T$ where $L$ is a lower triangular matrix.

Why is this useful? Well, behold: $$ \bar{y}^T \Sigma^{-1} \bar{y} = \bar{y} L^{-T}L^{-1} \bar{y} = \|L^{-1}\bar{y}\|^2. $$

Let us calculate this inverse $$ \begin{bmatrix} \bar{y}_1\\ \bar{y}_2 \end{bmatrix} = \begin{bmatrix} L_{11} & 0\\ L_{21} & L_{22} \end{bmatrix}\begin{bmatrix} x_1\\ x_2 \end{bmatrix} $$ This implies \begin{align*} x_1 &= L_{11}^{-1}\bar{y}_1 \\ x_2 &= L_{22}^{-1}(\bar{y}_2 - L_{21}x_1) \\ &= L_{22}^{-1}\bigl( y_2 - \underbrace{(\mu_2 + L_{21}L_{11}^{-1}\bar{y}_1)}_{=:\mu_{2\mid 1}}\bigr) \end{align*}

So we have \begin{align*} \bar{y}^T \Sigma^{-1} \bar{y} &= \|x_1\|^2 &+\quad& \|x_2\|^2 \\ &= \underbrace{\bar{y}_1^T L_{11}^{-T}L_{11}^{-1}\bar{y}_1}_{\text{marginal part}} &+\quad& \underbrace{(y_2- \mu_{2\mid 1})^T L_{22}^{-T}L_{22}^{-1}(y_2- \mu_{2\mid 1})}_{\text{conditional part}} \end{align*} Now we are almost done, we simply need to get back to the covariance matrix version.

For this consider $$ \Sigma = \begin{bmatrix} \Sigma_{11} & \Sigma_{12}\\ \Sigma_{21} & \Sigma_{22} \end{bmatrix} = \begin{bmatrix} L_{11}L_{11}^T & L_{11}L_{21}^T \\ L_{21}L_{11}^T & L_{21}L_{21}^T + L_{22}L_{22}^T \end{bmatrix} = LL^T $$ with the lower triangular block matrix $$ L = \begin{bmatrix} L_{11} & 0\\ L_{21} & L_{22} \end{bmatrix}. $$ From this we obtain \begin{align*} && L_{11}L_{11}^T &= \Sigma_{11} \\ \Sigma_{2\mid 1} &:=& L_{22}L_{22}^T &= \Sigma_{22} - L_{21}L_{21}^T \\ && & = \Sigma_{22} - \underbrace{L_{21}L_{11}^T}_{=\Sigma_{21}} \underbrace{L_{11}^{-T} L_{11}^{-1}}_{=\Sigma_{11}^{-1}} \underbrace{L_{11}L_{21}^T}_{=\Sigma_{12}} \\ \mu_{2\mid 1} &:=& \mu_2 + L_{21}L_{11}^{-1}\bar{y}_1 &= \mu_2 + \underbrace{L_{21}L_{11}^T}_{=\Sigma_{21}} \underbrace{L_{11}^{-T}L_{11}^{-1}}_{=\Sigma_{11}^{-1}}\bar{y}_1 \end{align*}

Putting everything together we have \begin{align*} (y-\mu)^T \Sigma^{-1} (y-\mu) &= \underbrace{(y_1-\mu_1)^T \Sigma_{11}^{-1} (y_1-\mu_1)}_{\text{marginal part}} + \underbrace{(y_2-\mu_{2\mid 1})^T \Sigma_{2\mid 1}^{-1} (y_2-\mu_{2\mid 1})}_{\text{conditional part}} \end{align*} with \begin{align*} \Sigma_{2\mid 1} &= \Sigma_{22} - \Sigma_{21}\Sigma_{11}^{-1}\Sigma_{12} \\ \mu_{2\mid 1} &= \mu_2 + \Sigma_{21}\Sigma_{11}^{-1}(y_1-\mu_1) \end{align*} Integrating out the marginal part leaves the conditional density (where we can ignore the constants because we know at the end we will get a density).

A note on the intuition

Notice that if $Y\sim \mathcal{N}(0,\Sigma)$, then $L^{-1}Y \sim \mathcal{N}(0, \mathbb{I})$. So we essentially removed all depdence using $L^{-1}$. Or starting from iid standard normal $X\sim\mathcal{N}(0, \mathbb{I})$ we can obtain our distribution $$Y=L X \sim \mathcal{N}(0,\Sigma).$$ Due to the lower triangular nature, only $X_1$ is used for $Y_1$. And by $L_{11}$ we can transform back and forth between $X_1$ and $Y_1$. So now the question is, what do we expect of $Y_2$ conditional on $Y_1$ (or equivalently $X_1$)? Well we know that $$ Y_2 = \underbrace{L_{21} X_1}_{\text{known}} + \underbrace{L_{22} X_2}_{\text{unknown}} $$ In essence $X_1$ is the old randomness and $X_2$ is the new randomness. So here the conditional expectation and conditional variance come from. That is why the conditional expectation is given by $$ \mathbb{E}[Y_2\mid Y_1] = L_{21}X_1 = L_{21}L_{11}^{-1}Y_1 $$ And why the conditional variance is given by $$ L_{22}L_{22}^T $$ which we then had to translate back into the covariance form.