Index: The Book of Statistical ProofsProbability Distributions ▷ Matrix-variate continuous distributions ▷ Matrix-normal distribution ▷ Marginal distributions

Theorem: Let $X$ be an $n \times p$ random matrix following a matrix-normal distribution:

$\label{eq:matn} X \sim \mathcal{MN}(M, U, V) \; .$

Then,

1) the marginal distribution of any subset matrix $X_{I,J}$, obtained by dropping some rows and/or columns from $X$, is also a matrix-normal distribution

$\label{eq:matn-marg-subs} X_{I,J} \sim \mathcal{MN}(M_{I,J}, U_{I,I}, V_{J,J})$

where $I \subseteq \left\lbrace 1, \ldots, n \right\rbrace$ is an (ordered) subset of all row indices and $J \subseteq \left\lbrace 1, \ldots, p \right\rbrace$ is an (ordered) subset of all column indices, such that $M_{I,J}$ is the matrix dropping the irrelevant rows and columns (the ones not in the subset, i.e. marginalized out) from the mean matrix $M$; $U_{I,I}$ is the matrix dropping rows not in $I$ from $U$; and $V_{J,J}$ is the matrix dropping columns not in $J$ from $V$;

2) the marginal distribution of each row vector is a multivariate normal distribution

$\label{eq:matn-marg-row} x_{i,\bullet}^\mathrm{T} \sim \mathcal{N}(m_{i,\bullet}^\mathrm{T}, u_{ii} V)$

where $m_{i,\bullet}$ is the $i$-th row of $M$ and $u_{ii}$ is the $i$-th diagonal entry of $U$;

3) the marginal distribution of each column vector is a multivariate normal distribution

$\label{eq:matn-marg-col} x_{\bullet,j} \sim \mathcal{N}(m_{\bullet,j}, v_{jj} U)$

where $m_{\bullet,j}$ is the $j$-th column of $M$ and $v_{jj}$ is the $j$-th diagonal entry of $V$; and

4) the marginal distribution of one element of $X$ is a univariate normal distribution

$\label{eq:matn-marg-elem} x_{ij} \sim \mathcal{N}(m_{ij}, u_{ii} v_{jj})$

where $m_{ij}$ is the $(i,j)$-th entry of $M$.

Proof:

1) Define a selector matrix $A$, such that $a_{ij} = 1$, if the $i$-th row in the subset matrix should be the $j$-th row from the original matrix (and $a_{ij} = 0$ otherwise)

$\label{eq:A} A \in \mathbb{R}^{\lvert I \rvert \times n}, \quad \text{s.t.} \quad a_{ij} = \left\{ \begin{array}{rl} 1 \; , & \text{if} \; I_i = j \\ 0 \; , & \text{otherwise} \end{array} \right.$

and define a selector matrix $B$, such that $b_{ij} = 1$, if the $j$-th column in the subset matrix should be the $i$-th column from the original matrix (and $b_{ij} = 0$ otherwise)

$\label{eq:B} B \in \mathbb{R}^{p \times \lvert J \rvert}, \quad \text{s.t.} \quad b_{ij} = \left\{ \begin{array}{rl} 1 \; , & \text{if} \; J_j = i \\ 0 \; , & \text{otherwise} \; . \end{array} \right.$

Then, $X_{I,J}$ can be expressed as

$\label{eq:XIJ} X_{I,J} = A X B$

and we can apply the linear transformation theorem to give

$\label{eq:XIJ-marg} X_{I,J} \sim \mathcal{MN}(A M B, A U A^\mathrm{T}, B^\mathrm{T} V B) \; .$

Finally, we see that $A M B = M_{I,J}$, $A U A^\mathrm{T} = U_{I,I}$ and $B^\mathrm{T} V B = V_{J,J}$.

2) This is a special case of 1). Setting $A$ to the $i$-th elementary row vector in $n$ dimensions and $B$ to the $p \times p$ identity matrix

$\label{eq:AB-row} A = e_i, \; B = I_p \; ,$

the $i$-th row of $X$ can be expressed as

$\label{eq:xi-marg} \begin{split} x_{i,\bullet} &= AXB = e_i X I_p = e_i X \\ &\overset{\eqref{eq:XIJ-marg}}{\sim} \mathcal{MN}(m_{i,\bullet}, u_{ii}, V) \; . \end{split}$ $\label{eq:xi-marg-trans} x_{i,\bullet}^\mathrm{T} \sim \mathcal{MN}(m_{i,\bullet}^\mathrm{T}, V, u_{ii})$ $\label{eq:xi-marg-trans-mvn} x_{i,\bullet}^\mathrm{T} \sim \mathcal{N}(m_{i,\bullet}^\mathrm{T}, u_{ii} V) \; .$

3) This is a special case of 1). Setting $A$ to the $n \times n$ identity matrix and $B$ to the $j$-th elementary row vector in $p$ dimensions

$\label{eq:AB-col} A = I_n, \; B = e_j^\mathrm{T} \; ,$

the $j$-th column of $X$ can be expressed as

$\label{eq:xj-marg} \begin{split} x_{\bullet,j} &= AXB = I_n X e_j^\mathrm{T} = X e_j^\mathrm{T} \\ &\overset{\eqref{eq:XIJ-marg}}{\sim} \mathcal{MN}(m_{\bullet,j}, U, v_{jj}) \end{split}$ $\label{eq:xj-marg-mvn} x_{\bullet,j} \sim \mathcal{N}(m_{\bullet,j}, v_{jj} U) \; .$

4) This is a special case of 2) and 3). Setting $A$ to the $i$-th elementary row vector in $n$ dimensions and $B$ to the $j$-th elementary row vector in $p$ dimensions

$\label{eq:AB-elem} A = e_i, \; B = e_j^\mathrm{T} \; ,$

the $(i,j)$-th entry of $X$ can be expressed as

$\label{eq:xij-marg} \begin{split} x_{ij} &= AXB = e_i X e_j^\mathrm{T} \\ &\overset{\eqref{eq:XIJ-marg}}{\sim} \mathcal{MN}(m_{ij}, u_{ii}, v_{jj}) \; . \end{split}$

As $x_{ij}$ is a scalar, this is equivalent to a univariate normal distribution as a special case of the matrix-normal distribution:

$\label{eq:xij-marg-norm} x_{ij} \sim \mathcal{N}(m_{ij}, u_{ii} v_{jj}) \; .$
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Metadata: ID: P343 | shortcut: matn-marg | author: JoramSoch | date: 2022-09-15, 11:41.