1
$\begingroup$

From wikipedia:

To obtain the marginal distribution over a subset of multivariate normal random variables, one only needs to drop the irrelevant variables (the variables that one wants to marginalize out) from the mean vector and the covariance matrix.

Is there any intuition of why this works out so well?

$\endgroup$
0
$\begingroup$

This should readily follow if you look at the expression for the moment generating function (M.G.F.) for a multivariate normal density. M.G.F. looks like:

$E_X(e^{t'X}) = e^{\mu't + 0.5t'\Sigma t}$ for any real $n$-vector $t$, where $X = (X_1, X_2,...,X_n)$ is multivariate normal $N_n(\mu,\Sigma)$ random variate [simple to derive this!]

You can see from the expression that the exponent $\mu't + 0.5t'\Sigma t$ is a scalar and moreover, if you want to find the M.G.F. of a $p$-subset ($p \ne 0$) of the random vector $X$, say, {${X_{i_1}, X_{i_2}, ..., X_{i_p}}$}, where, {$i_1,i_2,..,i_p$} is a permutation of {$1,2,..,n$}, then you just need to plug in $t_j=0$ for $j$ $\epsilon$ {$1,2,..,n$}-{$i_1,i_2,..,i_p$} in the expression of M.G.F. The simplified form reduces to a known M.G.F. of a uni/multi-variate normal and that M.G.F. uniquely determines a distribution confirms the nice property of the marginal density of multivariate normal distribution.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.