Marginal distribution over components of a normal-Wishart distribution

[hythlodaeus]

Hey,

I have a Gaussian-Wishart distribution (with hyperparameters {m, B, W, v}) over the mean and precision matrix of a bivariate Gaussian distribution. Would I be right in thinking that the marginal distributions for each component (i.e. distributions over mean and precision for univariate Gaussian distribution over each component) is simply given by a Gaussian-Gamma distribution with hyperparameters {m_i, B, v/2, 0.5*W_ii} (where v/2 is the shape parameter and 0.5*W_ii the inverse scale (rate) parameter). That's what I seem to get, but I'm a little unsure...

Thanks in advance!

[hythlodaeus]

So, I have:
f(mu,R|B,m,W,v) = Normal(mu|m,inv(B*R)).Wishart(R|W,v)

Using the property of the Wishart distribution that the inverse of R (the covariance matrix, C) has distribution:
C ~ Inverse-Wishart(V, v)
where V = W^-1

and the property of the inverse-Wishart distribution that if C and V are partitioned conformably with each other (in my case, partitioned into components for each of the 2 dimensions), then
C_11 ~ Inverse-Wishart(V_11, v-1)

Which for scalar C_11, V_11 reduces to
C_11 ~ Inverse-Gamma((v-1)/2, (V_11)/2)

Together with the property of the normal distribution that under such a partition (again, by component), if mu ~ Normal(m, C) then
mu_1 ~ Normal(m_1,C_11)

We obtain a distribution for mu_1, C_11:
f(mu_1,C_11|v,W,B,m) ~ Normal(mu_1|m_1, (1/B)*C_11).Inverse-Gamma(C_11|(v-1)/2, (V_11)/2)

I think...

[linadth]

Hi! Did you find concrete answers to your question?? I am struggling to find this marginal distribution also!! Thanks!