Given a Gaussian Likelihood, and a Normal-Wishart prior, what is the posterior?

Question

Given a Gaussian Likelihood : $$P(Y|\mu,\Lambda) = \prod_{t=1}^T P(y_t|\mu,\Lambda)$$

and a Normal-Wishart Prior: $$P(\Lambda) = Wishart(\Lambda|v_0,W_0)$$

what is the posterior distribution?

Answer 1

You know that

$$ f(\Lambda) = \frac{|\Lambda|^{\frac{\nu_0 - p - 1}{2}} exp(-\frac{1}{2}tr(W_0 \Lambda))}{2^{\nu_0 p /2} |W_0|^{-\nu/2} \Gamma_p (\nu_0/2)} $$

where $\Lambda, W_0$ : positive-definite. (Parameterization can yield somewhat different PDF, but that does not matter) Thus,

$$ \begin{aligned} \Lambda | \mathbf{y}_T &\propto \prod_i^T |2\pi \Lambda^{-1}|^{-1/2} exp(-\frac{1}{2}(y_i - \mu)' \Lambda^{-1} (y_i - \mu)) \times |\Lambda|^{\frac{\nu_0 - p - 1}{2}} exp(-\frac{1}{2}tr(W_0 \Lambda)) \\[10pt] &\propto |\Lambda|^{\frac{n+\nu_0 - p - 1}{2}} \exp\left(-\frac{1}{2}tr(nS\Lambda + W_0 \Lambda)\right) \\[10pt] &\sim Wishart(\nu_0 + n, W_0 + nS) \end{aligned} $$

where $S = \frac{1}{n} \sum (y_i - \mu)'(y_i - \mu)$