# For multivariate normal posterior with improper prior, why posterior is proper only if $n\geq d$

This is related to Gelman's BDA chapter 3 section 5's noninformative prior density for $$\mu$$.

Let $$\Sigma$$ be fixed positive definite symmetric matrix of size $$d$$ by $$d$$. Let $$y_1,\dots, y_n$$ be iid Gaussian r.v. with density normal $$N(\mu,\Sigma)$$. Let $$p(\mu)$$ be prior density of $$\mu$$ which is proportional to constant. Then posterior $$p(\mu|y_1,\dots, y_n)$$ is proportional to $$p(y_1,\dots, y_n|\mu)$$. In particular, $$p(\mu|y_1,\dots, y_n)$$ is proportional to $$|\Sigma|^{\frac{n}{2}}\exp(-\frac{1}{2}tr(\Sigma^{-1}S_0))$$ (3.11) where $$S_0=\sum_i(y_i-\mu)(y_i-\mu)^T$$.

The book says (3.11) is a proper posterior only if $$n\geq d$$ and if $$n, then $$S_0$$ is not of full rank.

Q1. Why proper posterior only if $$n\geq d$$ here? I have tried $$n=1,d=2$$ and I still find the posterior proper. The essential thing I see is that $$tr(\Sigma^{-1}S_0)$$ for this case's quadratic form remains to be $$\Sigma^{-1}$$ with $$(\mu_1-y_{11},\mu_2-y_{12})$$ where $$y_1=(y_{11},y_{12})$$.

Q2. What does it mean for $$S_0$$ is not of full rank for $$n? Consider points $$y_1=y_2=\dots=y_n$$. $$S_0$$ is of rank 1 in this case. Does the statement mean there is a dense open set of $$(y_1,\dots, y_n)\in R^{nd}$$ s.t. $$rk(S_0)$$ is full for $$n\geq d$$.

• @Xi'an Where can I find proof for Q1 and Q2 here? I do not think either seems trivial proof. Thanks. Commented Aug 2, 2022 at 12:38

Q¹: With a flat prior, and one single observation $$x$$, $$\pi(\mu|x)\propto\exp\{-(x-\mu)^\mathsf{T}\Sigma^{-1}(x-\mu)\big/2\}$$ which implies that the posterior is a multivariate $$\mathsf{N}_d(x,\Sigma)$$, definitely proper. Extends straightforwardly to more than one observation.
Q²: When $$n=d$$, $$S_0=\sum_{i=1}^d(y_i-\mu)(y_i-\mu)^\mathsf T$$is not of full rank iff there exists a non-zero $$\varepsilon$$ such that $$S_0\varepsilon=0$$, i.e.,$$\sum_{i=1}^d(y_i-\mu)(y_i-\mu)^\mathsf T\epsilon=\sum_{i=1}^d\{(y_i-\mu)^\mathsf T\epsilon\}(y_i-\mu)=0$$implying that (i) $$\epsilon$$ is orthogonal to all $$y_i-\mu$$'s which thus (ii) live in a subspace of dimension $$d-1$$ at most, a zero measure set. Extends straightforwardly to $$n>d$$.
Notice. While the 2014 edition does contain this$$-$$unnecessary when $$\Sigma$$ is known$$-$$ condition, the latest printing has removed it:
• The Q2's answer seems correct. I think this might be dumb. For your proof in Q1, how does it imply that $n\geq d$? It seems that the proof has shown that given any $d$, for any $n$ observation, $\pi(\mu|x)$ is always proper. This does not prove $n\geq d$ but $n\geq 1$. Commented Aug 2, 2022 at 12:59