For two multivariate Gaussians with covariance matrices $A$ and $B$ (of dimension $n$) and with equal means, the squared Hellinger distance is given by
$$
H^2 = 1 – \frac{[\det(A)\det(B)]^\frac{1}{4}}{\left[\det\left(\frac{A + B}{2}\right)\right]^{\frac{1}{2}}}
$$
How can I show that $H^2 \geq 0$? The fact is well-known for the Hellinger distance in general, but I wonder how I can prove it for this special case?
Here is what I have: $H^2 \geq 0$ gives
$$
\frac{[\det(A)\det(B)]^\frac{1}{4}}{\left[\det\left(\frac{A + B}{2}\right)\right]^{\frac{1}{2}}} \leq 1
$$
and therefore we have to show that
$$
\det{\left(\frac{A + B}{2}\right)} \geq \sqrt{\det(A)\det(B)}\tag{*}
$$
which is an interesting determinant inequality for positive semi-definite matrices in itself. It says that the determinant of an arithmetic mean of two matrices is larger than or equal to the geometric mean of their determinants.
The first relation I could think of was:
$$
\left(\sqrt{\det(A)} – \sqrt{\det(B)}\right)^2 \geq 0
$$
which leads to
$$
\det(A) + \det(B) – 2 \sqrt{\det(A)\det(B)} \geq 0
$$
and therefore
$$
\det(A) + \det(B) \geq 2 \sqrt{\det(A)\det(B)}.
$$
As the second relation we can apply the following inequality for positive semi-definite matrices (Abadir and Magnus, Matrix Algebra, p. 325):
$$
\det(A + B) \geq \det(A) + \det(B).
$$
This leads to
$$
\begin{align}
\det\left(\frac{A + B}{2}\right)
&= 2^{-n} \det(A + B)\\
&\geq 2^{-n} [\det(A) + \det(B)]\\
&\geq 2^{-n} 2 \sqrt{\det(A)\det(B)}
\end{align}
$$
and then I'm stuck since I can't get rid of the factor $2^{-n} 2$. Is there a mistake in my attempt? How can I prove the inequality (*) in a different way?
I think the inequality taken from the Abadir/Magnus book is too strong for this problem. Equality is reached for zero determinant of the sum or zero matrices (p.326), whereas we would need some inequality where equality is reached for identical matrices.
Best Answer
Hint.
If $A$ and $B$ are both positive definite, it suffices to prove that $$\ln \det \frac{A + B}{2} \ge \frac{1}{2}\ln \det A + \frac{1}{2}\ln \det B$$ which is true since log-determinant $X \mapsto \ln \det X$ is concave for positive definite matrices (See Page 74, [1]).
References
[1] Boyd and Vandenberghe, "Convex optimization". http://web.stanford.edu/~boyd/cvxbook/bv_cvxbook.pdf