# Division of Multivariate Normal Distributions

$$\newcommand{\vect}{\boldsymbol{\mathbf{#1}}} \newcommand{\nc}{\newcommand{#1}{#2}} \nc{\vx}{\vect{x}} \nc{\vmu}{\vect{\mu}} \nc{\vSigma}{\vect{\Sigma}} \nc{\vtheta}{\vect{\theta}}$$

# Question

Is the ratio of two multivariate normal distributions again a multivariate normal distribution?

# Context

I noticed that if we have two pdfs belonging to the Exponential Family of distributions, having the same sufficient statistics but different parameters, i.e. $$f(\vx) = \exp\left\{\vtheta_1^\top \phi(\vx) - A_1(\vtheta_1)\right\} \qquad g(\vx) = \exp\left\{\vtheta_2^\top \phi(\vx) - A_2(\vtheta_2)\right\}$$ Then dividing one by the other we get $$h(\vx) := \frac{f(\vx)}{g(\vx)} = \frac{ \exp\left\{\vtheta_1^\top \phi(\vx) - A_1(\vtheta_1)\right\}}{\exp\left\{\vtheta_2^\top \phi(\vx) - A_2(\vtheta_2)\right\}} = \exp\left\{(\vtheta_1 - \vtheta_2)^\top \phi(\vx) - A'(\vtheta_1 - \vtheta_2)\right\}$$ where $$A'(\vtheta_1 - \vtheta_2) = \log \int\exp\left\{(\vtheta_1 - \vtheta_2)^\top \phi(\vx)\right\}d\vx$$ So that the division is again in the exponential family of distributions. My question is: is this new pdf,resulting from the quotient of two multivariate normal distributions, also normally distributed?

• You can not in general expect the difference between two covariance matrices to be positive definite. Mar 12 '20 at 10:29
• The ratio of two pdfs hath no reason to be itself a pdf. Mar 12 '20 at 10:59
• @Forgottenscience Thank you for your answer! Does this mean that the result is going to be a member of the exponential family, but not necessarily a Normal distribution? Mar 12 '20 at 11:34
• You show yourself that it remain of the form of an exponential family, but there is no guarantee that it has any interesting properties. Mar 12 '20 at 12:03
• What the first two commenters are telling you can be illustrated with a simple example. Consider the univariate case where the Normal$(0,(1/3)^2)$ pdf is divided by the Normal$(0,1^2)$ pdf: the ratio, being proportional to $\exp(x^2) \ge 1,$ has an integral that diverges to infinity. Thus, this ratio cannot possibly represent any kind of probability distribution. The same problem will occur in any dimensions.
– whuber
Mar 12 '20 at 13:46