Suppose that we have a probability density function $\pi(x_1, \ldots, x_n)$ which is the density of a vector-valued random variable $X$ in $\mathbb{R}^n$. Here the density is proper, i.e., $\int_{\mathbb{R}^n} \pi(x_1, \ldots, x_n) dx_1 \ldots dx_n = 1 < \infty$.

Without any additional restrictions on the joint density, does this imply that any coordinate-wise marginal density of the form $$ \pi(x_i) = \int \pi(x_1, \ldots, x_n) dx_1 \ldots dx_{i-1} dx_{i+1} \ldots dx_n $$ is also proper, i.e., that $\int_{\mathbb{R}} \pi(x_i) dx_i = 1 < \infty$ for any $1 \leq i \leq n$?

The motivation for this question is that I am considering a hierarchical Bayesian model involving an improper prior, and although I know that the posterior density over all model parameters is proper, I have seemingly found a marginal that is not proper (although this may be an error in my analytic computation). Intuitively, I would expect that any marginal of a proper joint should also be proper?


1 Answer 1


The marginal density of a probability density is again a probability density (a.s.).

The same notations should not be used for different entities, so let me define $$\pi_i(x_i) = \int_{\mathbb R^{n-1}} \pi(x_1, \ldots, x_n) dx_1 \ldots dx_{i-1} dx_{i+1} \ldots dx_n$$ as the $i$-th marginal. Then $$\int_{\mathbb{R}} \pi_i(x_i) dx_i = \int_{\mathbb{R}}\int_{\mathbb R^{n-1}} \pi(x_1, \ldots, x_n) dx_1 \ldots dx_{i-1} dx_{i+1} \ldots dx_n dx_i\\ =\int_{\mathbb R^{n}} \pi(x_1, \ldots, x_n) dx_1 \ldots dx_{i-1} dx_i dx_{i+1} \ldots dx_n = 1$$ by Fubini's Theorem.

This applies to the Bayesian setting in the question: if the joint posterior is proper, then any posterior marginal or conditional is equally proper (a.s.).

  • 2
    $\begingroup$ +1. The "a.s." qualifier is essential, though! $\endgroup$
    – whuber
    May 10, 2023 at 19:45

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