The mean distribution versus distribution over mean of distribution over distributions?

Let $f$ be a distribution over distributions, i.e. $t \sim f$ and $x \sim t$ where $t$ is a distribution and $x$ is random variable. How to compute the mean distribution (denote it as $g(x)$) over $x$ using $f$ and $t$? Also, is this mean distribution different from the distribution over means, i.e. distribution over the mean value of each distribution $t$?

Solution 1:

I am not totally clear what your mean distribution is supposed to be. But here is a an example:

• Suppose $f$ is the Poisson distributions where their parameter $\lambda$ has an exponential distribution with rate and mean of $1$.

• $t$ is then a Poisson distribution with parameter $\lambda$ and so mean $\lambda$; the distribution of the means of $t$ is therefore the same as the distribution of $\lambda$, an exponential distribution with rate $1$.

• Given $\lambda$ and so $t$, you have $x$ having a Poisson distribution, while unconditionally $x$ has a geometric distribution on $0,1,2,\ldots$ with parameter $\frac12$.

A Poisson distribution with parameter $1$ is not the same as a geometric distribution with parameter $\frac12$ even when they both have the same mean of $1$; for the Poisson, the probability of seeing $0$ is $e^{-1}$ while for the geometric it is $\frac12$. Neither of those discrete distributions is the same as an exponential distribution, which is a continuous distribution.