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The Inverse-Wishart wikipedia article states that $ X \sim \operatorname{Wish}^{-1}(\Psi,\nu)$ if $X^{-1} \sim \operatorname{Wish}(\Psi^{-1},\nu),$ where $\Psi$ and $X$ are $p\times p$.

The mean of the Inverse-Wishart is given as $\frac{\Psi}{\nu-p-1}$, but the mean of the Wishart is $\nu(\Psi^{-1})$ (See here). Why don't these match up? Wouldn't the mean of the Inverse-Wishart be $\frac{\Psi}{\nu}$?

Is this just because of a difference in parameterizations?

aepound
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1 Answers1

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Let's look at the simpler case of positive univariate random variables. I assume it's already clear that in general for a positive random variable whose variance is not $0$, that $\text{Cov}(X,1/X)< 0$. If you're happy to accept that as the case, then

$\text{Cov}(X,1/X) = E(X.1/X) - E(X)E(1/X) = 1-E(X)E(1/X) <0\,,$ so

$E(X) E(1/X) > 1\,,$ or

$E(1/X)>1/E(X)$

or you could just use Jensen's inequality to the same end.

(In fact we can show $1/E(X)<E(1/X)\leq 2/E(X)$, a rather neat little fact)

Generally we should not expect means to "match up" in the way you anticipated.

Glen_b
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