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).

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

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.

This happens with univariate random variables as well and it's not a matter of how you parameterize it.

  I assume it's already clear that in general for a continuous positive random variable whose variance is not $0$, that $\text{Cov}(X,1/X)< 0$. If you accept that, then

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

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

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

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

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).

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