I stated this was due to stochasticity.
If I’m right, and I_{LZ}(X:Y) = I_{LZ}(E(X):Y) = 0 in general, then on average I_{LZ}(X:Y) > I_{LZ}(E(X):Y) should fail half the time.
The other control is when X=Y. In this case, I_{LZ}(X:Y) > I_{LZ}(E(X):Y) should fail less than half the time.
I’ll test this and get back to you.