TITLE: What is the mutual information $I(X;X)$?
QUESTION [6 upvotes]: $X$ is a random variable with normal distribution, assume $Y=X$, what is the mutual information $I(X;Y)$?
I guess that $h(Y|X)=0$ since when $X$ is known, $Y$ is completely known, so $$I(X;Y)=h(Y)-h(Y|X)=h(Y)=\frac{1}{2}\log 2\pi e\sigma^2$$ nat.
But, I was told I was wrong! and a numerical computation also shows that the value of $$I(X;Y) \neq \frac{1}{2}\log 2\pi e\sigma^2$$ Where is my mistake? Please help me out of this problem, thanks a lot! (Please note that $X$ and $Y$ are both continuous).

REPLY [7 votes]: For any r.v. $X$
$$I(X,X)=H(X).$$
To see this, put $Y=X$ in $I(X,Y)=H(X)-H(X|Y)$ and use 
$$H(X|X)=0.~~  (*)$$
In summary, the mutual information of $X$ with itself is just its self information $H(X)$, as $H(X|X)$, i.e. the residual average information carried by $X$ conditional $X$ is zero, i.e. $(*)$ holds.