# Joint entropy of a random variable with itself

If A and B are two independent random variables with $$n_A$$ and $$n_B$$ number of possible values, then the joint random variable AB would have $$n_A \times n_B$$ number of possible values and the joint entropy is simply additive

$$H(AB) = H(A) + H(B)$$ But what is $$H(AB)$$ if A and B are not independent, for example, if they are identical?

We can directly compute the case of when they are identical:

$$H(A, B) = \sum_{a, b} p(a, b) \log p(a,b) = \sum_{a = b} p(a, b) \log p(a,b)$$

where in the second equality we applied the assumption that $$A = B$$ with probability $$1$$. Then, since $$P(A=a, B=a) = P(A=a)$$ (since $$A=B$$), we have:

$$= \sum_a p(a)\log p(a) = H(A) = H(B)$$

But what is $$H(AB)$$ if A and B are not independent, for example, if they are identical?

dmh has posted the calculation which demonstrates that two identical distributions have the same entropy as either one of them does.

Entropy is a measure of "how much information". If we have two independent variables $$A$$ and $$B$$, then even if we know $$A$$, we do not know any information about $$B$$. So the joint entropy of the two variables is the sum of the entropies of the two variables. The amount of information in $$A$$ and $$B$$ is the sum of the amounts of information in either, since they are unrelated.

If we have two completely dependent variables $$C$$ and $$D$$, and we know $$C$$, the amount of information we gain by knowing the value of $$D$$ is nothing, because if we know $$C$$ already, we can exactly predict the value of $$D$$. Thus the entropy of $$C$$ and $$D$$ is equal to the entropy of $$C$$ alone. The amount of information in $$C$$ and $$D$$ together is exactly the same as the amount of information in just $$C$$ or just $$D$$.

In general (chain rule)

$$H(A,B) = H(A) + H(B|A)$$

where $$H(B|A)$$ is the conditional entropy of $$B$$ given $$A$$. If $$A$$ and $$B$$ are identical (more in general, if $$B=g(A)$$, i.e., it's deterministically given from $$A$$) then $$H(B|A)=0$$ (see)