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How can be proven that for random variables $A$ and $B$, and $C = A + B$,

$$H(C\mid A) = H(B\mid A).$$

Also, would it be possible to determine if $H(C)$ would be greater than $H(A)$?

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1 Answer 1

I guess $A$ and $B$ are discrete random variables. Since $C$ is a function of $A$ and $B$ one has $H(C \mid A) \leq H(A,B \mid A)$ (because this is even true that $H(C \mid A=a) \leq H(A,B \mid A=a)$ for every $a$). Moreover $H(A,B \mid A) = H(B \mid A)$ because it is easy to check that $H(A,B \mid A=a) = H(B \mid A=a)$ for every $a$. Finally $H(C \mid A) \leq H(B \mid A)$ and one similarly gets the converse inequality.

For your second question: no, take $B=-A$ then $H(C)=0$. I have not tried to check whether this is true under the independence assumption of $A$ and $B$.

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