Vtyjkyuk

Given $Z=X+Y$ where X and Y are two random variables, under what conditions does $H(Z)=H(X)+H(Y)$?

Notice $Z$ is a function of $(X,Y)$, therefore $H(Z)leq H(X,Y)$, and $H(X,Y)leq H(X)+H(Y)-I(X;Y)$. Therefore when $Z$ and $(X,Y)$ is a one-to-one function and $X$ is independent of $Y$.

My question is, how to find an example that satisfy this problem ? I have a simple example that let $X=0$ so that $Z=Y$, which satisfy the conditions. However, can anyone gives a more non-trivial example ? Thanks.

To construct a proper example, as you found out yourself in a very nice fashion, there is a bijection between $Z$ and $(X,Y)$ (I suppose that random variables are discrete of course).

Let $X'sim textBernoulli(frac 12)$ and $Ysim textBernoulli(frac 12)$. Set $X=2X'$ and take $Z=X+Y$. Then $Z$ is uniformly distributed over $0,1,2,3$ and $X$ and $Y$ are also uniformly distributed over $0,2$ and $0,1$. Therefore it is easy to see that $H(Z)=H(X)+H(Y)$.

In addition to the nice example @Arash gives. Assume the discrete uniform distribution. If $$X sim mathcal U_[a,b](n)$$and$$Y sim mathcal U_[c,d](m)$$
Let's define $$Z = X+Y$$
We know that the entropies of $X$ and $Y$ are $$H(X) = log n$$ and $$H(Y) = log m$$ therefore $$H(X) + H(Y) = log(n) + log(m) = log(nm)$$
This means that if $Z$ has $nm$ outcomes, then we can assert that $H(Z) = H(X) + H(Y)$. Many examples could be generated here to demonstrate the need of being bijective.

One Example out of MANY

Assume $X in lbrace 1,2,3rbrace$ and $Y in lbrace 0,10,20 rbrace$. It is easy to see that $$Z in lbrace 1,2,3,11,12,13,21,22,23 rbrace$$ with $$P(Z = z) = frac19$$
Is $H(Z) = ln(3times 3) = log 9$ ? Of course, because we $X+Y$ here is bijective.

Let's prove it: $$H(Z) = sumlimits_k=1^9 p_k log(frac1p_k) = sumlimits_k=1^9 frac19 log(9) =9frac19 log(9) = log(9)$$

Counter example

Now let's take the same distributions (with different outcomes) but show that $H(Z)$ will not reach $log(9)$ if there is no bijectivity. Let $X in lbrace 0,1,2 rbrace$ and $Y in lbrace 3,4,5 rbrace $. So $$Z in lbrace 3,4,5,6,7 rbrace$$
Therefore, it is clear there is no bijectivity, because $(1,3)$ and $(0,4)$ both give $Z = 4$, hence we can not decode $X,Y$ from $Z$.
Computing the entropy $$H(Z) = frac19 log 9 + frac29 log frac92 +frac39 log frac93 + frac29 log frac92 + frac19 log 9 < log(9)$$