say $X$ and $Y$ are two identical, independent and discrete Uniform random variables and $Z=X+Y$. I do not know more about the random variables.
Assuming $H(\cdot)$ to be the entropy of a random variable, I know that $H(Z) \geq \max\{H(X),H(Y)\}$. However, I wonder if a tighter bound exists given $X$ and $Y$ are discrete Uniform random variables?
any hint is appreciated.
On
Baudolino's solution was a good hint to derive the exact entropy of sum of two uniforms. I assume $X$ and $Y$ are two discrete, independent uniformly distributed r.v. I make a further assumption and assume that both $X$ and $Y$ are symmetric around zero and zero itself is a member of the r.v.; therefore $X$ and $Y$ have $2n+1$ members with equal pmf and $Z$ has $4n+1$ entries.
As the pmf of sum of two random variables is the convolution of pmfs:
$f_Z(z)=\int\limits_{-2n}^{2n} f_X(\tau)f_Y(z-\tau)\mathrm{d}z$
Consequently:
Prob($Z=0$) = $(2n+1)(\frac{1}{2n+1})^2$,
Prob($Z=-1$) = Prob($Z=1$) = $(2n)(\frac{1}{2n+1})^2$
Prob($Z=-2$) = Prob($Z=2$) = $(2n-1)(\frac{1}{2n+1})^2$
$\vdots$
Prob($Z=-2n$) = Prob($Z=2n$) = $(\frac{1}{2n+1})^2$
and so, $H(Z)=2(\frac{1}{2n+1})^2 \log (2n+1)$ + $\sum\limits_{t=1}^{n} (\frac{2}{2n+1})^2 (2n+1-t) \log ({2n+1})$
$H(Z)= (\frac{2}{2n+1})^2 \Big( \frac{1}{2}+\sum\limits_{t=1}^{n} (2n+1-t) \Big) \log ({2n+1})$
Here's my attempt. We're using this definition for $H(X)$, for a discrete r.v. $X$ with probabilities $p_1, p_2, \ldots, p_n$: $$ H(X)=-\sum_{i=1}^n p_i \log_2(p_i) $$
We know that the sum of two i.i.d. discrete uniform r.v. is a discrete triangular r.v., call it $Z$. $Z$ is symmetrical, so even though the support extends to $2n$ we need only to evaluate the sum until midway. We also know that the probabilities $z_i$ of $Z$ are $\frac{1}{n^2}, \frac{2}{n^2},\ldots,\frac{n}{n^2},\frac{n-1}{n^2},\ldots,\frac{1}{n^2}$. Hence, we have, $$ \begin{aligned} H(Z) &= -2\sum_{i=1}^{n-1} z_i \log_2(z_i) - z_n \log_2(z_n)\\ &= -2\sum_{i=1}^{n-1} \frac{i}{n^2} \log_2 \left( \frac{i}{n^2} \right) -\frac{n}{n^2}\log_2 \left( \frac{n}{n^2}\right) \\ &= - \frac{2}{n^2} \sum_{i=1}^{n-1} i [\log_2(i) - 2\log_2(n)] + \frac{\log_2(n)}{n}\\ &= - \frac{2}{n^2} \left(\sum_{i=1}^{n-1} i \log_2(i) - 2\log_2(n) \sum_{i=1}^{n-1} i \right) + \frac{\log_2(n)}{n}\\ &= - \frac{2}{n^2} \left[\sum_{i=1}^{n-1} \log_2(i^i) - 2\log_2(n) \frac{n(n-1)}{2} \right] + \frac{\log_2(n)}{n}\\ &= \frac{2(n-1) \log_2(n)}{n}- \frac{2}{n^2} \log_2 \left(\prod_{i=1}^{n-1} i^i \right) + \frac{\log_2(n)}{n}\\ &= \frac{(2n-1) \log_2(n)}{n}- \frac{2}{n^2} \log_2 \mathcal{H}(n-1). \end{aligned} $$ In the derivation, $\mathcal{H}(x)$ is the hyperfactorial function. And you'll probably need to adjust the formula if $n$ is odd.
As a check, if $n=6$ (a fair dice), then the information entropy associated with the event that observes the sum of two dice is about $3.2744$ (checked with WolframAlpha). I am not familiar with this area, so I do not know if this is the actual result, but it seems reasonable.