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The inequality is true for ordinary addition, and also for addition in a group. For any fixed $y$, the translation $x\mapsto x+y$ is injective, so it merely relabels the probability masses in the conditional distribution. Since $X,Y$ are independent random variables,Averaging over $y$ and using conditioning reduces entropy proves entropy monotonicity under independent addition:There is no requirement that $X+Y$ remain in the original finite alphabet. The proof also gives $H(X+Y)\ge H(Y)$ by symmetry. Independence matters: for a nonconstant $X$ and $Y=-X$, the sum is constant while $H(X)>0$.

For $N\sim \mathrm{Pois}(\lambda )$, the Poisson distribution givesThis information entropy is finite: ${\log }_2(n!)\le n^2$ for integer $n\ge 0$, while $\mathbb{E}{N}^2=\lambda +{\lambda }^2$.

Fix $0<{\lambda }_1<{\lambda }_2$. Take independent random variables $U\sim \mathrm{Pois}({\lambda }_1)$ and $V\sim \mathrm{Pois}({\lambda }_2-{\lambda }_1)$. Their convolution isby the binomial theorem. Thus $U+V$ has Poisson distribution with parameter ${\lambda }_2$. Conditioning on $V=v$ only translates $U$, so $H(U+V\mid V)=H(U)=F({\lambda }_1)$. The entropy monotonicity under independent addition yields $F({\lambda }_2)\ge F({\lambda }_1)$.

In fact the inequality is strict. The covariance $\mathrm{Cov}(U+V,V)=\mathrm{Var}(V)={\lambda }_2-{\lambda }_1$ is positive, so $U+V$ and $V$ are not independent random variables. Their information entropies are finite, and the equality criterion from part (a) excludes equality. Therefore the Poisson entropy monotonicity gives the stronger conclusion

The information entropy $F(\lambda )$ of a Poisson distribution is strictly increasing for $\lambda >0$. Given ${\lambda }_2>{\lambda }_1$, take independent $U\sim \mathrm{Pois}({\lambda }_1)$ and $V\sim \mathrm{Pois}({\lambda }_2-{\lambda }_1)$. Their sum has Poisson distribution with mean ${\lambda }_2$, so entropy monotonicity under independent addition gives $F({\lambda }_2)\ge F({\lambda }_1)$. All these information entropies are finite, and $\mathrm{Cov}(U+V,V)=\mathrm{Var}(V)>0$ excludes equality.

// Target: probability-and-statistics.bigb