Let $H$ and $K$ be independent random variables. They both follow the binomial distribution of $Binomial(n,p)$.
The question is: Prove or disprove that $H+K$ and $H-K$ are independent to each other.
I know that $H+K$ is $Binomial(2n,p)$, which can be proved with the help of moment-generating functions:
$$M_{H+K}(t)=M_H(t)M_K(t)=(pe^t+1-p)^{2n}$$
However, what is the meaning of $H-K$? Is it $Binomial(0,p)$? How to solve the whole question?
$H+K$ and $H-K$ are not independent. For $$\{H+K = 0, H-K = 0\} = \{H=0,K=0\}$$ and so $$ \mathbb P(H=0,K=0) = \mathbb P(H=0)\mathbb P(K=0) = (1-p)^{2n}. $$ Similarly, $$ \mathbb P(H+K=0) = \mathbb P(H=0,K=0) = (1-p)^{2n}. $$ However, \begin{align} \mathbb P(H-K=0) &= \sum_{i=0}^n \mathbb P(H=i, K = i)\\ &= \sum_{i=0}^n \mathbb P(H=i)\mathbb P(K=i)\\ &= \sum_{i=0}^n \left(\binom ni p^i (1-p)^{n-i}\right)^2. \end{align} So clearly $$ \mathbb P(H+K=0,H-K=0)\ne \mathbb P(H+K=0)\mathbb P(H-K=0). $$