As part of my homework in the "Theoretical Statistics" course I was asked the following question and I need some help with that.
Given two dices with $2n$ sides.
Let $1_A$ be indicator variables such that:
$A$ $=$ $\{$At least one of the dice shows the number n$\}$ $$1_A=\begin{cases}1 & x\in{A}\\ 0 & x\notin A \\ \end{cases}$$
Let $1_B$ be indicator variables such that:
$B$ $=$ $\{$The sum of the two dice is exactly $2n+1$$\}$
$$1_B=\begin{cases}1 & x\in{B}\\ 0 & x\notin B \\ \end{cases}$$
Prove/disprove:
The covariance of these two indicator variables is positive for every $n$
What I did so far:
Probability to get n is: $$\sum_{i=1}^2 (1-\frac{1}{2n})^{i-1}\cdot \frac{1}{2n} = \frac{1}{2n}+(1-\frac{1}{2n})\cdot\frac{1}{2n} $$ $$=\frac{1}{2n}+(\frac{1}{2n}-\frac{1}{4n^2}) $$ $$=\frac{1}{n}-\frac{1}{4n^2} = \frac{4n}{4n^2}-\frac{1}{4n^2} = \frac{4n-1}{4n^2}$$ $$P(A) = \frac{4n-1}{4n^2}$$
$$P(B) = \frac{1}{2n}$$ proof: $$P(p,n,s) = \frac{1}{s^n}\cdot \sum_{k=0}^{k_{max}}(-1)^k\cdot {n \choose k}{p-s\cdot k -1 \choose p-s\cdot k -k}$$ $$k_{max} = \lfloor \frac{p-n}{s}\rfloor$$ $$k_{max} = \lfloor \frac{2n+1-2}{n}\rfloor = 1$$
$$P(2n+1,2,2n) = \frac{1}{(2n)^n}\cdot \sum_{k=0}^{1}(-1)^k\cdot {n \choose k}{2n+1-2n\cdot k -1 \choose 2n+1-2n\cdot k -2}$$ $$\frac{1}{4n^2}\cdot [{2n \choose 2n-1}-2\cdot{0 \choose k}] = \frac{2n}{4n^2} = \frac{1}{2n}$$
$$P(A\cap B) = \frac{1}{2n^2}$$
$$P(A\cap B) - P(A)\cdot P(B) = \frac{1}{2n^2} - \frac{4n-1}{4n^2}\cdot\frac{1}{2n} = \frac{1}{8n^3} \gt 0 $$
Now - I only have to prove: $$P(A\cap B) = \frac{1}{2n^2}$$ Can someone help me with that?