Prove or disprove: covariance of two indicator variables is positive

Question

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?

Answer 1

Your calculations are overly complicated and confusing.

$A = \{\text{At least one of the $2n-$sided die shows the number }n\}$

The probability that a single die shows a particular face is $1/2n$. The probability that two die both show that face is $1/4n^2$. So the probability for event $A$ is is:

$\qquad\mathsf P(A) = 2\cdot\dfrac 1{2n}-\dfrac 1{4n^2} = \dfrac{4n-1}{4n^2}$

$B = \{\text{The sum of the two dice is exactly }2n+1 \}$

To obtain the favoured event, whatever the leftmost die shows, the other must show $2n+1$ minus that. The probability that one die shows $x$ and the other $2n+1-x$ is $1/4n^2$ for any $x$ in $[[1..2n]]$. So the probability for event $B$ is:

$\qquad\mathsf P(B) = \displaystyle\sum_{x=1}^{2n}\dfrac{1}{4n^2}=\dfrac{1}{2n}$

Now - I only have to prove:$\mathsf P(A\cap B)=\frac 1{2n^2}$ Can someone help me with that?

$A\cap B=\{\text{One of the two $2n-$sided die shows face $n$ while the other shows face $n+1$}\}$

As above, the probability that a single die shows a particular face is $1/2n$, so:

$\qquad\mathsf P(A\cap B)=2\cdot\dfrac 1{2n}\cdot\dfrac 1{2n}=\dfrac{1}{2n^2}$.

Answer 2

To have the sum to be $2n+1$ and at least one being $n$. We must have $X_1=n$, $X_2=n+1$ or $X_1=n+1, X_2=n$.

\begin{align} P(A \cap B) &=\frac2{(2n)^2}=\frac1{2n^2} \end{align}