jernplate8

## Answered question

2021-08-15

Let X be the number of 1’s and Y the number of 2’s that occur in n rolls of a fair die. Compute Cov(X, Y).

### Answer & Explanation

irwchh

Skilled2021-08-16Added 102 answers

Define ${X}_{i}$to be indicator random variable that is equal to one if and only if we have obtained 1 in the ith throw.
Similary, define ${Y}_{i}$ to be indicator random variable that is equal to one if and only if we have obtained 2 in ith throw. Therefore
$X=\sum _{i=1}^{n}{X}_{i},Y=\sum _{j=1}^{n}{Y}_{j}$
Using basic properties of the covariance, we have that
$Cov\left(X,Y\right)=Cov\left(X=\sum _{i=1}^{n}{X}_{i},Y=\sum _{j=1}^{n}{Y}_{j}\right)=\sum _{i}jCov\left({X}_{i},{Y}_{j}\right)$
Observe that for $i\ne j$ random variables${X}_{i}$ and ${Y}_{j}$ are independent.
Knowing what happened in ith throw does not change probabilities for jth throw.
Hence, in that case
$Cov\left({X}_{i},{Y}_{j}\right)=0$
So
$Cov\left(X,Y\right)=\sum _{i}Cov\left({X}_{i},{Y}_{i}\right)$
We have that
$Cov\left({X}_{i},{Y}_{i}\right)=E\left({X}_{i},{Y}_{i}\right)-E\left({X}_{i}\right)E\left({Y}_{i}\right)$
Observe that $P\left({X}_{i}{Y}_{i}=1\right)=P\left({X}_{i}=1,{Y}_{i}=1\right)=0$since it is impossible have two outcomes in a single throw simultaneously. Hence
$Cov\left({X}_{i},{Y}_{i}\right)=-\frac{1}{36}$
which implies
$Cov\left(X,Y\right)=-\frac{n}{36}$
Result: $Cov\left(X,Y\right)=-\frac{n}{36}$

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