jernplate8

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).

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

$Cov(X,Y)=Cov(X=\sum _{i=1}^{n}{X}_{i},Y=\sum _{j=1}^{n}{Y}_{j})=\sum _{i}jCov({X}_{i},{Y}_{j})$

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

$Cov({X}_{i},{Y}_{j})=0$

So

$Cov(X,Y)=\sum _{i}Cov({X}_{i},{Y}_{i})$

We have that

$Cov({X}_{i},{Y}_{i})=E({X}_{i},{Y}_{i})-E\left({X}_{i}\right)E\left({Y}_{i}\right)$

Observe that$P({X}_{i}{Y}_{i}=1)=P({X}_{i}=1,{Y}_{i}=1)=0$ since it is impossible have two outcomes in a single throw simultaneously. Hence

$Cov({X}_{i},{Y}_{i})=-\frac{1}{36}$

which implies

$Cov(X,Y)=-\frac{n}{36}$

Result:$Cov(X,Y)=-\frac{n}{36}$

Similary, define

Using basic properties of the covariance, we have that

Observe that for

Knowing what happened in ith throw does not change probabilities for jth throw.

Hence, in that case

So

We have that

Observe that

which implies

Result:

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