Get the solution to "Let X be the number of 1’s and..."

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).

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

C o v (X, Y) = C o v (X = \sum_{i = 1}^{n} X_{i}, Y = \sum_{j = 1}^{n} Y_{j}) = \sum_{i} j C o v (X_{i}, Y_{j})

Observe that for

i \neq 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

C o v (X_{i}, Y_{j}) = 0

C o v (X, Y) = \sum_{i} C o v (X_{i}, Y_{i})

We have that

C o v (X_{i}, Y_{i}) = E (X_{i}, Y_{i}) - E (X_{i}) E (Y_{i})

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

C o v (X_{i}, Y_{i}) = - \frac{1}{36}

which implies

C o v (X, Y) = - \frac{n}{36}

Result:

C o v (X, Y) = - \frac{n}{36}

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