 Solution

ID: past-exam-of-the-year-1-of-the-computer-science-course-of-the-university-of-oxford/2022/discrete-mathematics-continuous-mathematics-and-probability-resit/7/b/iii/solution

E [X] = p

and

E [Y] = p

. In the joint distribution,

X Y

is nonzero only at

(1, 1)

: when

X = 1

, the only possible values of

Y

are

0

and

1

. Hence

E [X Y] = p e^{- p}, Cov (X, Y) = p e^{- p} - p^{2} = p (e^{- p} - p) .

These quantities are finite; for example,

0 ⩽ X Y ⩽ 1

. The covariance changes sign when

p = e^{- p}

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