 Solution

ID: past-exam-of-the-computer-science-course-of-the-university-of-oxford/2015/probability-and-computing/1/g/solution

X

and

Y

permits factoring the expectation of the product

e^{tX} e^{t Y}

M_{X + Y} (t) = E [e^{t (X + Y)}] = E [e^{tX}] E [e^{t Y}] = (1 - p + p e^{t})^{n} (1 - p + p e^{t})^{m} = (1 - p + p e^{t})^{n + m} .

B (n + m, p)

X + Y \sim B (n + m, p) .

The common success probability

p

and independence are both used in this argument.

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