Past exam of the computer science course of the University of Oxford / 2015 / Probability and computing / 1 / g

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- Solution g

Solution

The independence of

X

and

Y

permits factoring the expectation of the product

e^{tX} e^{t Y}

. The moment-generating function of a binomial distribution from the previous part therefore gives

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} .

This is the moment-generating function of

B (n + m, p)

. Both moment-generating functions are finite on a neighbourhood of zero, so the uniqueness theorem for moment-generating functions identifies the probability distribution:

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

The common success probability

p

and independence are both used in this argument.

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