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 Incoming links: Moment-generating function of a binomial distribution

Past exam of the computer science course of the University of Oxford / 2015 / Probability and computing / 1 / f / Solution 2026-10-31

For

X

with binomial distribution

B (n, p)

, where

n

is a nonnegative integer and

0 \leq p \leq 1

, use its probability mass function and the binomial theorem:

M_{X} (t) = E [e^{tX}] = k = 0 \sum n e^{t k} (k n) p^{k} (1 - p)^{n - k} = k = 0 \sum n (k n) (p e^{t})^{k} (1 - p)^{n - k} = (1 - p + p e^{t})^{n}, t \in R .

This moment-generating function of a binomial distribution is finite for every real

t

, since the random variable has finite support. The formula also includes the degenerate cases

p = 0

,

p = 1

and

n = 0

.

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Past exam of the computer science course of the University of Oxford / 2015 / Probability and computing / 1 / g / Solution 2026-10-31

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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