Question

Recall that the binomial distribution with parameters n and p is governed by

P(Y = y) = \binom {n}{y} p^{y} (1-p)^{n-y}.

Let n be some known number, say n=1000. Then the pmf is

f_ p(y) = \binom {1000}{y} p^ y (1-p)^{1000 - y}.

Write this as an exponential family of the form

\displaystyle \displaystyle f_ p(y) = h(y) \exp \left( \eta (p) T(y) - B(p) \right)\qquad \text {where }\, h(y) = \binom {1000}{y},
then enter \eta (p)\, T(y)\, and B(p) below. To get unique answers, use 1 as the coefficient of y in T(y).

\eta (p) =\quad
unanswered

T(y) =\quad
unanswered

B(p) =\quad

Answer 1

To write the binomial distribution as an exponential family, we need to express the probability mass function (pmf) in the form:

f_p(y) = h(y) * exp(η(p) * T(y) - B(p))

Given that n = 1000, the pmf is:

f_p(y) = (1000 choose y) * p^y * (1-p)^(1000-y)

Comparing this to the exponential family form, we can identify the following components:

h(y) = (1000 choose y)

T(y) = y

B(p) = -1000 * log(1-p)

Therefore, the exponential family form is:

f_p(y) = (1000 choose y) * exp(η(p) * y - 1000 * log(1-p))

So, the values are:

η(p) = η(p)

T(y) = y

B(p) = -1000 * log(1-p)