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
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)
To rewrite the binomial distribution as an exponential family, we need to express the probability mass function in the form:
f_p(y) = h(y) * exp(η(p) * T(y) - B(p))
Given that h(y) is the binomial coefficient C(1000, y), we have:
h(y) = C(1000, y)
To find η(p), T(y), and B(p), let's expand the binomial coefficient and observe the pattern of powers of p and (1-p):
C(1000, y) = 1000! / (y! * (1000 - y)!)
= (1000 * 999 * 998 * ... * (1000 - y + 1)) / y!
= [1000 / y] * [999 / (y - 1)] * [998 / (y - 2)] * ... * [1000 - y + 1] * (1 / y!)
Now, let's express this in terms of exp(η(p) * T(y) - B(p)):
η(p) will be the natural log of the term multiplying the powers of p:
η(p) = ln([1000 / y] * [999 / (y - 1)] * [998 / (y - 2)] * ... * [1000 - y + 1])
T(y) will be y itself (since the coefficient of y in T(y) should be 1):
T(y) = y
B(p) will be the sum of the natural logs of the terms multiplying (1 - p):
B(p) = ln([1000 / y] * [999 / (y - 1)] * [998 / (y - 2)] * ... * [1000 - y + 1])
Now we can fill in the values:
η(p) = ln([1000 / y] * [999 / (y - 1)] * [998 / (y - 2)] * ... * [1000 - y + 1])
T(y) = y
B(p) = ln([1000 / y] * [999 / (y - 1)] * [998 / (y - 2)] * ... * [1000 - y + 1])