Let N be a random variable taking values 1, 2, . . . with known probabilities p1, p2, . . ., where Σpi = 1.Having observed N = n, perform n Bernoulli trials with success probability θ, getting X successes.(a) Prove that the pair (X, N ) is minimal sufficient and N is ancillary for θ.(b) Prove that the estimator X/N is unbiased for θ and has variance θ(1 − θ)E(1/N )

(a) To prove that the pair (X, N) is minimal sufficient and N is ancillary for θ, we need to use the Factorization Theorem and Basu's Theorem.

The Factorization Theorem states that a statistic T(X) is sufficient for θ if the joint pdf/pmf of X can be factorized into two functions g(T, θ) and h(X), where g does not contain X and h does not contain θ.

In this case, the joint pmf of X and N can be written as:

P(X=x, N=n) = P(N=n)P(X=x|N=n) = p_n * (n choose x) * θ^x * (1-θ)^(n-x)

This can be factorized into:

g(X, N, θ) = (n choose x) * θ^x * (1-θ)^(n-x) h(X, N) = p_n

Here, g contains θ and h does not, so by the Factorization Theorem, (X, N) is sufficient for θ.

Basu's Theorem states that if T is a complete sufficient statistic and U is a statistic that is independent of T, then U is ancillary.

In this case, N is independent of θ, so by Basu's Theorem, N is ancillary for θ.

(b) To prove that the estimator X/N is unbiased for θ and has variance θ(1 − θ)E(1/N), we need to calculate the expected value and variance of X/N.

The expected value of X/N is:

E(X/N) = E(E(X/N|N)) = E(θ) = θ

So X/N is an unbiased estimator of θ.

The variance of X/N is:

Var(X/N) = E(Var(X/N|N)) + Var(E(X/N|N)) = E(Nθ(1-θ)/N^2) + Var(θ) = θ(1-θ)E(1/N)

So the variance of X/N is θ(1 − θ)E(1/N).

(a) To prove that the pair (X, N) is minimal sufficient and N is ancillary for θ, we need to show that the joint distribution of (X, N) does not depend on θ and that the conditional distribution of X given N = n does not depend on the probabilities p1, p2, ...

The joint distribution of (X, N) is given by the product of the probability mass function of N and the binomial distribution of X given N = n. The probability mass function of N does not depend on θ, and the binomial distribution of X given N = n depends on θ but not on the probabilities p1, p2, ..., so the joint distribution of (X, N) does not depend on θ.

The conditional distribution of X given N = n is a binomial distribution with parameters n and θ, which does not depend on the probabilities p1, p2, ..., so N is ancillary for θ.

(b) To prove that the estimator X/N is unbiased for θ and has variance θ(1 − θ)E(1/N), we need to show that the expected value of X/N is θ and that the variance of X/N is θ(1 − θ)E(1/N).

The expected value of X/N is E[X/N] = E[E[X/N|N]] = E[θ] = θ, so X/N is an unbiased estimator for θ.

The variance of X/N is Var[X/N] = E[Var[X/N|N]] + Var[E[X/N|N]] = E[θ(1 − θ)/N] + Var[θ] = θ(1 − θ)E[1/N] + 0 = θ(1 − θ)E[1/N], so the variance of X/N is θ(1 − θ)E(1/N).