Question

Let X₁ , ... X , ... ∼ iid Poisson (θ). We wish to estimate the quantity g(θ) = (θ - 1)². For all values of θ, find a limiting distribution for g (X) = (X - 1 )²

Answer 1

To estimate g(θ) = (θ-1)², we need to find the limiting distribution for g(X) = (X-1)². The expected value of g(X) is (θ² - θ + 1) and the variance is (4θ + 3θ²).

Step-by-step explanation:

To estimate the quantity g(θ) = (θ-1)², we need to find the limiting distribution for g(X) = (X-1)², where X₁, X₂, ... ∼ iid Poisson(θ).

First, let's find the expected value of g(X):

1. Expected value of X is E(X) = θ (from Poisson distribution).
2. Expected value of (X-1)² is E((X-1)²) = E(X² - 2X + 1) = E(X²) - 2E(X) + 1.
3. Using the variance of a Poisson distribution, E(X²) = θ + θ² (from Poisson distribution).
4. Substituting the values, E((X-1)²) = (θ + θ²) - 2θ + 1 = θ² - θ + 1.

Next, let's find the variance of g(X):

1. Variance of X is Var(X) = θ (from Poisson distribution).
2. Variance of (X-1)² is Var((X-1)²) = Var(X² - 2X + 1) = Var(X²) + 4Var(X) + Var(1).
3. Using the variance of a Poisson distribution, Var(X²) = θ + 3θ² (from Poisson distribution).
4. Substituting the values, Var((X-1)²) = (θ + 3θ²) + 4θ + 0 = 4θ + 3θ².

Therefore, the limiting distribution for g(X) = (X-1)² is a Poisson distribution with mean (θ² - θ + 1) and variance (4θ + 3θ²).