Statistics, Delta Method, Central Limit Theorem

Question

Let {X_n} be a sequence of random variables such that \sqrt{n}(X_n-\theta) converges in distribution to N(0,\sigma^2). The function g(x) is differentiable at theta and we have g'(\theta)\neq 0, while the function h(x) is differentiable at theta and we have h'(\theta)= 0 and h''(\theta)\neq 0.

A. 2n(h(X_n)-h(\theta))/(\sigma^2 h''(\theta))converges in distribution to \chi^2(1).

B. 2n(h(X_n)-h(\theta))converges in distribution to.\sigma^2 h''(\theta)\chi^2(1).

Are A and B respectively, true or false?

Answer 1

The second-order delta method applies when you have a sequence of random variables {X_n}, and the sequence satisfies √(n) (X_n - θ) ---d---> N(0, σ²).

For a given function h and specific value of θ, suppose h'(θ) = 0 and h"(θ) ≠ 0 exists.

Then n(h(X_n) - h(θ)) ---d---> σ² (h"(θ) / 2) χ²(1).

Algebraic manipulation shows:

2n (h(X_n) - h(θ)) ---d---> σ² h"(θ) χ²(1), which shows B is true.

Further manipulation shows:

2n (h(X_n) - h(θ)) / (σ²h"(θ)) ---d---> χ²(1), which shows A is true as long as h"(θ) ≠ 0 (which it is, since it was given).

A and B are true.