Let X1,X2,…∼i.i.d.X . The distribution of X depends on a positive parameter θ , which is a function of the mean μ , i.e θ=g(μ) . You estimate θ by the estimator θˆ=g(X¯¯¯¯n) .
For which function g can the delta method be applied? Remember that θ>0 .
(Choose all that apply.)
1. g(x)=x^3
2.g(x)=sqrt(x)
3.g(x)=ln(x)
4.g(x)={x if x≤1, 2x−1 if x>1
5.g(x)=1/(x−1)
The delta method is a mathematical tool used for approximating the distribution of a function of a random variable. It is applicable when the function is differentiable and when the random variable converges in distribution to a specific distribution.
To determine which function 'g' can the delta method be applied to, we need to check if the function satisfies the conditions for the delta method:
1. g(x) = x^3
The function g(x) = x^3 is differentiable for all positive values of x. Hence, the delta method can be applied.
2. g(x) = sqrt(x)
The function g(x) = sqrt(x) is differentiable for positive values of x. Hence, the delta method can be applied.
3. g(x) = ln(x)
The function g(x) = ln(x) is differentiable for positive values of x. Hence, the delta method can be applied.
4. g(x) = {x if x ≤ 1, 2x - 1 if x > 1}
The function g(x) is not differentiable at x = 1 since it has a jump discontinuity. Therefore, the delta method cannot be applied.
5. g(x) = 1/(x - 1)
The function g(x) = 1/(x - 1) is not differentiable at x = 1 since it has a vertical asymptote. Therefore, the delta method cannot be applied.
In summary, the functions g(x) = x^3, g(x) = sqrt(x), and g(x) = ln(x) satisfy the conditions for the delta method and can be applied.