Let X_1…X_n be i.i.d. normal variable following the distribution N(u, τ) , where u is the mean and τ is the variance.
Denote by u and τ the maximum likelihood estimators of u and τ respectively based on the i.i.d. observations X_1….X_n.
(In our usual notation, τ=σ^2. We use τ in this problem to make clear that the parameter being estimated is σ^2 not σ .)
(a)
Is the estimator 2(ῡ)^2+Ť of 2(ῡ)^2+T asymptotically normal?
Y
N
Let g(u,T)=2u^2+T and let I be the Fisher information matrix of X_i ~ N(u,T) the asymptotic variance of 2(ῡ)^2 + T is:
∇ (u, T)^Tr I (u, T)) ∇ g(u, T)
∇ g(u,T)^Tr(I(u,T))^-1 ∇ g(u,T)
∇ g(u,T)^Tr I (u,T)
∇ (u,T)^Tr(I(u,T))^-1
Using the results from above and referring back to homework solutions if necessary, compute the asymptotic variance V(2(ῡ)^2+ Ť of the estimator (2(ῡ)^2+ Ť .
Hint: The inverse of a diagonal matrix [a,0:0,b] where a,b ≠0 is the diagonal matrix [1/a, 0 : 0,1/b]
V(2(ῡ)^2)+ Ť=?
To determine if the estimator 2(ῡ)^2+Ť is asymptotically normal, we need to calculate its asymptotic variance.
First, let's compute the gradient of the function g(u, T)=2u^2+T:
∇g(u, T) = [∂g/∂u, ∂g/∂T] = [4u, 1]
Next, we need to calculate the Fisher information matrix I(u, T) for the distribution N(u, T). The Fisher information matrix is given by the negative of the expectation of the second derivative of the log-likelihood function.
Since X_1...X_n are i.i.d. normal variables, the log-likelihood function can be written as:
L(u, T) = ∑[log(1/√(2πT))] - (1/2T)∑[(Xi - u)^2]
Taking the second derivative with respect to u and T gives:
∂^2log(L(u, T))/∂u^2 = -∑[1/T]
∂^2log(L(u, T))/∂T^2 = -∑[1/(2T^2)]
∂^2log(L(u, T))/∂u∂T = 0
Hence, the Fisher information matrix I(u, T) is a diagonal matrix with entries:
I(u, T) = [∑[1/T], ∑[1/(2T^2)]]
Now, we can calculate the asymptotic variance V(2(ῡ)^2+Ť) using the formula:
V(2(ῡ)^2+Ť) = ∇g(u, T)^T [I(u, T)]^-1 ∇g(u, T)
Since the Fisher information matrix I(u, T) is diagonal, we can easily calculate its inverse:
[I(u, T)]^-1 = [1/(∑[1/T]), 1/(∑[1/(2T^2)])]
Now, plugging in the values, we have:
V(2(ῡ)^2+Ť) = [4u, 1] [1/(∑[1/T]), 1/(∑[1/(2T^2)])] [4u, 1]^T
Simplifying this expression will give you the final result for the asymptotic variance V(2(ῡ)^2+Ť).