N(θ,θ): MLE for a Normal where mean=variance is
a) (sqrt(4*bar(X_n^2)+1)-1)/2
and In this problem, we apply the Central Limit Theorem and the 1-dimensional Delta Method.
First, compute the limit and asymptotic variance of V(X_n^2)
The limit to which (X_n^2) converges in probability, also known as its P -limit , is
b)
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The asymptotic variance V(X_n^2) of (X_n^2), which is equal to Var(X_1^2), is
c)
Now, write theta^ as the function of (X_n^2) you found in part (a),
theta^=g((X_n^2))
and give its first derivative,g'(x) ,
d)
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What can you conclude about the asymptotic variance V(theta^) of theta^?
V(theta^)=?
To find the MLE for a Normal distribution where the mean is equal to the variance, denoted as N(θ,θ), we first need to compute the limit and asymptotic variance of (X_n^2). Let's go through the steps:
Step 1: Compute the limit of (X_n^2):
To find the limit, we apply the Central Limit Theorem, which states that the sample mean converges to a normal distribution as the sample size increases. In this case, we consider the sample mean of (X_n^2), which converges to the population mean squared.
Let's denote the limit as L:
b) L = lim(n→∞) E(X_n^2) = θ^2
Step 2: Compute the asymptotic variance of (X_n^2):
The asymptotic variance of (X_n^2) can be found using the 1-dimensional Delta Method. It is equal to the population variance of X_1^2.
Let's denote the asymptotic variance as V(X_n^2):
c) V(X_n^2) = Var(X_1^2) = 2θ^4
Step 3: Write θ-hat as a function of (X_n^2) found in part (a):
To find the MLE θ-hat, we need to express it as a function of (X_n^2) found in part (a):
theta^ = g((X_n^2)) = (sqrt(4*bar(X_n^2)+1)-1)/2
Step 4: Find the first derivative of g(x):
To find the first derivative of g(x), we differentiate the function with respect to x:
d) g'(x) = (2x + 1)^(-1/2)
Finally, we need to determine the asymptotic variance V(theta^) of theta^:
To find V(theta^), we need to use the Delta Method. The formula for V(theta^) is:
V(theta^) = [g'(E(X_n^2))]^2 * V(X_n^2)
Let's substitute the values we computed earlier:
V(theta^) = [g'(θ^2)]^2 * V(X_n^2)
= [g'(θ^2)]^2 * 2θ^4
= [(2θ^2 + 1)^(-1/2)]^2 * 2θ^4
= (2θ^2 + 1)^(-1) * 2θ^4
Hence, the asymptotic variance V(theta^) of theta^ is given by:
V(theta^) = (2θ^2 + 1)^(-1) * 2θ^4