To find the asymptotic variance of the maximum likelihood estimator (MLE) of lambda, we need to use the Fisher Information.
The Fisher Information (I(lambda)) in this case is given by:
I(lambda) = -E(d^2(log(g(X)))/d(lambda)^2)
We can calculate the second derivative of the log-likelihood function:
d^2(log(g(X)))/d(lambda)^2 = -x^2 e^(-lambda x)
Taking the expectation of the second derivative, we have:
E(d^2(log(g(X)))/d(lambda)^2) = -integrate(x^2 e^(-lambda x) g(x) dx, 0, infinity)
= -integrate(x^2 lambda e^(-lambda x) dx, 0, infinity)
= -2/lambda^3
Therefore, the Fisher Information is:
I(lambda) = 2/lambda^3
The asymptotic variance of the MLE of lambda is the inverse of the Fisher Information:
V_{\hat{lambda}^{MLE}} = 1 / I(lambda)
= lambda^3 / 2
So, the asymptotic variance of the MLE of lambda is lambda^3 / 2.
The lifetime (in thousands of hours) X of a light bulb has pdf
g(x)= \lambda e^{-\lambda x}, \hspace{3mm} x\geq 0
for some unknown \lambda >0.
Find the asymptotic variance \displaystyle V_{\hat{\lambda }^{\text {MLE}}} of \hat{\lambda }^{\text {MLE}}.
V_{\hat{\lambda }^{\text {MLE}}}=\quad
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