Question

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

Answer 1

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.