- #1

rkwack

- 2

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**please help me! the variance of the score function??**

I have been stuck with this one particular question over this weekend, and

unluckily I could not figure it out by myself. If you can help me a bit with this problem,

I will really really appreciate it.

My question is about an exponential function, with its density function known as

f(x;theta) = (1/theta) e^(-x/theta) for all x>0.

where E(x) = theta, var(x) = theta^2

I set up a likelihood function, and I was able to get the equation for the

d ln(theta) / d(theta).

it was:

d ln(theta)/d(theta) = -T/(theta) + 1/{(theta)^2} sigma(t=1 to t=T) Xt

My question is, what is E( [d ln(theta) / d(theta)}^2]?

I know E(x^2) = var(x) + [e(x)]^2

since the score is zero, then what i am looking for is must be equal to var(x),

which is var(-T/(theta) + 1/{(theta)^2} sigma(t=1 to t=T) Xt)

it has bunch of stuffs inside, and i am really confused. the question didnt provide

any information of var(theta).. i mean i calculated MLE for theta, but i am not sure

if i can plug in this equation.

can someone please help me out with this one question?