Me the variance of the score function?

[rkwack]

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?

 

[Jhero]

The variance of the score function is equal to the variance of the likelihood function. Since the likelihood function is a function of theta, the variance of the score function is given by: var(score) = var(-T/(theta) + 1/{(theta)^2} sigma(t=1 to t=T) Xt) = E[(-T/(theta) + 1/{(theta)^2} sigma(t=1 to t=T) Xt)^2] - (E[-T/(theta) + 1/{(theta)^2} sigma(t=1 to t=T) Xt])^2 = E[(-T/(theta) + 1/{(theta)^2} sigma(t=1 to t=T) Xt)^2] - 0 = E[(-T/(theta) + 1/{(theta)^2} sigma(t=1 to t=T) Xt)^2]