Me the variance of the score function in exponential distribution

[rkwack]

please help me! the variance of the score function in exponential distribution

Homework Statement

My question is about 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

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

Homework Equations

As you see, the question provided the variance and expected value for X, but not Theta.

I calculated theta's MLE for the previous queston.. am i supposed to use it somehow?

The Attempt at a Solution

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.

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


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

and 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?

 

[mighty2000]

Hello!

First of all, great job on getting the equation for d ln(theta)/d(theta)! To find E([d ln(theta)/d(theta)]^2), we can use the definition of variance:

var(x) = E[(x - E(x))^2]

Since we know that E(x) = theta, we can rewrite this as:

var(x) = E[(x - theta)^2]

Now, we can use this same idea to find E([d ln(theta)/d(theta)]^2):

E([d ln(theta)/d(theta)]^2) = var(d ln(theta)/d(theta)) + [E(d ln(theta)/d(theta))]^2

We already know that E(d ln(theta)/d(theta)) = 0, so we just need to find the variance. Using the same definition as before, we get:

var(d ln(theta)/d(theta)) = E[(d ln(theta)/d(theta) - E(d ln(theta)/d(theta)))^2]

Simplifying this, we get:

var(d ln(theta)/d(theta)) = E[(d ln(theta)/d(theta))^2]

Now, we can use the equation you found for d ln(theta)/d(theta) to get:

var(d ln(theta)/d(theta)) = E[(-T/theta + 1/(theta)^2 * sigma(t=1 to t=T) Xt)^2]

Expanding this out and using the fact that E(Xt) = theta, we get:

var(d ln(theta)/d(theta)) = E[(T/theta)^2 + 2(-T/theta)(1/(theta)^2 * sigma(t=1 to t=T) Xt) + (1/(theta)^2 * sigma(t=1 to t=T) Xt)^2]

= (T/theta)^2 + (1/(theta)^4) * E[(sigma(t=1 to t=T) Xt)^2]

= (T/theta)^2 + (1/(theta)^4) * var(sigma(t=1 to t=T) Xt)

= (T/theta)^2 + (1/(theta)^4) * T * var(Xt)

= (T/theta)^2 + (1/(theta)^4) * T * theta^2

= (T/theta)^2 + T/(theta)^2

= (T/theta)^2 + 1

Therefore, we have:

E