# Me the variance of the score function in exponential distribution

• rkwack
In summary: E(d ln(theta)/d(theta))]^2= (T/theta)^2 + 1 + 0= (T/theta)^2 + 1In summary, the variance of the score function in an exponential distribution is equal to (T/theta)^2 + 1.
rkwack

## 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.

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

## 1. What is the exponential distribution?

The exponential distribution is a probability distribution that models the time between events in a Poisson process. It is often used to represent the time between occurrences of rare events, such as natural disasters or equipment failures.

## 2. How is the variance of the score function calculated in exponential distribution?

The variance of the score function in exponential distribution is equal to the reciprocal of the sample size. It can also be calculated by taking the square of the mean score function.

## 3. What is the score function in exponential distribution?

The score function in exponential distribution is a measure of how well a statistical model fits the data. It is calculated by taking the log of the likelihood function and differentiating it with respect to the model parameters.

## 4. How is exponential distribution used in real life?

The exponential distribution is commonly used in various fields such as finance, engineering, and medicine to model the time between events. It is also used in survival analysis to estimate the time until an event occurs.

## 5. What are some properties of the exponential distribution?

The exponential distribution has a constant failure rate, meaning that the probability of an event occurring remains the same regardless of how much time has passed. It is also a continuous distribution with a skewed right shape and takes only positive values.

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