Expected value of Ymin from an exponential

[rhyno89]

Homework Statement

So I have to prove that Ymin is an unbiased estimator for lambda from the distribution:

(1/lambda)e^(-x/lambda)

Homework Equations

The Attempt at a Solution

I kno to show that the estimator is unbiased requires that its expected value to equal the given parameter.

For Ymin I got: {n(1+e^(-x/lambda))}*((1/lambda)e^(-x/lambda)

I know that for taking the expected value of this is multiplying by x and then integrating but this seems WAY too complicated based on what we have done so far.

Any help would be great!

 

[mighty2000]

Hello,

To prove that Ymin is an unbiased estimator for lambda, we first need to understand what an unbiased estimator is. An unbiased estimator is a statistic that, on average, gives an estimate of the true parameter that is equal to the true parameter. In other words, the expected value of the estimator is equal to the true parameter.

In this case, we have the following estimator for lambda: Ymin = n(1+e^(-x/lambda))*(1/lambda)e^(-x/lambda). We want to show that the expected value of this estimator is equal to lambda.

To find the expected value, we need to multiply the estimator by x and then integrate over all possible values of x. This may seem complicated, but we can simplify it by using the fact that the expected value of a sum is equal to the sum of the expected values. In other words, we can break up the integral into two parts: one for n(1+e^(-x/lambda)) and one for (1/lambda)e^(-x/lambda).

Let's start with the first part: n(1+e^(-x/lambda)). We can rewrite this as n + n*e^(-x/lambda). The expected value of n is just n, since it is a constant. For the second term, we can use the fact that the expected value of e^(-x/lambda) is equal to 1/lambda. This is a well-known property of the exponential distribution. Therefore, the expected value of n(1+e^(-x/lambda)) is equal to n + n*(1/lambda) = n(1+1/lambda).

Now, let's look at the second part: (1/lambda)e^(-x/lambda). We can rewrite this as (1/lambda)*e^(-x/lambda). Again, we can use the fact that the expected value of e^(-x/lambda) is equal to 1/lambda. Therefore, the expected value of (1/lambda)e^(-x/lambda) is equal to (1/lambda)*(1/lambda) = 1/lambda^2.

Now, we can put these two parts together to find the expected value of Ymin:

E(Ymin) = E(n(1+e^(-x/lambda))*(1/lambda)e^(-x/lambda))