Is [(n+1)X(n)/n]^2 an Unbiased Estimator of Theta^2 in Uniform Distribution?

[EvLer]

Homework Statement

show that [(n+1)X(n)/n]² is not an unbiased estimate of theta² in unform distribution U[0,theta]

X(n) is the order statistic

The attempt at a solution

I was going to take E{[(n+1)X(n)/n]²} split it as
E[((n+1)/2)²] * E[X(n)²] and then show that it does not equal to theta².
Could some one please confirm if this is correct? or help me out with the direction?

 

[mighty2000]

Hello there,

Your approach seems to be on the right track. However, there are a few things that need clarification.

Firstly, it would be helpful to define the notation used in the expression. Is n the sample size? Is X(n) the nth order statistic? Is theta the parameter of the uniform distribution?

Assuming that is the case, let's proceed with your approach.

We know that the expected value of the sample mean is an unbiased estimator of the population mean. In this case, we are trying to show that [(n+1)X(n)/n]2 is not an unbiased estimator of theta2.

Using your approach, we can write:

E{[(n+1)X(n)/n]2} = E[((n+1)/n)2] * E[X(n)2]

= (n+1)2/n2 * E[X(n)2]

= (n2+2n+1)/n2 * E[X(n)2]

Now, we need to find the expected value of X(n)2. This can be done by using the formula for the expected value of the nth order statistic of a uniform distribution, which is given by:

E[X(n)2] = (n+1)/(n+2) * theta2

Substituting this in the previous expression, we get:

E{[(n+1)X(n)/n]2} = (n2+2n+1)/n2 * (n+1)/(n+2) * theta2

= (n2+2n+1)(n+1)/(n2(n+2)) * theta2

= (n3+3n2+2n+1)/(n2(n+2)) * theta2

= (n+1)(n+1)/(n+2) * theta2

= [(n+1)/n]2 * theta2

This is not equal to theta2, which means that [(n+1)X(n)/n]2 is not an unbiased estimator of theta2.

Hope this helps! Let me know if you have any further questions.