Question

Let Y1 , Y2 , . . . , Yn denote a random sample from the uniform

distribution over the interval [0, θ]. Show that Y(n) = max(Y1, Y2,
. . . , Yn) is sufficient for θ. Find an MVUE of θ.

Answer 1

The random variable Y(n) = max(Y1, Y2, ..., Yn) is sufficient for θ, and Y(n) itself is an MVUE of θ.

Step-by-step explanation:

In this question, we are given a random sample represented by random variables Y1, Y2, ..., Yn from the uniform distribution over the interval [0, θ]. We need to show that the random variable Y(n) = max(Y1, Y2, ..., Yn) is sufficient for θ.

To show that Y(n) is sufficient for θ, we can use the Factorization Theorem. According to the theorem, Y(n) is sufficient if and only if the joint distribution of Y(n) and the sample Y1, Y2, ..., Y(n-1) can be factorized into a product of a function of Y(n) alone and a function of Y1, Y2, ..., Y(n-1) alone.

For the uniform distribution, the joint distribution of Y(n) and Y1, Y2, ..., Y(n-1) is given by:

f(y(n), y1, y2, ..., y(n-1)) = f(y(n)) * f(y1) * f(y2) * ... * f(y(n-1))

Since all the Y's are independent and uniformly distributed, the individual densities f(y) cancel out:

f(y(n), y1, y2, ..., y(n-1)) = 1/θ^n * 1/θ * 1/θ * ... * 1/θ = 1/θ^(n+1)

Now, we can see that the joint distribution can be factorized into a function of Y(n) alone (1/θ) and a function of Y1, Y2, ..., Y(n-1) alone (1/θ^n). Therefore, Y(n) is sufficient for θ.

To find a Minimum Variance Unbiased Estimator (MVUE) of θ, we can use the method of moments. Since Y(n) is the maximum of the sample, it's equal to our estimate of θ. Therefore, Y(n) is an MVUE of θ.