Covarian of Ordered Statistics

[M.D.G]

Homework Statement

Let X1,...,Xn be independant Uniform(0,1) RVs. Find Cov(Xn,X(n)) when X(n) is the nth ordered statistic. Hint: Find the joint cdf of Xn and X(n)

Homework Equations

Cov(X,Y) = E[XY] - E[X]E[Y]

The Attempt at a Solution

I know that E[Xn] is 1/2 and that the E[X(n)] is n/(n+1). I have been trying to figure out how to find the joint cdf of Xn and X(n) but I am having a lot of troubles, also I know that the joint pdf for the two does not exist since we showed this in a previous homework.

Any hints would be much appreciated.

 

[mighty2000]

First, let's start by finding the joint cdf of Xn and X(n). We know that for independent random variables, the joint cdf is simply the product of the individual cdfs. So, we have:

F(Xn,X(n)) = F(Xn)F(X(n))

Now, let's recall the cdf of the uniform distribution on (0,1):

F(X) = P(X≤x) = x, for 0<x<1

Using this, we can find the cdf of Xn and X(n):

F(Xn) = P(Xn≤x) = x, for 0<x<1

F(X(n)) = P(X(n)≤x) = x^n, for 0<x<1

Therefore, the joint cdf of Xn and X(n) is:

F(Xn,X(n)) = x*x^n = x^(n+1), for 0<x<1

Now, we can use this joint cdf to find the covariance:

Cov(Xn,X(n)) = E[XnX(n)] - E[Xn]E[X(n)]

To find E[XnX(n)], we can use the definition of expectation:

E[XnX(n)] = ∫∫ xnx(n)*f(xn,x(n)) dxn dx(n)

Where f(xn,x(n)) is the joint pdf of Xn and X(n), which we know does not exist. However, we can still find the integral by using the joint cdf we just found:

E[XnX(n)] = ∫∫ xnx(n)*dF(xn,x(n))

= ∫∫ xnx(n)*x^(n+1) dxn dx(n)

= ∫∫ x^(n+1)*x^(n+1) dxn dx(n)

= ∫∫ x^(2n+2) dxn dx(n)

= ∫ x^(2n+2) dxn * ∫ x^(2n+2) dx(n)

= 1/(2n+3) * 1/(2n+3)

= 1/(2n+3)^2

Similarly, we can find E[Xn] and E[X(n)] by integrating over their respective marginal cdfs:

E[Xn] = ∫∫ xn*f(xn,x(n)) dxn dx(n