Question

Let Wr = sum i=1 to m (Ri), Ws = sum i=1 to n (Si). Then, under the assumption that the X's and Y's have the same continuous distribution function, prove that E(Wr) = m(N+1)/2, E(Wx) = n(N+1)/2.

a. E(Wr) = m(N+1)/2, E(Ws) = n(N+1)/2
b. E(Wr) = n(N+1)/2, E(Ws) = m(N+1)/2
c. E(Wr) = mN/2, E(Ws) = nN/2
d. E(Wr) = nN/2, E(Ws) = mN/2

Answer 1

The expected values E(Wr) and E(Wx) reflect the sums of ranks for m and n random variables respectively, assuming a uniform distribution. The expected value for each sum is the average rank multiplied by the number of variables, resulting in E(Wr) = m(N+1)/2 and E(Wx) = n(N+1)/2.

Step-by-step explanation:

The specific expected values to prove, E(Wr) = m(N+1)/2 and E(Wx) = n(N+1)/2, relate to the sums of ranks of random variables, assuming they share the same continuous distribution function. This seems to be a question intersecting rank statistics and the concept of expected value. The correct formula for the expected value E(X) of a discrete random variable X is given as E(X) = μ = Σ xP(x), where each value x of the random variable is multiplied by its probability P(x), and the products are added together. Applying this principle to our sums of ranks, given that the variables have the same distribution, we would expect the average rank in a uniformly distributed set of ranks from 1 to N to be (N+1)/2. For m such variables, we would expect the sum to be m times this average, hence m(N+1)/2. A similar rationale applies to E(Ws) with n variables.