# Wilcoxon Rank Sum Test – Advanced

Property 1: Suppose sample 1 has size n1 and rank sum R1 and sample 2 has size n2 and rank sum R2, then R1 R2 = n(n+1)/2 where n = n1 n2.

Proof: This is simply a consequence of the fact that the sum of the first n positive integers is $\frac{n(n+1)}{2}$. This can be proven by induction. For n = 1, we see that $\frac{n(n+1)}{2} = \frac{1(1+1)}{2}$ = 1 = n. Assume the result is true for n, then for n + 1 we have,  1 + 2 + … + n + (n+1) = $\frac{n(n+1)}{2}$ + (n + 1) = $\frac{n(n+1)+2(n+1)}{2}$$\frac{(n+1)(n+2)}{2}$

Property 2: When the two samples are sufficiently large (say of size > 10, although some say 20), then the W statistic is approximately normal N(μ, σ) where

Proof: We prove that the mean and variance of W = R1 are as described above. The normal approximation was proven in Mann & Whitney (1947) of Bibliography and we won’t repeat the proof here.

Let xi = the rank of the ith data element in the smaller sample. Thus, under the assumption of the null hypothesis, by Property 1

By Property 4a of Expectation

As we did in the proof of Property 1, we can show by induction on n that

From these it follows that

We can now calculate the following expectations:

Also where i ≠ j

By Property 2 of Expectation (case where i = j)

By Property 3 of Basic Concepts of Correlation when i ≠ j

By an extended version of Property 5 of Basic Concepts of Correlation

### 18 Responses to Wilcoxon Rank Sum Test – Advanced

1. fra says:

2. shark says:

thanks 😀

3. Nnamdi says:

Thanks. Good one

4. Orukpe Lewis says:

I’m grateful for dis better understanding

5. willy kiptoo says:

Thanks alot, you realy put it out on paper

6. Marcel says:

• Charles says:

Marcel,
I hadn’t. What languages did you have in mind?
Charles

• Marcel says:

• Charles says:

Marcel,
Thanks for the offer, but so far people haven’t been asking for translations. In any case, I’ll think about it.
Charles

7. Marcel says:

I have one question about this proof. You calculate the expectation E(rirj) for all j not equal i. I don’t understand why we could take this expectation as equivalent to E(rirj) for all j, i. In the covariance we have to use E(rirj) of all rangs, but you use the expectation for all j not equal i, why is it correct? Can you explain me this problem?

• Charles says:

Marcel,
I show E[ri rj] both where i = j (i.e. var(ri)) and where i is not equal to j.
Charles

• Marcel says:

I am grateful to you for your answer. But I wanted to say that you take the expectation E(rirj) with i not equal to j by the covariance cov(rirj). I don’t understand, why we can do this. I thought we need the expectation of all i and j (also of the double sum i*j).

• Charles says:

Marcel,
Thanks for clarifying things. There are two case: (1) where i = j and (2) where i is not equal to j. In case (1) cov(rirj) = var(ri), which is the described in your print-screen. In case (2) the formula is the one shown in your print-screen. I have just updated the referenced webpage to try to make this a bit clearer. Does it help?
Charles

• Marcel says:

Thank you very much!
I understand it now.

• Charles says:

Marcel,
Good to hear. Glad I could help.
Charles

8. Jack says:

It’s 2Σ i!=j to n1 [-(n+1)/12].I write wrong. There should’t ‘2’ in there i think .

• Charles says:

Jack,
You need to sum all the terms cov(x_i,x_j) where i not equal to j. Note that each such covariance is repeated twice, once for cov(x_i,x_j) and once for cov(x_j,x_i). Thus, if you assume that the sum is where i < j, then you need to double the result. Another way to look at this is to determine how many pairs there are for the indices 1 to n1 where the indices are not equal. The answer is n1(n1-1), which is the value used in the proof. This is the same as 2 times n1(n-1)/2, the later being the number of pairs where the first index is less than the second index. To make this much clearer and more accurate, I have now replaced the lower limit of the summation symbol by i < j (instead of i not equal to j). Thanks for bringing this issue to my attention. Charles

9. Jack says:

oh ，now i understand it！thanks your respond