Mann-Whitney(-Wilcoxon Rank Sum) Distribution

Let's say we have a sample size of 4 in our first category and 3 in the second, i.e. \(n_1 = 4, n_2=3\). We also have a U-statistic of 5.

The U value is the sum of the ranks, minus the minimum possible sum. With 4 cases out of 4+3=7, the minimum would be the sum of the four lowest ranks, i.e. 1+2+3+4 = 10. If we add this to the requested U value of 5, we were looking for the number of ways to get a sum of 15, with four ranks.

These could be 1+2+5+7, 1+3+4+7, 1+3+5+6, 2+3+4+6, so only four options. We could simply each time try to find all the possible combinations, but the recursive function will do the same.

Let's go over all the steps. We begin with filling out our values

\(f_{4, 3}(5) = f_{4, 3-1}(5) + f_{4-1, 3}(5 - 3)= f_{4, 2}(5) + f_{3, 3}(2)\)

Lets focus on the last term each time. First \(f_{3, 3}(2)\), we can use the formula again:

\(f_{3, 3}(2) = f_{3, 3-1}(2) + f_{3-1, 3}(2 - 3)= f_{3, 2}(2) + f_{2, 3}(-1)\)

For \(f_{2, 3}(-1)\) we can use the first case since U < 0, we get

\(f_{2, 3}(-1) = 0\)

Plug this in \(f_{3, 3}(2)\) to get:

\(f_{3, 3}(2) = f_{3, 2}(2) + 0 = f_{3, 2}(2)\)

for \(f_{3, 2}(2)\) we use the formula again:

\(f_{3, 2}(2) = f_{3, 2-1}(2) + f_{3-1, 2}(2 - 2)= f_{3, 1}(2) + f_{2, 2}(0)\)

Focus on the last term: \(f_{2, 2}(2)\), we can use the formula again:

\(f_{2, 2}(0) = f_{2, 2-1}(0) + f_{2-1, 2}(0 - 2)= f_{2, 1}(0) + f_{1, 2}(-2)\)

The \(f_{2, 1}(0)\) has a 1, so will be 1, and since \(f_{1, 2}(-2)\) has U < 0, it will be 0. So we can simplify:

\(f_{2, 2}(0) = f_{2, 1}(0) + f_{1, 2}(-2)\) = 1 + 0 = 1

Plug this back in \(f_{3, 2}(2)\) to get:

\(f_{3, 2}(2) = f_{3, 1}(2) + f_{2, 2}(0)= f_{3, 1}(2) + 1\)

But \(f_{3, 1}(2)\) also has a 1, so will be 1:

\(f_{3, 2}(2) = f_{3, 1}(2) + 1= 1 + 1=2\)

Since we established that \(f_{3, 3}(2) = f_{3, 2}(2)\) we also now have:

\(f_{3, 3}(2) = f_{3, 2}(2) = 2\)

Which we can now plug back in \(f_{4, 3}(5)\):

\(f_{4, 3}(5) = f_{4, 2}(5) + f_{3, 3}(2) = f_{4, 2}(5) + 2\)

We turn our focus on the remaining term \(f_{4, 2}(5)\):

\(f_{4, 2}(5) = f_{4, 2-1}(5) + f_{4-1, 2}(5 - 2)= f_{4, 1}(5) + f_{3, 2}(3)\)

Focus on the last term: \(f_{3, 2}(3)\), we can use the formula again:

\(f_{3, 2}(3) = f_{3, 2-1}(3) + f_{3-1, 2}(3 - 2)= f_{3, 1}(3) + f_{2, 2}(1)\)

Focus on the last term: \(f_{2, 2}(1)\), we can use the formula again:

\(f_{2, 2}(1) = f_{2, 2-1}(1) + f_{2-1, 2}(1 - 2)= f_{2, 1}(1) + f_{1, 2}(-1)\)

The \(f_{2, 1}(1)\) has a 1, so will be 1, and the \(f_{1, 2}(-1)\) has a U < 0, so will be 0. We can therefor simplify:

\(f_{2, 2}(1) = f_{2, 1}(1) + f_{1, 2}(-1)= 1 + 0 = 1\)

Plug this in \(f_{3, 2}(3)\):

\(f_{3, 2}(3) = f_{3, 1}(3) + f_{2, 2}(1)= f_{3, 1}(3) + 1\)

The \(f_{3, 1}(3)\) has a 1, so will be 1. We can therefor simplify:

\(f_{3, 2}(3) = f_{3, 1}(3) + 1= 1 + 1=2\)

Plug this in \(f_{4, 2}(5)\):

\(f_{4, 2}(5) = f_{4, 1}(5) + f_{3, 2}(3)= f_{4, 1}(5) + 2\)

The last one we need is that \(f_{4, 1}(5)\), but \(5 > 4\times 1\), so this will be 0:

\(f_{4, 1}(5) = 0\)

plugging this into \(f_{4, 2}(5)\)

\(f_{4, 2}(5) = f_{4, 1}(5) + 2 = 0 + 2 = 2\)

Finally we can complete \(f_{4, 3}(5)\):

\(f_{4, 3}(5) = f_{4, 2}(5) + 2 = 2 + 2 = 0\)

So the formula indeed shows there are 4 ways to get a U-value of 5, if we have 4 cases in the first category and 3 in the second.

The total number of combinations, is given by the binomial coefficient:

\(C(n, n_1) = nCr(n, n_1) = \binom{n}{n_1} = \frac{n!}{n_1!\times\left(n - n_1\right)!}\)

with \(n=n_1+n_2\).

In our example we get:

\(C(7, 4) = nCr(7, 4) = \binom{7}{4} = \frac{7!}{4!\times\left(7 - 4\right)!}= \frac{7!}{4!\times\left(3\right)!} = \frac{7\times 6\times 5\times 4\times 3\times 2\times 1}{4\times 3\times 2\times 1\times\left(3\times 2\times 1\right)} = \frac{7\times 6\times 5}{3\times 2\times 1} = \frac{210}{6} = 35\)

So the probability of \(f_{4, 3}(5)\) is:

\(\frac{f_{4, 3}(5)}{35}=\frac{4}{35}\)

The distribution deals with ranks. So the scores are combined, then ranked, then the rank for each category is placed back. For example if we had three scores 1, 2, 7 in the 'national' category, and two scores of 5 and 9 in the 'international' category, we combine each of them, we have the following:

The ranks for 'national' are then 1, 2, and 4. The ranks for international are 3, and 5.

This gives a sum of ranks of 1+2+4 = 7 for the national companies, and 3+5=8 for the international ones.

The smallest possible sum of ranks of \(i\) scores would be:

\(R_{min} = 1 + 2 + ...+ i = \sum_{k=1}^i k = \frac{i\times\left(i+1\right)}{2}\)

If we set \(n = i + j\), i.e. the total number of cases, **the maximum sum of ranks** for i cases would be:

\(R_{max} = n + (n - 1) + (n - 2) + ... + (n - i + 1) = \sum_{k=0}^{i-1} \left(n - k\right) = \frac{\left(2\times n - i + 1\right)i}{2}\)

In the most simple case we have only one score in each of the two categories. Combining all scores in one 'large' pool then gives the possible ranks in this pool as 1 and 2 only. Picking one of these means each has a count (frequency) of 1.

Let \(R\) be the sum of ranks, and \(f_{i,j}(R)\) the frequency of a sum of ranks \(R\) if there are \(i\) scores in the first category, and \(j\) in the second.

Our first found option is then simple:

If we have a sample of 1 in the first category and 2 in the second we have as options:

If we have a sample of 1 in the first category and 3 in the second, our total sample size would be 4.

If we combine them, our ranks could be 1, 2, 3, or 4. Our single item in the 1st category can be any of these, so there is only one way to have a sum of ranks (\(R\)) value of 1, 2, 3, or 4 in this situation:

In general we could say:

\(f_{1,j}(R) = \begin{cases} 1 & \text{ if } R_{min} \leq R \leq R_{max}\\ 0 & \text{ if } R < R_{min} \text{ or } R > R_{max} \end{cases}\)

But \(R_{min}\) will of course be 1 in the case of \(f_{1,j}\), and \(R_{max}\) is simply \(j + 1\):

\(f_{1,j}(R) = \begin{cases} 1 & \text{ if } 1 \leq R \leq j+1\\ 0 & \text{ if } R < 1 \text{ or } R > j+1 \end{cases} \)

If we switch things, and look at \(f_{2,1}(R)\), we can have as possible combinations:

• 12, with a sum of 3
• 13, with a sum of 4
• 23, with a sum of 5

So we get:

For \(f_{3,1}(R)\) we get the same combinations as \(f_{2,1}(R)\) but now need to add one more, so:

• 12 can become 123, 124, with rank sums 6 and 7 resp
• 13 can become 134, with rank sum 8
• 23 can become 234, with rank sum 9

Again the same as \(f_{1,3}\), but shifted to the right.

In general if we shift \(f_{i,1}(R)\) to the left we get the same as \(f_{1,i}\). In general we can therefor conclude:

\(f_{i,1}(R) = f_{1,i}(R - R_{min} + 1)\)

Lets see what happens if we have 2 in the first category and 2 in the second, i.e. \(f_{2,2}(R)\).

With \(f_{2,1}(R)\), we had:

• 12, with a sum of 3
• 13, with a sum of 4
• 23, with a sum of 5

But now because the additional case in the second category we also get:

• 14, with a sum of 5
• 24, with a sum of 6
• 34, with a sum of 7

Note that the distribution of the additional cases is the same as \(f_{1,2}\), but shifted the \(R\).

So we get:

Where the last row is the sum of the previous two.

Lets add another case to the second category, i.e. \(f_{2,3}\).

Again, we have the same options as with \(f_{2,2}\):

• 12, with a sum of 3
• 13, with a sum of 4
• 23, with a sum of 5
• 14, with a sum of 5
• 24, with a sum of 6
• 34, with a sum of 7

But now also add:

• 15, with a sum of 6
• 25, with a sum of 7
• 35, with a sum of 8
• 45, with a sum of 9

So we add another two rows to our table:

Again the last row is the sum of the previous two.

In general the following pattern seems to emerge:

\(f_{2, j}(R) = f_{2, j-1}(R) + f_{1, j}(R -j - 2)\)

The \(f_{2, j-1}(R)\) will eventually boil down to \(f_{2, 1}(R)\), for which we can use our generic \(f_{i,1}\) formula.

Lets add a third case to the first category and look at \(f_{3,2}\). The options of \(f_{2,2}\) were:

• 12, with a sum of 3
• 13, with a sum of 4
• 23, with a sum of 5
• 14, with a sum of 5
• 24, with a sum of 6
• 34, with a sum of 7

We can add a new one now to each:

• 123, 124, 125, 134, 135, with sums 6, 7, 8, 8, 9
• 234, 235, 245, with sums 9, 10, 11
• 345, with sum 12

But this is the same as we saw for \(f_{2, 3}(R)\) just shifted to a higher starting point.

\(f_{3, 2}(R) = f_{2, 3}(R - 3)\)

Lets look at \(f_{4,2}\), we get as options:

• 1234, 1235, 1236, 1245, 1246, 1256, with sums 10, 11, 12, 12, 13, 14
• 1345, 1346, 1356, with sums 13, 14, 15
• 1456, with sum 16
• 2345, 2346, 2356, with sums 14, 15, 16
• 2456, with sum 17
• 3456, with sum 18

Again, the same as \(f_{2,4}\) but shifted

So we also have:

\(f_{3, 2}(R) = f_{2, 3}(R - 3)\)

\(f_{4, 2}(R) = f_{2, 4}(R - 7)\)

The shift is the difference in minimum, or difference in maximum, i.e. \(R_{max_j} - R_{max_i} = R_{min_j} - R_{min_i}\)

This could be as one equation:

\(\Delta R = \frac{(j-i)(j+i+1)}{2}\)

So in general:

\(f_{i,j}(R) = f_{j,i}(R - \Delta R)\)

Now lets focus on \(f_{3,3}\).

The \(f_{3,2}\) had as options:

• 123, 124, 125, 134, 135, with sums 6, 7, 8, 8, 9
• 234, 235, 245, with sums 9, 10, 11
• 345, with sum 12

These are of course also included in \(f_{3,3}\). The new additional ones would be:

• 126, 136, 146, 156, with sums 9, 10, 11, 12
• 236, 246, 256, with sums 11, 12, 13
• 346, 356, with sums 13, 14
• 456 with sum 15

Now let's have a look at the options from \(f_{2,3}\), these were:

• 12, 13, 14, 15
• 23, 24, 25
• 34, 35
• 45

Note how these are the same as the ones we need to add, but just missing the 6 at the end.

\(f_{3,3}(R) = f_{3,2}(R) + f_{2,3}(R - 6)\)

How about \(f_{4,3}\). Of course the same options as \(f_{4,2}\):

• 1234, 1235, 1236, 1245, 1246, 1256, with sums 10, 11, 12, 12, 13, 14
• 1345, 1346, 1356, with sums 13, 14, 15
• 1456, with sum 16
• 2345, 2346, 2356, with sums 14, 15, 16
• 2456, with sum 17
• 3456, with sum 18

Now let's look at \(f_{3, 3}\):

• 123, 124, 125, 126, with sums 6, 7, 8, 9
• 134, 135, 136, with sums 8, 9, 10
• 145, 146, with sums 10, 11
• 156, with sum 12
• 234, 235, 236, with sums 9, 10, 11
• 245, 246, with sums 11, 12
• 256, with sum 13
• 345, 346, with sums 12, 13
• 356, with sum 14
• 456, with sum 15

We need to shift this with 7, so we get:

The general pattern that seems to emerge is:

\(f_{i,j}(R) = f_{i, j-1}(R) + f_{i-1, j}(R - (i+j))\)

Instead of using the sum of the ranks, it is more common to use the Mann-Whitney-Wilcoxon U statistic. This is defined as:

\(U = R - R_{min} = R - \frac{i\times\left(i + 1\right)}{2}\)

Let's start by looking at the conversion for \(f_{1,j}(R)\), which had as formula:

\(f_{1,j}(R) = \begin{cases} 1 & \text{ if } 1 \leq R \leq j+1\\ 0 & \text{ if } R < 1 \text{ or } R > j+1 \end{cases}\)

Let's rewrite our \(f_{1,j}(R)\) by for \(U\). We need to subtract \(R_{min}\) everywhere. With \(f_{1,j}(R)\) the \(R_{min} = 1\), so we subtract 1 everywhere:

\(f_{1,j}(U) = \begin{cases} 1 & \text{ if } 0 \leq U \leq j \\ 0 & \text{ if } U < 0 \text{ or } U > j \end{cases}\)

Next the: \(f_{i,1}(R) = f_{1,i}(R - R_{min} + 1)\)

We already concluded earlier that the distribution of \(f_{i,1}(R)\) is just the distribution of f_{1,i}(R) but then shifted. Since with \(U\) we want to always start at 0, we can simply use:

\(f_{i,1}(U) = f_{1,i}(U)\)

Now for the 'big one':

\(f_{i,j}(R) = f_{i,j}(U + R_{min}) = f_{i, j-1}(U + R_{min}) + f_{i-1, j}(U + R_{min} - (i+j))\)

If we focus on \(f_{i-1, j}(U + R_{min} - (i+j))\), we can rewrite:

\(R_{min} - (i+j) = \frac{i\times\left(i + 1\right)}{2} - (i + j) = \frac{i\times\left(i + 1\right)}{2} - i - j = \frac{i\times\left(i + 1\right) - 2i}{2} - j = \frac{i^2 + i - 2i}{2} - j = \frac{i^2 - i}{2} - j = \)

So we have:

\(f_{i-1, j}\left(R - (i+j)\right) = f_{i-1, j}\left(U + \frac{i^2 - i}{2} - j\right)\)

To obtain this using \(U\), we can subtract \(R_{min}\) as we did before:

\(R_{min} = \frac{i\times\left(i+1\right)}{2}\)

However, for this part, it would be for \(i - 1\), so we get:

\(R_{min} = \frac{(i - 1)\times\left((i - 1)+1\right)}{2} = \frac{(i - 1)\times i}{2}= \frac{i^2 - i}{2}\)

So we actually get:

\(f_{i-1, j}\left(U + \frac{i^2 - i}{2} - j - \frac{i^2 - i}{2}\right) = f_{i-1, j}\left(U - j\right)\)

So finally, we can write:

\(f_{i,j}(U) = f_{i, j-1}(U) + f_{i-1, j}(U - j)\)

A final observation would be that both \(f_{i,1}(U)\) and \(f_{1,j}(U)\) will return a 1 if \(U\) is in the range of possible \(U\) values and otherwise 0. Actually any \(U\) outside this range will be impossible so will have a frequency of 0.

The maximum sum of ranks was:

\(R_{max} = \frac{\left(2\times n - i + 1\right)i}{2}\)

Therefor:

\(U_{max} = R_{max} - R_{min} = \frac{\left(2\times n - i + \right)i}{2} - \frac{i\times(i+1)}{2} = \frac{i}{2}\left(2\times n - i + 1 - (i+1)\right)= \frac{i}{2}\left(2n - 2i\right) = \frac{2i\left(n - i\right)}{2}= i\left(n - i\right)\)

And since \(n = i + j\) we also have \(j = n - i\):

\(U_{max} = i\times j\)

The distribution could therefor be written as:

\(f_{i,j}(U) = \begin{cases} 0 & \text{ if } U < 0 \text{ or } U > i\times j \\ 1 & \text{ if } (i=1 \text{ or } j=1) \text{ and } 0 \leq U \leq i\times j \\ f_{i, j-1}(U) + f_{i-1, j}(U - j) & \text{ else } \end{cases} \)