I'm reading the book Probability Theory: The Logic of Science by Jaynes. While I'm reading chapter 3, on page 56, it says:

Although the hypergeometric distribution $h(r)$ appears complicated, it has some surprisingly simple properties. The most probable value of $r$ is found to within one unit by setting $h(r') = h(r' − 1)$.

I'd like to know why is that? Is it because the relationship between the hypergeometric distribution and hypergeometric function?


You are looking for the mode $r^\prime$. The hypergeometric distribution is unimodal, in the sense that the most likely values cluster together, in such a way that the two most likely values are adjacent. That is why this method works.

For example, consider a hypergeometric distribution based on an urn with 10 red balls and 10 green balls, from which you sample 5 balls at random without replacement. Rounded to four places, the probability of getting $X = k$ red balls is:

      k :      0      1      2      3      4      5
    h(k): 0.0163 0.1354 0.3483 0.3483 0.1354 0.0163

In this case, two adjacent values $k= 2$ and $3$ are tied for 'most likely', with probability 0.3483.

Using the hypergeometric formula (with parameters 10, 10, and 5), if you set $h(k) = h(k-1)$ and solve for $k,$ you will get the integer value $k = 3$.

In other examples, you might not get an integer value, but the value will be within 1 of the mode. This will be true whether there is a 'double-mode' as in my example, or a single most likely value of $k$.

You might want to try it for an urn with 5 red balls and 3 black ones, from which you draw 5 balls. In this case, integer $k = 3$ is a unique mode. What noninteger value near 3 do you get when you solve the equation? (Express $h(\cdot)$ in terms of factorials; there is a lot of cancelling.)

      k :      0      1      2      3      4      5
    h(k): 0.0000 0.0000 0.1786 0.5357 0.2679 0.0179

You might want to look at the Wikipedia article on 'hypergeometric distribution; which has a formula for the mode in terms of the total number of balls in the urn, the number of red balls, and the number being sampled.

  • $\begingroup$ By unimodal, do you mean having only one global maximum (the strict definition of unimodal) or only one local maximum? It would be better to explain that the reason for the equation $h(r')=h(r'-1)$ is that the hypergeometric distribution first strictly increase than strictly decrease as the variable increases. As a matter of fact, it is more accurate to solve for $h(r')\ge h(r'-1)$ as you may not have a solution for $h(r')=h(r'-1)$. $\endgroup$ – Hans Jun 27 '15 at 18:22
  • 1
    $\begingroup$ @Hans: Totally reasonable comment. I was trying to address the exact Question asked, without getting into related details.Although not a traditional definition of anything, I hoped my phrase "two most likely values are adjacent" would do the job intuitively. As noted, a solution to $h(r^\prime) = h(r^\prime - 1)$ is not necessarily an integer (hence not a possible point of positive hypergeometric probability), but the Question allows $\pm 1$ "wiggle room". Of course a more refined argument is in terms of $h(r^\prime) / h(r^\prime - 1).$ $\endgroup$ – BruceET Jun 27 '15 at 20:07
  • 1
    $\begingroup$ The ratio $h(r')/h(r'-1)$ is exactly what I am implying. It is not a hard computation and it will be a short, clean and clear explanation, much better than the tedious numerical computation which leaves, at least me, feeling a bit beating around the bush and unsure of the exact mechanism involved. From there you can directly explain (prove) why the real value solution $h(r')=h(r'-1)$ must be within $1$ of the true integer solution. $\endgroup$ – Hans Jun 27 '15 at 20:21
  • $\begingroup$ @Hans: I'm happy enough with my answer (admittedly to a question that's intuitive and a bit imprecise).You are always welcome to post your own more elegant Answer to a related question if you please (which would essentially be a duplicate of another answer on this site, as I recall). I doubt if that is what the OP is looking for, but I can't know that for sure. And please do see the Wikipedia article. $\endgroup$ – BruceET Jun 27 '15 at 20:25
  • $\begingroup$ Sorry for my rushed comment with unintended possible implication that your solution is inferior. I do not mean that. I just thought the direct proof added to your answer would make it more elegant. OP is reading Jaynes' book which is a deep book, so I would think he has the sophistication to appreciate the mathematics involved in this derivation. Perhaps linking to that answer you mentioned which has the mathematical derivation would be helpful and save both of us effort. $\endgroup$ – Hans Jun 27 '15 at 20:32

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.