Discrete and Normal Distribution

Date: 01/10/2001 at 23:54:54
From: David Jordan
Subject: Birthday problem in a normal distribution

The traditional "birthday problem" of finding the probability that two 
or more people share a birthday assumes a uniform probability. How 
would one solve the problem if the distribution were a normal one, as 
in trying to find the probability that any two years had the same 
number of inches of rainfall?

Date: 01/11/2001 at 15:27:48
From: Doctor Schwa
Subject: Re: Birthday problem in a normal distribution

Hi David,

If it's really a normal distribution, it's a CONTINUOUS variable, so 
the probability of getting the same number (say, the 14.3687294... 
inches of rain we had here last year) is zero.

Of course, what you mean is "what do you do when the variable is 
discrete but approximately normal?"

Unfortunately the normality doesn't help much. What you need to do is, 
for each DISCRETE value of X, compute P(X)^2 and add them up. That is, 
P(1)*P(1) is the probability that both years had 1 inch of rainfall, 
P(2)*P(2) for 2 inches, and so on. So adding them gives the 
probability that both years had the same.

You need a DISCRETE probability distribution, not a continuous one, 
for this problem to make sense.


- Doctor Schwa, The Math Forum
  http://mathforum.org/dr.math/   

Date: 01/12/2001 at 05:57:16
From: Doctor Mitteldorf
Subject: Re: Birthday problem in a Normal distribution

David,

Here are some more thoughts on the problem of two years with the same 
number of inches of rainfall.

The obvious way to formulate a discrete-distribution problem with this 
continuous distribution is to ask for the probability that the 
rainfall for two years is within some tolerance. Let's assume that you 
want to look for pairs of years in which the rainfall is within 1" of 
being the same. We also assume that the number of inches of rain is 
normally distributed. (Footnote: normal distributions go from 
-infinity to +infinity; clearly, it's meaningless to talk about 
rainfall that is less than 0"; but often the left tail end of the 
normal distribution decreases fast enough that the probability is 
already very close to zero before the number of inches becomes 
negative. So I'll talk about integrals from -infinity to +infinity, 
and you'll know what I mean.)

Let's review the shared birthday problem: For any two people, the 
probability that they DON'T share a birthday is 364/365. Add a third 
person, and the probability that he doesn't share a birthday with 
either of them is 363/365. Continuing in this way, you can see that 
with N people, the probability that no 2 of them share a birthday is 
given by:

            364 * 363 * 362 * ... * (366-N)
     P(N) = -------------------------------
                       365^(N-1)

Returning to the rainfall problem: What is the probability that two 
years don't have the same number of inches?  You can find this by 
integrating over the two Normal distributions, linking the integrals 
together to assure that they are at least 1" apart. Let's assume that 
year 2 has a greater rainfall than year 1. We'll write N(x) for the 
particular normal distribution that applies to this region, with the 
correct mean and standard deviation. Then the double integral is:

      inf   inf
      INT { INT [N(x)N(y) dy] dx}
     -inf   x+1

Take this double integral and multiply by 2 to correct for the 
assumption that year 2 has greater rainfall than year 1 (since it's 
just as likely that year 1 has the larger rainfall).

This double integral can't be done analytically, but it's an easy task 
numerically. Now let's look at a third year. We'll assume that the 
amount of rain is ordered: year 3 > year 2 > year 1. Then the 
probability that all three years are different by more than an inch is 
a triple integral:

      inf   inf   inf
      INT ( INT { INT [N(x)N(y)N(z) dz] dy} dx)
     -inf   x+1   y+1

Take this triple integral and multiply by 6 to correct for the 
assumption that the years are ordered year 3 > year 2 > year 1. There 
are 6 such orderings, all equally likely.

Again, this triple integral can be evaluated numerically with no 
difficulty. And it is not difficult to generalize to the case of N 
years: we'll have N-1 nested integrals, and the result will be divided 
by N!.

There is a practical problem, however, in the fact that large-
dimensional integrals quickly become intractable. A 6-dimensional 
integral is a real challenge, and numerical evaluation of a 10-D 
integral is not conceivable. I am confident that there are methods to 
surmount this problem, which offer good approximations to these 
multiple integrals over the normal function, but for these you'll need 
a more experienced statistician than myself.

- Doctor Mitteldorf, The Math Forum
  http://mathforum.org/dr.math/