Associated Topics || Dr. Math Home || Search Dr. Math

### 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/
```
Associated Topics:
College Statistics

Search the Dr. Math Library:

 Find items containing (put spaces between keywords):   Click only once for faster results: [ Choose "whole words" when searching for a word like age.] all keywords, in any order at least one, that exact phrase parts of words whole words

Submit your own question to Dr. Math
Math Forum Home || Math Library || Quick Reference || Math Forum Search