Wednesday, November 25, 2009

Birthday Paradox Variance

First a message from David Johnson for proposals on locations for SODA 2012 both in and outside the US.

Here's an interesting approach to the birthday paradox using variances.

Suppose we have m people who have birthdays spread uniformly over n days. We want to bound m such that the probability that there are are at least two people with the same birthay is about one-half.

For 1 ≤ i < j ≤ m, let Ai,j be a random variable taking value 1 if the ith and jth person share the same birthday and zero otherwise. Let A be the sum of the Ai,j. At least two people have the same birthday if A ≥ 1.

E(Ai,j) = 1/n so by linearity of expectations, E(A) = m(m-1)/2n. By Markov's inequality, Prob(A &ge 1) ≤ E(A) so if m(m-1)/2n ≤ 1/2 (approximately m ≤ n1/2), the probability that two people have the same birthday is less than 1/2.

How about the other direction? For that we need to compute the variance. Var(Ai,j) = E(Ai,j2)-E2(Ai,j) = 1/n-1/n2 = (n-1)/n2.

Ai,j and Au,v are independent random variables, obvious if {i,j}∩{u,v} = ∅ but still true even if they share an index: Prob(Ai,jAi,v = 1) = Prob(The ith, jith and vth person all share the same birthday) = 1/n2 = Prob(Ai,j=1)Prob(Ai,v=1).

The variance of a sum of pairwise independent random variables is the sum of the variances so we have Var(A) = m(m-1)(n-1)/2n2.

Since A is integral we have Prob(A < 1) = Prob(A = 0) ≤ Prob(|A-E(A)| ≥ E(A)) ≤ Var(A)/E2(A) by Chebyshev's inequality. After simplifying we get Prob(A < 1) ≤ 2(n-1)/(m(m-1)) or approximately 2n/m2. Setting that equal to 1/2 says that if m ≥ 2n1/2 the probability that everyone has different birthdays is at most 1/2.

If m is the value that gives probability one-half that we have at least two people with the same birthday, we get n1/2 ≤ m ≤ 2n1/2, a factor of 2 difference. Not bad for a simple variance calculation.

Plugging in n = 365 into the exact formulas we get 19.612 ≤ m ≤ 38.661 where the real answer is about m = 23.

Enjoy the Thanksgiving holiday. We'll be back on Monday.

14 comments:

  1. Thanksgiving was approx. a month ago!

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  2. well, the birthday problem aint really new neither is the approach so i aint sure wat this post aint about

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  3. Seems David Johnson is behind for about a year, next SODA in Austin is SODA 2010 ;-)

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  4. cool way of using basically the same inequality but with the unknown variable once in the numerator and once in the denominator

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  5. that should be obvious...

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  6. The chebyshev's inequality has greater and equal in the probability statement. Is it simply OK to remove the equality here?

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  7. Lance, you have some typos in the second part of the argument. In particular, you should be proving an upper bound on Prob(A = 0), not Prob(A >= 1) again.

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  8. last anon: all right, all right, but give us a break. it's thanks-giving. instead of typo-giving you should be giving something else.

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  9. hey waowao, this is supposed to be a mathematically and scientifically centered blog, having errors in math is really bad and worth pointing out whatever day of the week it is

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  10. hey lance something you could explain a little bit. Why the need to advertise for wolfram and all the clunky wolfram alpha utilities.
    I find it a little bit disappointing ... Seems like Wolfram and Co clearly won you over to his side.

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  11. Would this not be dedicated to "Thanksgiving holiday", then I would ask "Lance -- what is the message?"

    Birthday paradox is no paradox -- it is just counting. We count with weights, forget what we count -- and here is a "paradox" ... I wonder how people (also my students) find this "strange". Markov, Chebysachev = trivial counting. Chernoff = a bit mmore delicate counting. Nothing more.

    Still, was nice to read.

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  12. Would this not be dedicated to "Thanksgiving holiday", then I would ask "Lance -- what is the message?"

    Birthday paradox is no paradox -- it is just counting. We count with weights, forget what we count -- and here is a "paradox" ... I wonder how people (also my students) find this "strange". Markov, Chebysachev = trivial counting. Chernoff = a bit mmore delicate counting. Nothing more.

    Still, was nice to read :-)

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  13. This post is so exciting it makes my penis hurt!

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