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#1
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Discovered series result
Sometimes a probability problem affords one solution by some general mathematical technique, and another solution purely from probability considerations, and this provides insight into a general mathematical problem. I just stumbled upon such a problem which enables me to sum the following series:
sum{n = k to infinity} C(n,k)*x^n = (x^k)/(1-x)^(k+1) for 0 < x < 1. At first glance this may appear to be a more familiar series, until you realize that k is held constant, and the sum is over n. Do not confuse this with the more familiar binomial expansion in which n is constant and the sum is over k. Can you prove this? Is this series result well-known? Will I be famous? [img]/images/graemlins/smile.gif[/img] |
#2
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Re: Discovered series result
This is my kind of problem .
It seems like a very interesting result and I haven't seen it anywhere . I could be wrong but it looks as if this problem relates to gamblers ruin . Substitute x=q and 1-x = p |
#3
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Re: Discovered series result
Differentiate the power series for 1/(1-x) k times and multiply both sides by x^k/k!
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#4
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Re: Discovered series result
Jason , there has to be a better way .
I always try to look for elegant solutions to complex problems . This one has a nice solution to it . |
#5
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Re: Discovered series result
Okay then. FWIW, that solution is elegant to me. It's 4 lines long on my scratch paper and I think it would make a nice example for one of my undergraduate classes. I'm sure Bruce's probabilistic interpretation is neat too. I like your posts, so I look forward to seeing what you come up with.
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#6
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Re: Discovered series result
sum{n = k to infinity} C(n,k)*x^n =
= {reindex} = sum{n = 0 to infinity} C(n+k,k)*x^(n+k) = x^k sum{n = 0 to infinity} C(n+k,k)*x^n = x^k sum{n = 0 to infinity} C(n+k,(n+k)-k)*x^n = x^k [sum{n = 0 to infinity} C(n+k,n)*x^n ] I hadn't seen the sum in the brackets before but it was listed at Wikipedia to be: sum{n = 0 to infinity} C(n+k,n)*x^n = 1/(1-x)^(k+1) The stated result follows. |
#7
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Re: Discovered series result
I couldn't think of a probabilistic argument to show this .
Bruce , do you mind sharing to us your solution? |
#8
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Re: Discovered series result
Suppose I have a biased coin with probability x (0 < x < 1) of coming up heads. I flip it and count the number of tails after each flip until I get k+1 tails. On average, how many times will I flip the coin after a count of exactly k tails?
Clearly this answer does not depend on k. None of the flips matter until I hit k tails. Once I do, the probability is x that I get only one flip, x*(1-x) that I get two flips and so on. The expected number is x + x*(1-x) + x*(1-x)^2 + . . . = 1/(1-x). I can also represent this amount by summing the probability of having exactly k tails after n flips for n = 0 to infinity since the expectation of a sum is the sum of the expectations. This probability is clearly zero for n < k, so I have: sum{n = k to infinity} C(n,k)*x^(n-k)*(1-x)^k which must equal 1/(1-x). If I multiply both sides by [x/(1-x)]^k, I get BruceZ's result. |
#9
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Re: Discovered series result
Aaron your posts are always very insightful and I always look forward to what you have to say .
However , x+x*(1-x)+x*(1-x)^2+.... = x/(1-(1-x)) which is just a geometric series with first term x and a common ratio of (1-x) . The Rhs is 1 and not 1/(1-x) like you mentioned . |
#10
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Re: Discovered series result
Good catch jay_shark. I actually solved it with x and (1-x) reversed, then screwed up the transcription. The real series should be 1 + x + x^2 + . . . which does indeed equal 1/(1-x). You're sure to get one flip with k tails, if that is a head (probability x) you'll get a second, if the first two are heads (probability x^2) you'll get a third, and so on.
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