D**u
发帖数: 204 | 1 Two candidates (R and L) are for the 2008 election. In the election, voters
are in a single line and are going to vote one by one. After each voter
makes the vote, the other voters immediately knows who he/she voted for.
Voters tend to stay with the "winner's side", if at the moment the
candidates have x and y votes respectively, the voter will vote R with
probability x/(x+y), and vote L with probability y/(x+y).
Suppose R and L initially have x0 and y0 votes each (x0 and y0 are far
smaller than | H****h
发帖数: 1037 | 2 期望不变。过程是鞅。计算分布有难度。
voters
【在 D**u 的大作中提到】
: Two candidates (R and L) are for the 2008 election. In the election, voters
: are in a single line and are going to vote one by one. After each voter
: makes the vote, the other voters immediately knows who he/she voted for.
: Voters tend to stay with the "winner's side", if at the moment the
: candidates have x and y votes respectively, the voter will vote R with
: probability x/(x+y), and vote L with probability y/(x+y).
: Suppose R and L initially have x0 and y0 votes each (x0 and y0 are far
: smaller than
| D**u
发帖数: 204 | 3 Maybe starting with a few examples like (x0,y0)=(1,1), or (1,2) can be
helpful on the distribution.
【在 H****h 的大作中提到】
: 期望不变。过程是鞅。计算分布有难度。
:
: voters
| H****h
发帖数: 1037 | 4 你知道答案？
【在 D**u 的大作中提到】
: Maybe starting with a few examples like (x0,y0)=(1,1), or (1,2) can be
: helpful on the distribution.
| D**u
发帖数: 204 | 5 yes.
【在 H****h 的大作中提到】
: 你知道答案？
| H****h
发帖数: 1037 | 6 那我就等答案了。呵呵。
【在 D**u 的大作中提到】
: yes.
| D**u
发帖数: 204 | 7 再等两天, 答案还是比较有趣的.
【在 H****h 的大作中提到】
: 那我就等答案了。呵呵。
| D**u
发帖数: 204 | 8 答案.
(1) x0/(x0+y0), easy to prove.
(2) Beta distribution:
f(t; x0,y0) = 1/Beta(x0,y0) * (1-t)^(x0-1) * t^(y0-1)
= gamma(x0+y0)/(gamma(x0) * gamma(y0)) * (1-t)^(x0-1) * t^(y0-1).
Easy to check that f(t; x0,y0) satisfies the recursive formula:
f(t;x0,y0) = x0/(x0+y0) * f(t;x0+1,y0) + y0/(x0+y0) * f(t;x0,y0+1)
It is relatively easy to prove when x0 and y0 are integers. For non-integers
x0 and y0, I only have a guess that it's Beta distribution but no
theoretical proof (only have numerical "proof").
【在 D**u 的大作中提到】
: 再等两天, 答案还是比较有趣的.
| H****h
发帖数: 1037 | 9 这是你做的科研问题吗？看来一个问题的突破关键是要找到不变量。
那个递推公式明显在x0和y0为非整数情形也成立。你指的困难是什么？
最后你得出的结论似乎是对于任何的t，f(t;xn,yn)是一个鞅。
怎么再进一步？
integers
【在 D**u 的大作中提到】
: 答案.
: (1) x0/(x0+y0), easy to prove.
: (2) Beta distribution:
: f(t; x0,y0) = 1/Beta(x0,y0) * (1-t)^(x0-1) * t^(y0-1)
: = gamma(x0+y0)/(gamma(x0) * gamma(y0)) * (1-t)^(x0-1) * t^(y0-1).
: Easy to check that f(t; x0,y0) satisfies the recursive formula:
: f(t;x0,y0) = x0/(x0+y0) * f(t;x0+1,y0) + y0/(x0+y0) * f(t;x0,y0+1)
: It is relatively easy to prove when x0 and y0 are integers. For non-integers
: x0 and y0, I only have a guess that it's Beta distribution but no
: theoretical proof (only have numerical "proof").
| D**u
发帖数: 204 | 10 This is not my research topic. It's a brain teaser (for (x0,y0)=(1,1),
which is the original "Polya's urn problem")
given by a friend, and I made a guess for general (x0,y0).
The difficulty I have is that I can not prove that the Beta distribution
is indeed the answer for non-integer (x0,y0), though it is very ease to
check that Beta distribution satisfies the recursive formula (which is
a formula the answer must satisfy).
I did some numerical test, and Beta distribution matches the numerical
so
【在 H****h 的大作中提到】
: 这是你做的科研问题吗？看来一个问题的突破关键是要找到不变量。
: 那个递推公式明显在x0和y0为非整数情形也成立。你指的困难是什么？
: 最后你得出的结论似乎是对于任何的t，f(t;xn,yn)是一个鞅。
: 怎么再进一步？
:
: integers
|
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