Probability rank-n

  • Thread starter hugomcp
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  • #1
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Hello
Let y1,..yn be drawn from a normal distribution width known parameters.
Let x be drawn from another normal distribution with known parameters.
If the set {y1,...,yn, x} is ordered, what is the probability that x appears in the "k" position?
 

Answers and Replies

  • #2
mathman
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It depends very much on the parameters of these distributions. You need to be a little more specific. The simplest example is all having the same distribution, then the probability is the same for all positions.
 
  • #3
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It depends very much on the parameters of these distributions. You need to be a little more specific. The simplest example is all having the same distribution, then the probability is the same for all positions.
I know that the probability depends on the parameters of both distributions. If both have the same distribution, it will give an uniform. If x has a higher mean, it will give an exponential distribution.

I am looking for a formula that, taking into account the gaussian parameters of both distributions, computes the desired probability.
 
  • #4
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P(X appears in kth position) where k is an element from {0,1,2,....,n}
= P(X higher than k number of Yi)
= (n C k) [P(X> Yi)]k[P(X< Yi)]n-k
 
  • #5
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P(X appears in kth position) where k is an element from {0,1,2,....,n}
= P(X higher than k number of Yi)
= (n C k) [P(X> Yi)]k[P(X< Yi)]n-k
Thanks for your answer, but it may not be that simple. If that formula was true, then such probability for distributions with the sama parameters will give a normal, instead of a uniform, which I actually confirmed by simulation
 
  • #6
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Have you found the answer then?
 
  • #7
525
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P(X appears in kth position) where k is an element from {0,1,2,....,n}
= P(X higher than k number of Yi)
= (n C k) [P(X> Yi)]k[P(X< Yi)]n-k
Not quite correct - the events {X>Yi} are only conditionally independent, however conditioning on X will get a similar expression

P(X higher than k number of Yi)
[tex]=\int_{-\infty}^{\infty}f(x)(^nC_k)F(x)^k(1-F(x))^{n-k}dx[/tex]
where F is the cdf of the Yi and f is the pdf of X, which reduces to 1/(n+1) when X has the same distribution as Y.
 

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