Mathematics desk | ||
---|---|---|
< December 27 | << Nov | December | Jan >> | December 29 > |
Welcome to the Wikipedia Mathematics Reference Desk Archives |
---|
The page you are currently viewing is an archive page. While you can leave answers for any questions shown below, please ask new questions on one of the current reference desk pages. |
Given positive integers n, m and real , suppose I have iid exponential variables (rate 1) . What is the probability p that there exists an integer index such that ?
Intuition: The variables represent time between successive events in a Poisson process. An index i as described represents the start of a lucky streak, where events happen within a span of time T (assumed to be lower than the average time it should take). I want to find the probability of finding such a streak.
If it matters, in my application .
The probability v that a specific index starts a streak is easy to calculate using the Poisson or Erlang distribution. This gives a bound of . But because of the correlations between the different indices, I can't figure out a way to calculate the exact probability.
If a symbolic expression can't be found, I will also appreciate methods to calculate it numerically or simulate it which improve upon the most naive simulation (my T is such that the event is fairly rare, so a naive simulation will take a lot of tries to reach sufficient precision). -- Meni Rosenfeld ( talk) 00:34, 28 December 2015 (UTC)
The probability that is .
The requested probability is .
For T=24, m=48, and n=2(m+1)=98 we have
Do you agree?
Bo Jacoby ( talk) 15:24, 1 January 2016 (UTC).
The stochastic variable has mean value and standard deviation . For T=24 and m=48 we have .
Assuming that Z has approximately a standard normal distribution:
The gamma distribution gave
The normal distribution approximation must be rejected.
erf =: (1 H. 1.5)@*:*2p_0.5&*%^@:*: NB. error function n01cdf=: -:@>:@erf@%&(%:2) NB. cumulative normal dist 0j15":p1=.n01cdf -2*%:3 0.000266002752570 -.50^~-.p1 0.0132138 50*p1 0.0133001 0j15":p2=.-.(+/(T^k)%!k=.i.48)*^-T=.24 0.000010428432773 -.50^~-.p2 0.000521288
Bo Jacoby ( talk) 14:41, 2 January 2016 (UTC).
(Removed duplicate question. See WP:RDS#Which algorithm to use. -- ToE 13:39, 28 December 2015 (UTC))
Mathematics desk | ||
---|---|---|
< December 27 | << Nov | December | Jan >> | December 29 > |
Welcome to the Wikipedia Mathematics Reference Desk Archives |
---|
The page you are currently viewing is an archive page. While you can leave answers for any questions shown below, please ask new questions on one of the current reference desk pages. |
Given positive integers n, m and real , suppose I have iid exponential variables (rate 1) . What is the probability p that there exists an integer index such that ?
Intuition: The variables represent time between successive events in a Poisson process. An index i as described represents the start of a lucky streak, where events happen within a span of time T (assumed to be lower than the average time it should take). I want to find the probability of finding such a streak.
If it matters, in my application .
The probability v that a specific index starts a streak is easy to calculate using the Poisson or Erlang distribution. This gives a bound of . But because of the correlations between the different indices, I can't figure out a way to calculate the exact probability.
If a symbolic expression can't be found, I will also appreciate methods to calculate it numerically or simulate it which improve upon the most naive simulation (my T is such that the event is fairly rare, so a naive simulation will take a lot of tries to reach sufficient precision). -- Meni Rosenfeld ( talk) 00:34, 28 December 2015 (UTC)
The probability that is .
The requested probability is .
For T=24, m=48, and n=2(m+1)=98 we have
Do you agree?
Bo Jacoby ( talk) 15:24, 1 January 2016 (UTC).
The stochastic variable has mean value and standard deviation . For T=24 and m=48 we have .
Assuming that Z has approximately a standard normal distribution:
The gamma distribution gave
The normal distribution approximation must be rejected.
erf =: (1 H. 1.5)@*:*2p_0.5&*%^@:*: NB. error function n01cdf=: -:@>:@erf@%&(%:2) NB. cumulative normal dist 0j15":p1=.n01cdf -2*%:3 0.000266002752570 -.50^~-.p1 0.0132138 50*p1 0.0133001 0j15":p2=.-.(+/(T^k)%!k=.i.48)*^-T=.24 0.000010428432773 -.50^~-.p2 0.000521288
Bo Jacoby ( talk) 14:41, 2 January 2016 (UTC).
(Removed duplicate question. See WP:RDS#Which algorithm to use. -- ToE 13:39, 28 December 2015 (UTC))