Let X1, X2, ..., Xn be a random sample of size n for the standard exponential distribution and X1:n, X2:n, ..., Xi:n be the corresponding order statistics, where X1:n 0 : Dn < t}, for some positive number t. What's wrong with the following derivation?
P(N=n) = P(Tn-1 > t, Dn t) ... P(Dn-1 > t) P(Dn < t)
Now, it is easy to compute the right-hand side of the above equation, but, I found that the summation of P(N=n) over all strictly positive integers n is not equal to 1. Is there any thing wrong?
Thanks in advance for your help.
I think the problem is that Tn-1 ( == D1:n-1 ) and Dn are not statistically independent. Why don't you just change your argument slightly:
P(N> n-1; t) = P(D1:n-1 > t), P(N>n; t) = P(D1:n > t),
P(N=n; t) = P(N>n-1; t) - P(N>n; t).
Simulation agrees very well with this expression, see the attached file. But many trials are required when t is small.
Dear Kåre,
Thanks so much for your answer including the attached document about simulation. I should make it clear that in this new question I am asking about the exponential spacings Di themselves, not their order statistics Di:n, . Your answer is really good for the spacings Di, as they are independent and exponentially distributed with parameters (n-i+1), where \lambda =1 here.
What I am trying to do is to find the distribution of the minimum of spacings conditioned to reach below a fixed positive number t; i.e., I need to compute the distribution of min{D1, D2, ..., DN}, where N = inf{n>0: Dn < t}. In order to do that, I have to compute the probability density function of $N$.
I did the following:
P(N=n) = P( D1 > t, D2 > t, ..., Dn-1 > t, Dn < t)
= P(D1 > t)P(D2 > t) ... P(Dn-1 > t) P(Dn < t).
I could compute P(N=n), but unfortunately, when I run the sum over all positive integers n, I do not get one. The summation is in fact less than 1. I don't know why?
If we solve this problem, then we (you Joachim and me) can collaborate and write an article or paper about the minimum (maximum) spacings conditioned to hit a certain positive number.
Thanks so much again for your times and efforts.
Best Regards'
Mohamed Riffi
Let us continue this discussion by means which are better suited for mathematics.
Dear Kåre,
I have sent a private message for you.
Best Regards'
Mohamed
Dear Mohamed,
Probably, there is a misleading gap in your formulation of the problem:
1. By the introductory definition D1 = X{1:n} - X{0:n} \equiv min{X1, X2, . . . Xn} which depends on n
2. On the other hand I am getting an impression, that you are using: Dk = X{k:k} - X{k-1:k}, for every ntural k. In particular this would give D1= X{1:1} - X{0:1} \equiv X1, which is independent of n.
Hopefully this helps,
Best regards, Joachim
Dear Followers,
driven by the remarks by Mohamed I. and Kåre, I think we need the following:
Reformulation of the notation: Let X1, X2, ..., Xn, ... be an infinite sequence of iid standard exponential random variables. For every given natural n >= 1
-- let X1:n < X2:n < ... < Xn:n denote the corresponding order statistics of the
first n elements of the sequence.
-- define D1:n = X1:n, Di:n = Xi:n - Xi-1:n, i=2,3, ..., n; the random variables
D1:n, D2:n, ..., Dn:n are called the spacings between successive order statistics.
-- define Tn := min{D1:n, D2:n, ..., Dn:n}, for n>=1, T0=\infty (the shortest n-th space)
-- define N_t := inf{n: Tn 0.
Reformulation of the problem: For every t>0, find the pd of N_t.
Obvious Proposition: With probability 1 the following holds:
-- the sequence Tn is (weakly) decreasing
-- the function N_t is piecewise constant right continuous and decreasing
-- {N_t 0.
Calculations: Thus it sufficies to calculate the probability, that {Tn t} = \exp[ - (n + (n-1) + . . . +2 + 1)t ] = \exp[- n(n+1)t/2]
Pr{N_t=n} = \exp[- n(n-1)t/2] - \exp[- n(n+1)t/2]
= 2 \exp[-n^2 t/2] \sinh[ nt/2 ]
Is this OK? Joachim
Dear Joachim Domsta,
Thanks a million for your answer.
I am now sure that your calculations are correct and logical. I have sent you a private message concerning this discussion.
Best Regards'
Mohamed
Joachim> Is this OK?
Yes indeed; and in agreement with the simulation results I included in the first post.
Dear Followers,
Thanks go to Kåre for his confirming simulations. Concerned with the formulation of Mohamed's problem, after careful analysis of all texts, I am (almost) convinced, that one of the aims is to calculate the following distribution functions parmetrized by t, for t >0 (the notation of my previous answer is applied),
F_t(s) :=P{T_{N_t} t AND T_n 0.
This is a really hard problem, whenever closed formulas are sought. Obviously, the calculations are interesting for s= t the sum equals 1.
Regards, Joachim
Dear Joachim,
You are right. The result might not be written in closed form. We are trying now to set our theoretical model, which should be flexible enough to get nice results.
Thanks a lot for your great ideas.
Best Regards'
Mohamed
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