Surely not, if I understand your question correctly. For large n, the probability that D_{1:n} equals D_{1:n-1} must be very large. Even though each spacing is independent. Just run a simulation, and watch.

Dear Kåre Olaussen,

Thank you so much.  Your note makes sense.

For precise answer precise formulation is required. Thus, let me try the following assumptions:

Let X1, X2, X3, . ..   be iid with EXP(1) as their common pd,  i.e. with cdf F(x) = 1 - e-x, for  x > 0. And let  us define D_{1:k} :=  min{X1, X2,..., Xk}  for  k=1,2,3, ... Thus, for example   D_{1:1} = X1  and  D_{1:2} = min{X1,X2}.  and for the joint probability distribution  of  the pair D_{1:1}, D_{1:2} we have

P{D_{1:1}y} = P{ X1y } = P{ X1y, X2>y } = 0, whenever  0< x < y

Simultaneously, for the product probabilities we have

P{D_{1:1}y }= (1 - e-x ) * e-2y   is NOT  zero.

This proof exteds suitably to the general case that  for every non-degenerate common p.d (i.e. not concentrated at one point,   the minimals   D_{1:k}  and  D_{1:K} are not independent, for arbitrary  pair  of naturals  k  and  K.

However: are the assumption the same as they are meant by Mohamed I. ?

Regards

Dear Joachim Domsta,

The assumption is the same except for the exponential parameter, which is in my case is lambda, but this makes no difference.

You really gave a clear cut answer to my question.

hanks a million for your cooperation.

Best Regards'

Mohamed.

Obviously, no parameters even the exponetial type counts. Only the assumption that the  pd of each term in the sequence  X1 X2 ... is NOT concentrated at one point This is the simplest argument (now after a correction):

P{ D_{1;n-1} > D_{1,n}  } >0 and P{ D_{1;n-1} => D_{1,n}  } =1 in the indicated cases.

Regards, Joachim

Just one remark: In general, if the lengths tn of the spacings are independently distributed, then

P(D_{1; n} > t) = P(D_{1; n-1} > t) * P(tn > t)

Nice!

Kåre Olaussen

What do you mean by the lengths of the spacings?

It is well known that the exponential spacings D_i are independent and exponentially distributed with parameters (n-i+1) \lambda, for i = 1, 2, ..., n, where \lambda is the parameter of the underlying exponential distribution.  My question is: how is your  t_n related to these D_i's?

To All.

Dear Followers, probably a misunderstanding appeared due to lack of commonly accepted definition of the notion introduced by 

"D1:n-1 is the minimum of spacings between exponential order statistics"

My answer was related to the following one:  D_{1,n}:= min{ t_i ; i = 1,2,...,n}, where  t_i   are iid exponential. Seemingly, this has been used also by Kåre Olaussen. From the last answer I've got an impression that Mohammad I. Riffi uses another definition.

@ Mohamed I. 

Can you please provide your definition?  This should be helpful for all of us:)

Regards, Joachim

Sorry, I probably used a mixed notation. In my notation

D_{1; n} = min { tk | 1 0.

Mohamed> exponentially distributed with parameters (n-i+1) \lambda

I don't understand where and how (n-i+1) enters.

Dear All,

Please pardon me!  It seems that my question was not clear enough.  In fact, my question was about the minimum of the order statistics of the spacings D_i, where D_1 = X_{1:n},  D_2 = X_{2:n} - X_{1,n}, ... and so forth.

Now, I denoted the order statistics of these spacings as D_{1:n}, D_{2,n}, ... etc.  Therefore, D_{1:n} = min{D_1, D_2, ..., D_n}.

Now, it seems, when sequentially sampling, that D_{1:n-1} and D_{1, n} are dependent.  And that's what I meant by my question.

Thanks so much for your time and efforts.

@ Kåre Olaussen, @ Mohamed I. Riffi

I am almost understanding the definition by Mohamed I. The following theorem is a basic one for all reliabilitists: 

  If    t1, t2, t3, . . . tn  are iid exponential with parameter  \lambda, then the joint distribution of order statistics     0

@ Joachim Domst

Let X1,  X2, ..., Xn  be a random sample of size n and X1:n, X2:n, ..., Xi:n be the corresponding order statistics, where X1:n< X2:n< ... < Xn:n.

Let Di = Xi:n - Xi-1:n be the distance between Xi:n and Xi-1:n for i=1, 2, 3, ..., n.

By convention, X0:n=0 and, consequently, D1 = X1:n.

The random variables D1, D2, ..., Dn are called the spacings between successive order statistics.

Now,  D1:n, D2:n, ..., Di:n are the order statistics of the spacings D1, D2, ..., Dn.  So, my question is about the dependence/independence between the D1:1, D1:2, D1:3, ..., D1, n.  This means that we conduct sequential sampling and at each draw we look at the minimum of the spacings.  We stop this process when for the first time the minimum of the spacings becomes less than a positive number t.

I hope that my question is clear now.

@Mohamed: Sorry for completely misunderstanding your question!

Consider a generated sequence of n independent identically exponential distributed elements. As you already said, according to the Rényi representation mentioned by Joachim, the spacings Dk are also independent exponentially distributed with parameters (n+1-k)λ, with D1:n the minimum of these spacings. Now consider the first n-1 elements of the same original sequence. The corresponding spacings dk of this sequence is identical to the Dk, with the exception that two adjacent spacings Dj, Dj+1 for some random j have been added. This implies that D1:n-1 >= D1:n, which means that they cannot be independent.

I ran a simulation displaying the joint distribution. The probability that       D1:n-1 = D1:n is quite high (and I guess computable), demonstrating a strong correlation.

I hope I understood you correctly this time(?).

Added: I conjecture that Prob( D1:n-1 != D1:n) = 2/(n+1) 

Dear Kåre and Mohamed I.

Now everything is clear. The answer is given by Kåre, since  D_{1,n-1} >= D_{1,n} with probability 1, and the pd of the original iid rv-s is not concentrated at one point. 

The conjecture by Kåre is very interesting and (if tru) should hold for arbitrary continuous probability distribution of the iid-s.

Best regards, Joachim

Joachim> should hold for arbitrary continuous probability distribution of the iid-s

The conjecture fails (miserably for small n) when I replace the exponential distribution with a flat [0,1) distribution. I will not rule out that some universality may emerge for large n.

Dear Kåre, 

Yes, you are right! I have mixed some situations in my head:(

Thanks for quick setting me straight !

Joachim

Dear Kåre and Joachim,

You are right.  I understood this from your first answer to my question.

Thank you both for your help.

Best Regards'

Mohamed Riffi