# Thread: First order Statistics problem

1. ## First order Statistics problem

Hi There

I have random variables X1,X2,X3,...,XN which are I.I.D. Each of these random variables are Erlang Distributed(Special case of Gamma Distribution). Random variable X=min{X1,...,XN}. Now the sample size N changes with each iteration according to the following rule

If Yn<=X<Yn+1 Then
N=2^n
where n is an element of {1,2,3,4}.............................(1)
Initially, N is set to 2, N=2. Yn and Yn+1 are constants that partition the domain of X. X is defined between 0 and Infinity. If X falls into a certain range say Y2 and Y3, then the sample size for the next iteration is changed from N=2 to N=2^2=4. So if X=min{X1,X2} initially, and we find that this X value lies between Y3 and Y4, the next sample size for the next iteration becomes N=2^3=8 according to Rule (1). The problem is to find the CDF and hence the PDF of Random variable X.

I need to solve this problem in order for me to get results for my masters dissertation. Thanks
Regards

2. ## Re: First order Statistics problem

will you be citing MHF in your bibliography? :P

3. ## Re: First order Statistics problem

ofcourse i will, i am not doing a Maths degree,its a Masters in Eng(Wireless Comms)...so this is a small part of a bigger problem...so i needed someone to help me go past this obstacle and then i can continue solving the other problems that follow...i am just stuck on this part. What i have done so far(which i suspect is wrong) is the following

I said the CDF for X is as follows P(X<c)=1-P(X>=c). But P(X>=c)=(P(min{X1,X2}>=c) Or P(min{X1,X2,..,X4}>=c) Or P(min{X1,X2,...,X8}>=c) Or P(min{X1,X2,....,X16}>=c)),....(2)

but since X1,X2,X3,...,XN are I.I.D, then P(min{X1,X2}>=c)=P(X1>=c)*P(X2>=c), the same is true for the other terms. Since Random values X1,X2,X3,...,XN are defined between 0 and Infinity, we can integrate the Erlang Distribution PDF for X1 between the lower limit c and the upper limit Infinity which gives us an Upper Incomplete Gamma Function. Since I.I.D the first term P(X1>=c)*P(X2>=c)=(Upper Gamma)^2, thus the CDF is the sum of the Upper Gamma Functions raised to the power 2,4,8 and 18 respectively for each term in (2). and then since the random value ranges for each term are between 0 and infinity, we will find that in the case of (2), the PDF integral over 0 and Inifnity range will yield the value of 4 instead of 1. So i normalised the CDF summation by 4. so

F(X)=p(X<c)=1-(Sum of GammaInc^2+GammaInc^4+GammaInc^8+GammaInc^16)/4. The PDF is just the derivative of this with respect to X. I am not too confident with this answer tho...