The sample range is the difference between the maximum and minimum.
It is easy..
sample range = max - min
Let Fx is the CDF
Distribution of Min is 1- (1- Fx)^n
Distribution of Max is Fx^n
Expected range = E(max) - E(min)
The function for the expected value of a sample range is given on the following page.
http://www.jmu.edu/docs/sasdoc/sasht...hapc/sect9.htm
I'd be interested to know which distribution is used in the calculation of this expectation and how you get from the general formulation of an expectation (Integral x*f(x)dx, -%INF, %INF) to the final result.
An original paper by Tippett(1925) is on JSTOR, but unfortuneately I have no access to it.
The sample range is the difference between the maximum and minimum.
It is easy..
sample range = max - min
Let Fx is the CDF
Distribution of Min is 1- (1- Fx)^n
Distribution of Max is Fx^n
Expected range = E(max) - E(min)
In the long run, we're all dead.
There is one more formula for expectation .
E(X) = integral ( [1- Fx] dx )
This you can prove using integration by parts.
Use this formula in the above you will get the expression for d2
In the long run, we're all dead.
I did know what a sample range was, but I'm curious about the distributions of min and max. Where did they come from? Also I think your signs are different than the ones in the link. The integrand there is
As I mentioned in my original post an expectation should look like an integral from minus infinity to plus infinity of the independent variable x times the pdf of the distribution. In the formula you gave it is hard to see the explicit presence of x.Code:1 - (1 - Fx)^n - (Fx)^n which is not equal to (Fx)^n - (1 - (1 - Fx)^n)
Last edited by Papabravo; 10-15-2008 at 11:49 PM.
The distribution part is not difficult.
Let X,.X2..Xn are the iid(indpendent and identically distributed) r.v.s
with CDF Fx
let Ma is the maximum
P[Ma <=x ]= P[X1<=x,.....Xn<=x] = product(i=1:n)(P[Xi<=x]) = Fx^n ( identical)
For deriving Minimum, use the Reliability function
Let Mi is the minimum
P(Min>x) = P[X1>x,.....Xn>x]
etc.
See this link for some more information
http://en.wikipedia.org/wiki/Order_statistic
I think you didn't read my second post.
There is one more formula for expectation .
E(X) = integral ( [1- Fx] dx )
This you can prove using integration by parts.
Use this formula in the above you will get the expression for d2
In the long run, we're all dead.
Of course I saw your second post. The information on the order statistics was useful. Still I can't get from that formulation of the pdf for the minimum and the maximum to the expression in the original link. Here's the problem:
In the orginal link the integrand involved only constants and powers of the normal CDF. The integrand you write for the expected value has the independant variable x, the pdf f(x), and powers of the CDF F(x) and the complement (1-F(x)). That is the remaining step that I'm missing. There is also the matter of the constant n! /(r-1)!*(n-r)! in the initial expression for the expectation where r = 1 for the minimum and r = n for the maximum.
Let
F1(x) is the CDF of Min &
F2(x) is the CDF of Max
E(Min)=(Integral ( [1-F1(x) ] dx, -%INF, %INF)
E(Max)=(Integral ( [1-F2(x) ] dx, -%INF, %INF)
E(Range) =E(max)-E(min)
= (Integral ( [1-F2(x) -1 + F1(x) ] dx, -%INF, %INF)
= (Integral ( [F1(x) - F2(x) ] dx, -%INF, %INF)
I have already explaind the formulation of CDF of Min and Max
ie F1(x) = 1-{1 - F(x)}^nCode:let Ma is the maximum P[Ma <=x ]= P[X1<=x,.....Xn<=x] = product(i=1:n)(P[Xi<=x]) = Fx^n ( identical) For deriving Minimum, use the Reliability function Let Mi is the minimum P(Min>x) = P[X1>x,.....Xn>x] etc.
F2(x) = F(x)^n
Replace this, you will get the answer.
you have to offer a beer for more explanation.
In the long run, we're all dead.
Problem is I still don't believe that your espressions for E(Min) and E(Max) are correct. That being the case the rest of your argument is hand waving to fit the result. Thanks for trying, but I guess I'll have to dig out a copy of Tippett.
Tweet |