Help with a cdf question

1. Oct 20, 2010

sneaky666

X1,X2,...,Xn is a sample from a distribution with cdf F. Show
FX(i)(x)= $$\sum$$$$\stackrel{n}{j=i}$$ ($$\stackrel{n}{j}$$) Fj(x)(1-F(x))(n-j)

How would I start on this?

Last edited: Oct 20, 2010
2. Oct 21, 2010

sneaky666

Fx(n)(x) = Px(n)(X(n)<=x)
Fx(n)(x) = Px1...xn(max{x1,...xn}<=x)
Fx(n)(x) = Px1...xn(x1<=x,...,xn<=x)
Fx(n)(x) = Px1(x1<=x)*...*Pxn(xn<=x)
Fx(n)(x) = Fx1(x)*...*Fxn(x)
Fx(n)(x) = Fx1(x)*...*Fxi(x)*...*Fxn(x)
Fx(n)(x) = Fx1(x)*...*Fx(n-1)(x)*Fxi(x)
Fxi(x) = Fx(n)(x) / Fx1(x)*...*Fx(n-1)(x)
Fxi(x) = (Fx(1)(x))n / (Fx(1)(x))(n-1)
Fxi(x) = Fx(1)(x)

not sure where to go from here...

Last edited: Oct 21, 2010
3. Oct 21, 2010

vela

Staff Emeritus
You might get more responses if you say what the notation X(n) means. Others might be familiar with the notation, but I have no clue what the problem is about.

4. Oct 21, 2010

sneaky666

example of what X(n) means

X1=0.5 X(1) = 0.1 (so it is increasing...)
X2=0.7 X(2) = 0.2
x3=0.96 X(3) = 0.45
x4=0.45 X(4) = 0.5
x5=0.2 X(5) = 0.7
x6=0.1 X(6) = 0.96

Fx(n)(X) = Px(n)(Xn<=X)

so a cumulative distribution function