# 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