What is the formula for understanding sample distribution in CDF?

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sneaky666
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X1,X2,...,Xn is a sample from a distribution with cdf F. Show
FX(i)(x)= [tex]\sum[/tex][tex]\stackrel{n}{j=i}[/tex] ([tex]\stackrel{n}{j}[/tex]) Fj(x)(1-F(x))(n-j)


How would I start on this?
 
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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...
 
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sneaky666 said:
X1,X2,...,Xn is a sample from a distribution with cdf F. Show
FX(i)(x)= [tex]\sum[/tex][tex]\stackrel{n}{j=i}[/tex] ([tex]\stackrel{n}{j}[/tex]) Fj(x)(1-F(x))(n-j)


How would I start on this?

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