What is the formula for understanding sample distribution in CDF?

[sneaky666]

X₁,X₂,...,X_n is a sample from a distribution with cdf F. Show
F_X(i)(x)= \sum\stackrel{n}{j=i} (\stackrel{n}{j}) F^j(x)(1-F(x))^(n-j)


How would I start on this?

 

[sneaky666]

F_x(n)(x) = P_x(n)(X_(n)<=x)
F_x(n)(x) = P_x1...xn(max{x₁,...x_n}<=x)
F_x(n)(x) = P_x1...xn(x₁<=x,...,x_n<=x)
F_x(n)(x) = P_x1(x₁<=x)*...*P_xn(x_n<=x)
F_x(n)(x) = F_x1(x)*...*F_xn(x)
F_x(n)(x) = F_x1(x)*...*F_xi(x)*...*F_xn(x)
F_x(n)(x) = F_x1(x)*...*F_x(n-1)(x)*F_xi(x)
F_xi(x) = F_x(n)(x) / F_x1(x)*...*F_x(n-1)(x)
F_xi(x) = (F_x(1)(x))ⁿ / (F_x(1)(x))^(n-1)
F_xi(x) = F_x(1)(x)

not sure where to go from here...

 

[vela]

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.

 

[sneaky666]

X₁,X₂,...,X_n is a sample from a distribution with cdf F. Show
F_X(i)(x)= \sum\stackrel{n}{j=i} (\stackrel{n}{j}) F^j(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

F_x(n)(X) = P_x(n)(X_n<=X)

so a cumulative distribution function