Let F(x)= Sum j=0 to infinity of (1-1/j+1) I [j,j+1) (x).....?

Question

Is this a distribution function? How can you tell?

Answer 1

If I understand your notation...

I [j,,j+1) (x) is the "characteristic function", which takes the value 1 if x is in the interval [j, j+1) and takes the value zero otherwise.

Also, (1-1/j+1) means [1-1/(j+1)].

If both of those are right, then good news: you have a Cumulative Distribution Function (which is any nondecreasing function which tends to zero on the left and one on the right).

We can break F into its components. Say F(x) = Sum f_j(x), where f_j(x) = [1-1/(j+1)] I [j, j+1) (x).

The graph of f_j(x) is just the line y=0, with a bump of height [1-1/(j+1)] at the interval [j, j+1). Imagine a short hat with an infinitely wide brim. With j=0, the height is [1-1/(0+1)]=0. With j=1, the height is [1-1/(1+1)]=1/2. As j increases, the height of hat #j--which is [1-1/(j+1)] = j/(j+1)--tends to 1.

Your final function, F(x), is described by the equation
F(x) = Floor(x)/Floor(x+1) for x>0; 0 otherwise. This is a nondecreasing function that approaches 0 as x goes to negative infinity, and approaches 1 as x goes to positive infinity.