# cumulative distributed function example

by xeon123
Tags: cumulative, distributed, function
 P: 81 I was looking to a video about cumulative distribution function (http://www.youtube.com/watch?v=658WXDkhU_w) and he show the following function: $\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ | 1/4, 0 \leq x \leq1 \\ f(x) =<(x^3)/5, 1 \leq x \leq 2 \\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ |0, otherwise.$ At minute 8:45, he presents the cumulative distribution as: $\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ | 0, x \leq 0 \\ F(x) = < \frac{1}{4}x, 0 \leq x \leq 1 \\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ | \frac{1}{20}(x^4+4), 1 \leq x \leq 2 \\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ | 1, \ x \geq 2$ I don't understand why F(x) is 1 for $x \geq 2$, if f(x) is 0, otherwise. Why? BTW, I hope that that my functions are legibles, because I don't know how to put big curly brackets.
P: 3,173
 Quote by xeon123 I don't understand why F(x) is 1 for $x \geq 2$, if f(x) is 0, otherwise. Why?
Look at a simpler example. Suppose f(x) = 1/2 when x = 1 or x = 2 and f(x) = 0 otherwise. The value of the cumulative distribution F(x) would be 1 at x = 3 because F(3) gives the probability that x is equal or less than 3. The condition that x is equal or less than 3 includes the cases x = 1 and x = 2.
 P: 81 I understand what you said, but the probability of happening 3 is 0, because it's not defined in f(x). For me, F(3) should never be defined.
P: 3,173

## cumulative distributed function example

 Quote by xeon123 F(3) should never be defined.
Do you mean "should" in some moral or religious sense? Mathematics would only care about you opinion if you could show some logical contradiction in the standard definition of cumulative distribution function.

 because it's not defined in f(x).
It is defined in the domain of f(x). f(3) = 0. That's part of the "f(x) = 0 otherwise" clause.
Math
Emeritus
If f(x) is the "probability density function" then $F(X)= \int_{-\infty}^X f(x)dx$ is the probability that x is less than or equal to X.