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Cumulative distributed function example

by xeon123
Tags: cumulative, distributed, function
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xeon123
#1
May10-12, 07:03 AM
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:


[itex] \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ | 1/4, 0 \leq x \leq1 \\
f(x) =<(x^3)/5, 1 \leq x \leq 2 \\
\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ |0, otherwise.[/itex]

At minute 8:45, he presents the cumulative distribution as:


[itex]
\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ | 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
[/itex]

I don't understand why F(x) is 1 for [itex]x \geq 2 [/itex], 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.
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Stephen Tashi
#2
May10-12, 10:39 AM
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P: 3,300
Quote Quote by xeon123 View Post

I don't understand why F(x) is 1 for [itex]x \geq 2 [/itex], 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.
xeon123
#3
May11-12, 03:54 AM
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.

Stephen Tashi
#4
May11-12, 07:27 PM
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P: 3,300
Cumulative distributed function example

Quote Quote by xeon123 View Post
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.
HallsofIvy
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May11-12, 07:48 PM
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Quote Quote by xeon123 View Post
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.
You seem to be thinking that "F(3)" is the probability that x is equal to 3. That is not the case. F(X) is the probability that x is less than or equal to 3. Since, by the definition of f(x), x must be less than or equal to 2, x therefore must be less than or equal to 3. F(x)= 1 for any number larger than or equal to 2.

If f(x) is the "probability density function" then [itex]F(X)= \int_{-\infty}^X f(x)dx[/itex] is the probability that x is less than or equal to X.


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