Cumulative distributed function example

AI Thread Summary
The discussion centers on the cumulative distribution function (CDF) and its relationship with the probability density function (PDF). The confusion arises from why F(x) equals 1 for x ≥ 2, despite f(x) being 0 for values outside its defined range. It is clarified that F(x) represents the probability that x is less than or equal to a certain value, not the probability of x being equal to that value. Thus, since all values of f(x) are accounted for up to x = 2, F(x) reaches 1 for any x greater than or equal to 2. Understanding this distinction is crucial for grasping the behavior of cumulative distribution functions.
xeon123
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I was looking to a video about cumulative distribution function () and he show the following function:\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ | 1/4, 0 \leq x \leq1 \\<br /> f(x) =&lt;(x^3)/5, 1 \leq x \leq 2 \\<br /> \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ |0, otherwise.

At minute 8:45, he presents the cumulative distribution as:<br /> \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ | 0, x \leq 0 \\<br /> F(x) = &lt; \frac{1}{4}x, 0 \leq x \leq 1 \\<br /> \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ | \frac{1}{20}(x^4+4), 1 \leq x \leq 2 \\<br /> \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ | 1, \ x \geq 2<br />

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.
 
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xeon123 said:
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.
 
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.
 
xeon123 said:
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.
 
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xeon123 said:
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 F(X)= \int_{-\infty}^X f(x)dx is the probability that x is less than or equal to X.
 
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