Cumulative distribution function

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SUMMARY

The forum discussion centers on the calculation of cumulative distribution functions (CDFs) for two given density functions: a) \( f(x) = 3(1-x^2) \) for \( x \in (0,1) \) and b) \( g(x) = x^{-2} \) for positive real numbers. The user identifies that the CDF derived from the first density function yields a value of 2 at \( x = 1 \), violating the properties of a CDF. Similarly, the second density function leads to an infinite integral, which also contradicts the requirements for a valid probability density function (PDF). The user's teaching assistant (TA) insists there are no errors in the questions posed.

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island-boy
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Once again, I'm having a disagreement with my TA regarding a problem set he gave us.

Here is the exact question, as written:
Find the distribution function associated with the following density functions:
a) [tex]f(x) = 3(1-x^2)[/tex] , x an element of (0,1)
b)[tex]g(x) = x^{-2}[/tex], x an element of positive real numbers

here's where my problem lies.
For a)
if you solve for the cdf, you get:
[tex]F(x) = \int_{0}^{x} 3(1-t^2) dt = [3t - t^3]_{0}^{x} = 3x - x^3[/tex] for x element (0,1)

however, this value of F(x) takes the value of 2 when x = 1, which violates the property of a cdf! Also, if you take the integral of the density from 0 to 1, you will get 2! again a violation of the property of a density function, as the integral should be between 0 and 1.

for b)
if you solve for the cdf, you get
[tex]F(x) = \int_{0}^{x} t^{-2} dt = [-t^{-1}]_{0}^{x} = - 1/x + \infty[/tex] for x element positive real

which again is greater than 1 for any value of x, positive real.

and if you get the integral of the density function from 0 to infinity (as the density function is defined for all positive real), you get infinity! which is not between 0 and 1.

the TA, however said that there is NOTHING wrong with the questions, even after he inspected it.

Am I insane to think that the questions are wrong? or am I not seeing something obvious?

help please. I'm going crazy.
 
Last edited:
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anyone have an idea? thanks.
 

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