- #1
shan
- 57
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I've gotten a weird answer after doing the problem but I'm stuck as to where I messed up.
The density function is this:
[tex]f_{X} (x) = \frac{1}{6}x[/tex] for [tex]0<x\leq2[/tex]
[tex] = \frac{1}{3}(2x-3)[/tex] for 2<x<3
and 0 otherwise
And the question is to find the distribution function.
So integrating for the first part from 0 to x:
[tex]\int \frac{1}{6}u du = \frac {1}{6} \frac{x^2}{2} = \frac{x^2}{16}[/tex]
for 0<x<=2
I have a big problem with the second, part. This is what I did (integrating from 2 to x):
[tex]\int \frac{1}{3} (2u-3) du = \frac{1}{3} [u^2-3u] = \frac{1}{3} (x^2-3x - (2^2-6)) = \frac{x^2}{3} - x + \frac{2}{3}[/tex]
for 2<x<3
I know my answer for the second part is wrong as when x=2, the distribution function = 0 and when x=3, the distribution function = 2/3. But the distribution shouldn't be broken up like that at x=2 and supposedly at x=3, it should = 1. So did I forget to do something to the end points or did I not integrate properly?
The density function is this:
[tex]f_{X} (x) = \frac{1}{6}x[/tex] for [tex]0<x\leq2[/tex]
[tex] = \frac{1}{3}(2x-3)[/tex] for 2<x<3
and 0 otherwise
And the question is to find the distribution function.
So integrating for the first part from 0 to x:
[tex]\int \frac{1}{6}u du = \frac {1}{6} \frac{x^2}{2} = \frac{x^2}{16}[/tex]
for 0<x<=2
I have a big problem with the second, part. This is what I did (integrating from 2 to x):
[tex]\int \frac{1}{3} (2u-3) du = \frac{1}{3} [u^2-3u] = \frac{1}{3} (x^2-3x - (2^2-6)) = \frac{x^2}{3} - x + \frac{2}{3}[/tex]
for 2<x<3
I know my answer for the second part is wrong as when x=2, the distribution function = 0 and when x=3, the distribution function = 2/3. But the distribution shouldn't be broken up like that at x=2 and supposedly at x=3, it should = 1. So did I forget to do something to the end points or did I not integrate properly?
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