Determining the distribution function

[shan]

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:
f_{X} (x) = \frac{1}{6}x for 0<x\leq2
= \frac{1}{3}(2x-3) 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:
\int \frac{1}{6}u du = \frac {1}{6} \frac{x^2}{2} = \frac{x^2}{16}
for 0<x<=2

I have a big problem with the second, part. This is what I did (integrating from 2 to x):
\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}
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?

 

[Hurkyl]

The distribution function is given by:

F_X(x) := \int_{-\infty}^x f_X(x) \, dx

So when you computed \int_2^x f_X(x) \, dx, you computed the wrong thing, and there's no reason you should have gotten the right answer.

P.S. 6*2 is not 16.

 

[shan]

P.S. 6*2 is not 16.

whoops sorry, typo that first part is
\int_0^x \frac{1}{6}u du = \frac {1}{6} \frac{x^2}{2} = \frac{x^2}{12}

The distribution function is given by:

F_X(x) := \int_{-\infty}^x f_X(x) \, dx

So when you computed \int_2^x f_X(x) \, dx, you computed the wrong thing, and there's no reason you should have gotten the right answer.

which shows I don't really understand what I'm doing but now that you mentioned it...

is the second part then given by \int_0^2 \frac{1}{6}u du + \int_2^x \frac{1}{3} (2u-3) du? ie I forgot to add the first part of f(x)?

 

[Hurkyl]

Right. Of course, there should also be a \int_{-\infty}^0 f_X(u) \, du component as well. (But you know it's zero, so I suppose that's why you left it out)

 

[shan]

Thank you very much for your help :)