# Distribution Funcion

1. Feb 16, 2008

### kuahji

Find the distribution function of the random variable X whose probability density is given by

f(x)= x/2 for x < x $$\leq1$$
1/2 for 1< x $$\leq2$$
(3-x)/2 for 2 < x $$\leq3$$
0 elsewhere

Ok, so I know have to take the integral of each piece, but I'm having a hard time figuring out the limits of the integrals.

F(x)= 0 for x $$\leq0$$
x^2/4 for 0 < x $$\leq1$$
1/4(2x-1) 1 < x $$\leq2$$
1/4(6x-x^2-5) for 2 < x < 3
1 for x$$\geq3$$

Which is about what I got when I took the "indefinite" integrals of each piece (again, not sure how to go about setting the limits), but for example in this section 1/4(6x-x^2-5) for 2 < x < 3 where is the -5 coming from? Any help would be appreciated.

2. Feb 16, 2008

### jhicks

I suppose you mean cumulative distribution function? The cumulative distribution function is given as $$F(x)=\int_{-\infty}^x f(t)dt$$. Because it's a piecewise function, the constants are coming out of summing all the parts of the density less than the current piece you're integrating over. For the part you appear to have questions with, if you were to find F(x) where 2<x<3, you would find F(3) (definite integral) and then add it to $$\int_{2}^x f(t)dt$$

(Post isn't updating quickly for me so I don't know if this turned out right with my rampant typos)

Last edited: Feb 16, 2008
3. Feb 16, 2008

### kuahji

Right, that much was in the text, but I'm still a bit confused.

x^2/4 for 0 < x $$\leq1$$ I believe from the original step I just take the integral from 0 to x.

But for the next step, I'm not sure what to do. I mean I know its suppose to add up to one, but figuring out the second and consecutive steps is where I'm having the problem.

4. Feb 16, 2008

### jhicks

The cumulative distribution function keeps track of the total probability found between $$-\infty$$ and x. In a sense it is a "running total". If you want to find F(x) for 1<x<2, you have F(1)=1/4, then $$F(x)= \int_{-\infty}^x f(t)dt = \int_{-infty}^{1}f(t)dt+\int_{1}^{x}f(t)dt = \frac{1}{4}+\int_{1}^{x}\frac{1}{2}dt = \frac{1}{4}+\frac{x}{2}-\frac{1}{2} = \frac{1}{4}(2x-1)$$

Edit: fixing typos

Last edited: Feb 16, 2008
5. Feb 16, 2008

### kuahji

Ok, I see how it works now. Thanks for the help, it was really appreciated. ^_^