# cumulative distribution function problem

#### bryce09

##### New Member
I have this question I am trying to get through but i keep coming into trouble. The question is:

Show that the cumulative distribution function from a uniform distribution of the random variable is Fx(y) = (y-a) / (b-a) for some a < y < b

I've started the question but have become stuck going from here

y
∫ x/(b-a) dx
a

when i integrate i come up with y2-a2/2(b-a). When i'm pretty sure it should come out as (y/b-a)-(a/b-a) which would then solve to y-a/b-a

any advice as to what i'm forgetting to do would help me alot, thanks.

#### vinux

##### Dark Knight
What is the definition of cumulative distribution function?
what is the density function of uniform distribution?

#### bryce09

##### New Member
Is this right? excuse my notation, I hope you understand it

Fx(y)=
= ∫a^y dx/(b-a)
= [x/(1(b-a))]a^y
= [y/(b-a)- a/(b-a)]
= (y-a)/(b-a)

Mod edit: I think this is what you were going for?
$$\int_a^y \frac{1}{b-a}dx$$
$$=\frac{x}{b-a} |_a^y$$
$$=\frac{y}{b-a} - \frac{a}{b-a}$$
$$=\frac{y-a}{b-a}$$

#### vinux

##### Dark Knight
You are right. You could use tex format using $$[/math ]. eg: [math***] \int_{a}^{y}\dfrac{1}{b-a} dx[/math***]. deleting *** gives \( \int_{a}^{y}\dfrac{1}{b-a} dx$$.\)