Math Help - A question about PDF of Uniform Distribution

1. A question about PDF of Uniform Distribution

Hi, this is probably ultra trivial, but here it goes

We know that $\int_{-\infty }^{\infty}f(x)dx=\int_{-\infty}^{c}f(x)dx+\int_{c}^{\infty}f(x)dx$, where $\int_{-\infty}^{c}f(x)dx=\lim_{a->-\infty}\int_{a}^{c}f(x)dx$etc.

Now pdf for uniform dist is defined as:

$f(x)=\left\{\begin{matrix}
\frac{1}{b-a}\: \! ( \: a\leq x\leq b)\\
0 \; (x>b \; or \; x<
a)
\end{matrix}\right.$

so I apply the above formula, which yields 1+1 = 2

i.e. $\lim_{a->-\infty}\int_{a}^{c}f(x)dx = \lim_{a->-\infty}\left ( \frac{c-a}{b-a} \right )= (L'Hopital)\lim_{a->-\infty}\left ( \frac{1}{1} \right )=1$
and similarly 1 for the other one as well.

But all the probability books I'm checking says that $\int_{-\infty }^{\infty}f(x)dx=1$
Can anybody explain me what am I doing wrong?

Thanks

2. Originally Posted by ichoosetonotchoosetochoos
$f(x)=\left\{\begin{matrix}
\frac{1}{b-a}\: \! ( \: a\leq x\leq b)\\
0 \; (x>b \; or \; x<
a)
\end{matrix}\right.$

$\int_{-\infty }^{\infty}f(x)dx=1$
Evaluate $\int_{ - \infty }^a {f(x)dx} + \int_a^b {f(x)dx} + \int_b^\infty {f(x)dx}$

3. ok that was quite a dumb question, got it now

4. this rests on the assumption that a is bigger than minus infinity and b is smaller than infinity though

5. Originally Posted by ichoosetonotchoosetochoos
this rests on the assumption that a is bigger than minus infinity and b is smaller than infinity though
Look at the definition of uniform distribution very carefully.