A question about PDF of Uniform Distribution

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February 21st 2011, 09:05 AM
ichoosetonotchoosetochoos

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
February 21st 2011, 09:42 AM
Plato

Quote:

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 $\int_{ - \infty }^a {f(x)dx} + \int_a^b {f(x)dx} + \int_b^\infty {f(x)dx}$
February 21st 2011, 09:47 AM
ichoosetonotchoosetochoos

ok that was quite a dumb question, got it now :)
February 21st 2011, 09:49 AM
ichoosetonotchoosetochoos

this rests on the assumption that a is bigger than minus infinity and b is smaller than infinity though
February 21st 2011, 09:52 AM
Plato

Quote:

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.