Please derive the probability density function(pdf) for a continuous
Uniform Distribution

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Request for Question Clarification by mathtalk-ga on 15 Mar 2005 15:53 PST

Hi, bu1234-ga:

In the simplest case the "probability space" for a continuous uniform
distribution would be a finite interval of real numbers, but the idea
of a uniform distribution could be easily extended to any bounded,
measurable open subset of a higher-dimensional space, e.g. a ball of
finite radius in 3 dimensions.

Are you interested in just the one-dimensional (bounded interval)
case, or do these higher-dimensional analogs also enter into your
studies?

regards, mathtalk-ga

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Clarification of Question by bu1234-ga on 15 Mar 2005 18:18 PST

I am interested in the derivation of the pdf of the continuous Uniform
Distribution bounded by the interval 0<x<theta.

I know the pdf to be: f(x)=1/theta.

Please prove that f(x)=1/theta.

Thank you.

Hi, bu1234-ga:

To say a distribution is "uniform" means that equal size subintervals
have an equal probability as "events".  For a continuous distribution
this means that the probability density function is a constant; it
must be equal at all points of the domain, in this case all points of
the interval (0,theta).

Suppose the constant value of the pdf f(x) is c.  We prove that c =
1/theta as a consequence of the probability measure of the entire
interval being 1.

That is:

  Pr[X in (0,theta)] = INTEGRAL f(x) dx OVER (0,theta)

                     = INTEGRAL c dx OVER (0,theta)

                     = c * (theta - 0)

                     = 1

Therefore c = 1/theta is necessary.

Let me know if further clarification of these points would be helpful.


regards, mathtalk-ga

Just as I suspected.  Your verbal answer was clear and concise.  thank you.