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Q: Probability and Statistics ( Answered ,   0 Comments )
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 Subject: Probability and Statistics Category: Science > Math Asked by: bu1234-ga List Price: \$10.00 Posted: 15 Mar 2005 13:19 PST Expires: 14 Apr 2005 14:19 PDT Question ID: 495190
 ```Please derive the probability density function(pdf) for a continuous Uniform Distribution``` 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``` 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
 ```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```
 bu1234-ga rated this answer: `Just as I suspected. Your verbal answer was clear and concise. thank you.`