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January 3rd, 2017, 05:59 AM   #1
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Prove CDF for Discrete Uniform Distributions

We have Discrete Uniform Distributions; how can prove CDF for this?
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Last edited by skipjack; January 12th, 2017 at 04:45 AM.
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January 12th, 2017, 04:24 AM   #2
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$\displaystyle \sum_{s=a}^x \frac{1}{b-a+1} = \frac{1}{b-a+1} \sum_{s=a}^x 1 = \frac{1}{b-a+1} \cdot \frac{(x+a)(x-a)}{2} = \frac{(x+a)(x-a)}{2(b-a+1)} $
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