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Math Help - Finding a probability function R (the remainder when divided by a distribution)

  1. #1
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    Finding a probability function R (the remainder when divided by a distribution)

    Hi everyone, I am really unsure of how to do this question.

    Let X have a geometric distribution with f(x) = p(1-p)^x for x = 0,1,2,....
    Find the probability function of R, the remainder when X is divided by 4.

    I don't know where to start, but I'm suspecting that you would construct a geometric series with r, and find an expression for f(r)... Not really sure.

    Any insight would be helpful.
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  2. #2
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    Re: Finding a probability function R (the remainder when divided by a distribution)

    If 0\le r < 4, then P(R = r) = P(X = r) + P(X = 4 + r) + P(X = 8 + r) + \dots = p(1 - p)^r + p(1 - p)^{4+r} + p(1 - p)^{8+r} + \dots, which is a geometric series.
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  3. #3
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    Re: Finding a probability function R (the remainder when divided by a distribution)

    So taking that geometric series, I can find the sum of the geometric series,which will express P (R = r)?

    That is, we have P (R = r) = p(1-p)^r/ 1-((1-p)^4+r)) ?

    THanks!
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  4. #4
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    Re: Finding a probability function R (the remainder when divided by a distribution)

    Quote Originally Posted by KelvinScale View Post
    That is, we have P (R = r) = p(1-p)^r/ 1-((1-p)^4+r)) ?
    There is no +r in the exponent: P(R = r) =\frac{p(1-p)^r}{1-(1-p)^4}.
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