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Math Help - Expectation a RV with negative binomial distribution

  1. #1
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    Expectation a RV with negative binomial distribution

    Hi,
    In a derivation for the expectation of an RV with a geometric distribution I have the following:


    E(X) = \sum^{\infty}_{k=r} k \binom{k-1}{r-1}p^{r} (1-p)^{k-r} = \frac{r}{p} \sum^{\infty}_{k=r} \binom{k}{r} p^{r+1} (1-p)^{k-r} = \frac{r}{p}


    I have several questions.

    1. What sorcery took place with the binomial coefficient and k, such that k disappeared, r appeared as a factor and its now \binom{k}{r} rather than \binom{k-1}{r-1} ?


    2. Presumably the logic is that \sum^{\infty}_{k=r} \binom{k}{r} p^{r+1} (1-p)^{k-r} = 1. Can anyone show me how to prove this and what class of distribution is this, not binomial right?

    Many thanks in advance. MD.
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  2. #2
    Senior Member
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    Re: Expectation a RV with negative binomial distribution

    For your first question..

    k\binom{k-1}{r-1}p^r

    =\frac{r}{p}\frac{k}{r}\binom{k-1}{r-1}p^{r+1}

    =\frac{r}{p}\frac{k}{r}\frac{(k-1)!}{(k-r)!(r-1)!}p^{r+1}

    =\frac{r}{p}\frac{k!}{(k-r)!r!}p^{r+1}

    =\frac{r}{p}\binom{k}{r}p^{r+1}

    I haven't answered your second question but if you're trying to find E(X) for a geometric distribution I think there is an easier way.

    \displaystyle E(X)=\sum_{i=1}^\infty i p q ^{i-1}

    \displaystyle =p\sum_{i=1}^\infty \frac{d}{dq}q^i

    \displaystyle =p \frac{d}{dq} \sum_{i=1}^\infty q^i

    etc.
    Thanks from Mathsdog
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