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    This is a silly question, but I'm struggling to find anything online.

    Does anyone have a copy of the OCR S1 textbook to hand? I want to check if it has a proof of the expected value of the geometric distribution in it.

    I seem to remember an elegant proof based off the common ratio of a geometric sequence, however the only proofs I can find online that use that idea also use differential calculus. Long story short, the only place I can think of where I would've seen that proof would've been in the OCR S1 textbook, which I no longer own a copy of.

    So, naturally, I'm here to ask if that's even the case, what proof does the book have?
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    (Original post by BecauseFP)
    This is a silly question, but I'm struggling to find anything online.

    Does anyone have a copy of the OCR S1 textbook to hand? I want to check if it has a proof of the expected value of the geometric distribution in it.

    I seem to remember an elegant proof based off the common ratio of a geometric sequence, however the only proofs I can find online that use that idea also use differential calculus. Long story short, the only place I can think of where I would've seen that proof would've been in the OCR S1 textbook, which I no longer own a copy of.

    So, naturally, I'm here to ask if that's even the case, what proof does the book have?
    I will write one by tomorrow evening if you cannot find it or no one else provides it
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    (Original post by BecauseFP)
    This is a silly question, but I'm struggling to find anything online.

    Does anyone have a copy of the OCR S1 textbook to hand? I want to check if it has a proof of the expected value of the geometric distribution in it.

    I seem to remember an elegant proof based off the common ratio of a geometric sequence, however the only proofs I can find online that use that idea also use differential calculus. Long story short, the only place I can think of where I would've seen that proof would've been in the OCR S1 textbook, which I no longer own a copy of.

    So, naturally, I'm here to ask if that's even the case, what proof does the book have?
    I think the following is most elegant

    Let X be geometrically distributed with parameter p (and to remove anambiguity, X is the number of trials*until* successt, so X>0).

    On the first trial, we have success with probability p (in which case we need no extra trials). Otherwise the number of additional trials before success is geometrically distributed with probability p.

    Thus E[X] = 1 + (1-p)E[X], and so ,E[X] = 1/p.
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    (Original post by BecauseFP)
    This is a silly question, but I'm struggling to find anything online.

    Does anyone have a copy of the OCR S1 textbook to hand? I want to check if it has a proof of the expected value of the geometric distribution in it.

    I seem to remember an elegant proof based off the common ratio of a geometric sequence, however the only proofs I can find online that use that idea also use differential calculus. Long story short, the only place I can think of where I would've seen that proof would've been in the OCR S1 textbook, which I no longer own a copy of.

    So, naturally, I'm here to ask if that's even the case, what proof does the book have?
    If DFranklin's proof suffices for your course is all good, if not I can post an elementary one
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    (Original post by TeeEm)
    If DFranklin's proof suffices for your course is all good, if not I can post an elementary one
    The elementary one sounds interesting,
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    (Original post by BecauseFP)
    The elementary one sounds interesting,
    here you go
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    (Original post by BecauseFP)
    The elementary one sounds interesting,
    Another alternative:
    Let X \sim Geo(p)
    Then: E[X]=\displaystyle\sum_{r=1}^{\infty  } rP(X=r)
    But we know that P(X=r)=(1-p)^{r-1}p.
    \Rightarrow E[X]=\displaystyle\sum_{r=1}^{\infty  } r(1-p)^{r-1}p=p\displaystyle\sum_{r=1}^{ \infty} r(1-p)^{r-1}
    Let q=1-p and sub in:
    E[X]=p\displaystyle\sum_{r=1}^{\inft  y} rq^{r-1}
    Notice that:
    \displaystyle\sum_{r=1}^{\infty} rq^{r-1}=\dfrac{d}{dq} \left( \displaystyle\sum_{r=1}^{\infty} q^r \right).
    Aside:
    Spoiler:
    Show
    This term by term differentiation of the infinite sum can be shown to be valid provided the sum:
    \displaystyle\sum_{r=1}^{\infty} q^r converges, which it does for |q|<1.
    Hence we have:
    E[X]=p\dfrac{d}{dq} \left( \displaystyle\sum_{r=1}^{\infty} q^r \right)
    Using the formula for the sum of a geometric series (which applies as |q|<1.
    E[X]=p\dfrac{d}{dq} \left( \dfrac{1}{1-q} \right) =p\dfrac{1}{(1-q)^2}
    Recalling that p=1-q \Rightarrow q=1-p gives us that:
    E[X]=p\dfrac{1}{p^2}=\dfrac{1}{p}. \ \ \ \square
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    Nothing wrong with either of these proofs, but I fail to see why they are more elementary than one that's literally 1 line of calculation (and a little more explanation).
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    (Original post by DFranklin)
    Nothing wrong with either of these proofs, but I fail to see why they are more elementary than one that's literally 1 line of calculation (and a little more explanation).
    In my case I wrote the proof as I needed one to add to my own resources.
    (Now I have two more proofs to add.)
    The OP will decide what is complicated and what is elementary.
    Besides offering many alternatives educates many others watching this thread.
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    (Original post by DFranklin)
    Nothing wrong with either of these proofs, but I fail to see why they are more elementary than one that's literally 1 line of calculation (and a little more explanation).
    What she said:

    (Original post by TeeEm)
    In my case I wrote the proof as I needed one to add to my own resources.
    (Now I have two more proofs to add.)
    The OP will decide what is complicated and what is elementary.
    Besides offering many alternatives educates many others watching this thread.
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    (Original post by HeavisideDelts)
    What she said:
    she?
    I thought I was a he ever since I was a little boy....
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    (Original post by TeeEm)
    she?
    I thought I was a he ever since I was a little boy....
    don't lie to us caitlyn
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    To the two posters above, I don't think DFranklin is saying that the alternatives are wrong or useless, just that he can't see why they are more 'elementary' than his, which I'm agreeing with, the one line proof is very elegant.
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    I forgot to post back and say thanks... rep all round.
 
 
 
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