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    Completely stuck on this question and was wondering if anyone could give me some help. I've tried to find E[Nk] but it's proving hard to turn it into the answer shown.
    If I use the general Geometric Distribution formula for E[Nk] then I lose all my i's, however I've also worked through it using the summation of i x P(Nk=i) but I just get stuck T_T

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    So you really want to prove that if X ~ Geo(p), then E[X] = 1/p.

    E[X] = \displaystyle \sum_{i=1}^{\infty} i \cdot p(1-p)^{i-1}.

    Hint: we have \displaystyle \sum_{i=0}^{\infty} x^i = \frac{1}{1-x} -- try differentiating both sides with respect to x. The same idea will work for the variance.
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    I can only calculate E[Nk] as being summation from i = 1 to infinity of just "i" as opposed to 1/i (I'm guessing it's 1/i because than E[N] would just be "n" lots of them.
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    I really don't know what you mean.
    I'll try and work through the first part:

    !
    We have N = N_1 + ... + N_k \implies \mathbb{E}[N] = \mathbb{E}[N_1 + ... + N_k] = \mathbb{E}[N_1] + ... + \mathbb{E}[N_k] by linearity of expectation.
    So I said \displaystyle \sum_{i=0}^{\infty} x^i = \frac{1}{1-x}. Then differentiating both sides yields \displaystyle \sum_{i=1}^{\infty} ix^{i-1} = \frac{1}{(1-x)^2}. Now, if we choose x = 1-p, this gives \sum i(1-p)^{i-1} = p^{-2} \implies \sum ip(1-p)^{i-1} = \frac{1}{p}. Hooray! So letting p = p_k = (n+1-k)/n, we get
    E[N] = 1/p_1 + 1/p_2 + ... 1/p_n = \displaystyle \sum_{i=1}^n \frac{1}{p_i} = \sum_{i=1}^n \frac{n}{n+1-i} = n \sum_{i=1}^n \frac{1}{n+1-i} = .... Can you finish this off, and the variance too?
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    (Original post by Glutamic Acid)
    I really don't know what you mean.
    I'll try and work through the first part:

    !
    We have N = N_1 + ... + N_k \implies \mathbb{E}[N] = \mathbb{E}[N_1 + ... + N_k] = \mathbb{E}[N_1] + ... + \mathbb{E}[N_k] by linearity of expectation.
    So I said \displaystyle \sum_{i=0}^{\infty} x^i = \frac{1}{1-x}. Then differentiating both sides yields \displaystyle \sum_{i=1}^{\infty} ix^{i-1} = \frac{1}{(1-x)^2}. Now, if we choose x = 1-p, this gives \sum i(1-p)^{i-1} = p^{-2} \implies \sum ip(1-p)^{i-1} = \frac{1}{p}. Hooray! So letting p = p_k = (n+1-k)/n, we get
    E[N] = 1/p_1 + 1/p_2 + ... 1/p_n = \displaystyle \sum_{i=1}^n \frac{1}{p_i} = \sum_{i=1}^n \frac{n}{n+1-i} = n \sum_{i=1}^n \frac{1}{n+1-i} = .... Can you finish this off, and the variance too?
    I had all of this working in my notes, so I'm getting somewhere but this was as far as I got. The question seems to indicate that n \sum_{i=1}^n \frac{1}{n+1-i} = n \sum_{i=1}^n \frac{1}{i} but how can that be right?
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    (Original post by TomLeigh)
    I had all of this working in my notes, so I'm getting somewhere but this was as far as I got. The question seems to indicate that n \sum_{i=1}^n \frac{1}{n+1-i} = n \sum_{i=1}^n \frac{1}{i} but how can that be right?

    Write it out and you'll see: it's the same sum, but from different sides.
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    Doh :P
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    For the variance are we able to use the simple formula of 1-p/p^2 or should we look more closely at E[N^2]-(E[N])^2
 
 
 
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