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# Probability - Random Variables Watch

1. 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

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

E[X] = .

Hint: we have -- try differentiating both sides with respect to x. The same idea will work for the variance.
3. 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.
4. I really don't know what you mean.
I'll try and work through the first part:

!
We have by linearity of expectation.
So I said . Then differentiating both sides yields . Now, if we choose x = 1-p, this gives . Hooray! So letting p = p_k = (n+1-k)/n, we get
E[N] = 1/p_1 + 1/p_2 + ... 1/p_n = . Can you finish this off, and the variance too?
5. (Original post by Glutamic Acid)
I really don't know what you mean.
I'll try and work through the first part:

!
We have by linearity of expectation.
So I said . Then differentiating both sides yields . Now, if we choose x = 1-p, this gives . Hooray! So letting p = p_k = (n+1-k)/n, we get
E[N] = 1/p_1 + 1/p_2 + ... 1/p_n = . 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 but how can that be right?
6. (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 but how can that be right?

Write it out and you'll see: it's the same sum, but from different sides.
7. Doh :P
8. 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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Updated: December 9, 2010
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