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We are given $\xi$ random discrete variable, which takes values $> 1$.
$P(\xi = k) = \frac C {k \cdot (k+1) \cdot (k+2)}$, where C is some constant. I'm obliged to find expected value of $\xi$.
$E(\xi) = \sum_{k=2}^{\infty} k \cdot P(\xi = k) = \sum_{k=2}^{\infty} \frac C {(k+1) \cdot (k+2) }$

Clearly, this series converges: Let's take $\sum_{k=2}^{\infty} \frac 1 {k^2}$
$\lim_{k \to \infty} \frac {k^2} {(k+1) (k+2)} = C$, so $E(\xi)$ converges as well as $\sum_{k=2}^{\infty} \frac 1 {k^2}$.

Now find expected value. $\sum_{k=2}^{\infty} \frac C {(k+1)(k+2) } = \sum_{k=2}^{\infty} \frac C {(k+1)} - \frac C {(k+2)}$
$a_1 = \frac C 3 - \frac C 4$
$a_2 = \frac C 4 - \frac C 5$
$a_{n-1} = \frac C {n+1} - \frac C {n+2}$
$a_{n} = \frac C {n+2} - \frac C {n+3}$
Hence $S_n = \frac C 3 - \frac C {n+3}$ and $S = \lim_{k \to \infty} \frac C 3 - \frac C {n+3} = \frac C 3 = E(\xi)$
I would appreciate if you guys check my solution. I'm not confident with it because task contain indication that result should be integer or real number, but I have a dependency on C here. It looks like it isn't possible to get rid of C here, but still this indication got me confused. Thank you for your time.

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  • $\begingroup$ I'm not an expert at all. I was just wondering if $\sum_{k=2}^{\infty} P(\xi = k)$ must equal 1. Then you could get rid of C $\endgroup$ – Thanos Darkadakis Apr 3 '14 at 8:27
  • $\begingroup$ @ThanosDarkadakis I took some ideas from here: math.stackexchange.com/questions/350363/…. You are implying that sum of probabilities of all probable outcomes equal 1? I can agree with that. But how it affects expected value? $\endgroup$ – wf34 Apr 3 '14 at 8:34
  • $\begingroup$ If the sum of all probabilities is 1, then you can find out that C=12 $\endgroup$ – Thanos Darkadakis Apr 3 '14 at 10:20
  • $\begingroup$ @ThanosDarkadakis The sum of all probabilities is always 1. that's what 'all' stands for, right?. I'm also not an expert and I'm having trouble understanding your reply. Would you please provide more explicit comment on how we get C=12? $\endgroup$ – wf34 Apr 3 '14 at 10:35
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$P(\xi = k) = \frac C {k \cdot (k+1) \cdot (k+2)}=\frac{C}{2}(\frac{1}{k}-\frac{2}{k+1}+\frac{1}{k+2}) $

  • $P(\xi=2)=\frac{C}{2}(\frac{1}{2}-\frac{2}{3}+\frac{1}{4}) $
  • $P(\xi=3)=\frac{C}{2}(\frac{1}{3}-\frac{2}{4}+\frac{1}{5}) $
  • $P(\xi=4)=\frac{C}{2}(\frac{1}{4}-\frac{2}{5}+\frac{1}{6}) $
  • $P(\xi=5)=\frac{C}{2}(\frac{1}{5}-\frac{2}{6}+\frac{1}{7}) $
  • ...

If you add all this (in order to get $\sum_{k=2}^{\infty} P(\xi = k)$) you will see that you can remove the diagonals (because they equal zero. ex $\frac14 - \frac24 +\frac14$). This leaves you with:

$\frac{C}{2}(\frac12-\frac23+\frac13)=\frac{C}{12}$

Since $\sum_{k=2}^{\infty} P(\xi = k)=1$, then $\frac{C}{12}=1$, so C=12.

So your result $\frac{C}{3}=\frac{12}{3}=4$, which is an integer.

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