How do I compute the summation at the end:
$$E(x) = \sum_{x=1}^\infty x.P(X=x) = \sum_{x=1}^{\infty} x \left(\frac{5}{6}\right)^{x-1}$$
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http://en.wikipedia.org/wiki/Arithmetico-geometric_sequence#Sum_to_infinite_terms – lab bhattacharjee Oct 20 '13 at 15:06
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Many methods here. – David Mitra Oct 20 '13 at 15:09
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Sorry, rewriting my comment. Assuming $P(X=x)$ is a probability function, it can't be the case that $P(X=x)=\left(\frac 56\right)^{x-1}$ since then $P(X=1)=1$ and $P(X=2)=5/6$. – Thomas Andrews Oct 20 '13 at 15:21
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Recall the geometric series: $$\sum_{x=0}^{\infty} a^x = \dfrac1{1-a}$$ Differentiate the above with respect to $a$ on both sides to get $$\sum_{x=0}^{\infty} xa^{x-1} = \dfrac1{(1-a)^2}$$ Now plug in $a=\dfrac56$ to get what you want.
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Thanks a lot. I have been trying geometric method but haven't thought of differentiating. – Joyce Oct 20 '13 at 15:07