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Suppose $X$ is a discrete random variable with possible values $\{1, 2, 3,\dots\}.$ Further, suppose the p.m.f is $$c\left(\frac{1}{x}-\frac{1}{x+1}\right)\enspace\text{s.t. $c > 0$}$$ Find c and $E[X].$

Idea:

We have $$1=\sum_{x=1}^{\infty}c\left(\frac{1}{x}-\frac{1}{x+1}\right)=c\left(\sum_{x=1}^{\infty}\frac{1}{x}-\sum_{x=1}^{\infty}\frac{1}{x+1}\right)$$ But since $\sum_{x=1}^{\infty}\frac{1}{x}$ is a harmonic series, diverges. Thus, there is no value for $c$.

Since it diverges, $E[X]$ does not exist.

Questions:

Is it possible for c not to exist? Did I do a mistake?

Update:

$$1=\sum_{x=1}^{\infty}c\left(\frac{1}{x}-\frac{1}{x+1}\right)=c\sum_{x=1}^{\infty}\left(\frac{1}{x}-\frac{1}{x+1}\right)$$ by telescoping series we have $$1=c\cdot 1$$

So, our p.m.f is $$\left(\frac{1}{x}-\frac{1}{x+1}\right)$$ But, $$E[X]=1\cdot\frac{1}{2}+2\cdot\frac{1}{6}+3\cdot\frac{1}{12}+4\cdot \frac{1}{20}+\dots=\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\dots$$ But that diverges. So, $E[X]$ doesn't exist.

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Hint: Evaluate the series up to a finite $N$ first, then take the limit as $N\to \infty$. The series up to a finite $N$ will be a telescoping series, i.e. most of the terms will cancel, making it easy to evaluate.

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  • $\begingroup$ I see it, I forgot about that series! I will post the solution in a bit but I am seeing that $c=1$? $\endgroup$ – Tomás Palamás Mar 4 '19 at 6:18
  • $\begingroup$ Since $\sum\limits_{k=1}^{N}\left(\frac{1}{k} - \frac{1}{k+1}\right) = 1 - \frac{1}{N+1}\to 1$, you are right. :) $\endgroup$ – Minus One-Twelfth Mar 4 '19 at 6:32
  • $\begingroup$ Thanks. I updated the OP. Is it possible for the expected value not to exist? $\endgroup$ – Tomás Palamás Mar 4 '19 at 6:37
  • $\begingroup$ Yes, it is possible. $\endgroup$ – Minus One-Twelfth Mar 4 '19 at 7:46
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You can write $$\sum a_n-b_n=\sum a_n-\sum b_n$$only if at least one of $\sum a_n$ or $\sum b_n$ is bounded. In this case$$\sum_{x=1}^{\infty}{1\over x}-{1\over x+1}{=\left(1-{1\over 2}\right)\\+\left({1\over 2}-{1\over 3}\right)\\+\left({1\over 3}-{1\over 4}\right)\\+\left({1\over 4}-{1\over 5}\right)\\+\cdots\\=1}$$therefore $$c=1$$ and the rest is easy.

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