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Let $p_j\ge0,\ j=1,2,3,\dots,$ and suppose $\sum_j p_j=1.$ Is there a simple proof that $$\sum_{j=1}^\infty{jp_j}\tag{1}$$ converges? My question arises from the answer to this question. Consider a Markov chain with state space $\{1,2,3,\dots\}.$ If the chain is is state $1,$ it transitions to state $j$ with probability $p_j.$ If it is in state $j>1$ then it always transitions to state $j-1$. The chain is irreducible and aperiodic, so it has a unique stationary distribution. The sum $(1)$ arises in computing the stationary probabilities, so it must converge.

I've been trying unsuccessfully to find a more direct proof. There isn't any way to apply standard tests (root test, ratio test, Gauss's test) and I haven't any other ideas. (It's equivalent to the statement that if N is a random variable that takes positive integer values, then $E(N)$ exists, but I don't see how that helps. In fact, my intuition would be that this statement is false.)

EDIT

It has been amply shown that the statement is false. I would like to know the error in the linked question.

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    $\begingroup$ Isn't this math.stackexchange.com/a/520619/42969 a counter-example? $\endgroup$ – Martin R Dec 5 '18 at 16:24
  • $\begingroup$ You are right. What is the error in the linked question? $\endgroup$ – saulspatz Dec 5 '18 at 16:27
  • $\begingroup$ I don't see an argument in the linked question for why this statement should be true, just a claim in a comment that "this problem shows" it to be true. Maybe if you provided more detail as to why you think the problems shows it to be true then someone could find your error. $\endgroup$ – MartianInvader Dec 5 '18 at 17:52
  • $\begingroup$ @MartianInvader The stationary distribution exists by standard theorems on Markov chains, and the OP of the original question has shown how to calculate the the stationary probabilities. If my transformation of the series is correct, then it arises in the formula for those probabilities, and it must converge. So, it must be that the transformation is wrong, or there is an error in the original problem. $\endgroup$ – saulspatz Dec 5 '18 at 17:56
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Not all aperiodic, irreducible Markov processes have a stationary distribution. This is only true for finite state spaces. For infinite spaces, you need the process to be positive recurrent, meaning the expected time to return to a state is finite. Here, starting from $1$, the expected time to return to $1$ is $\sum jp_j$. Therefore, your proof goes in circles; in order for the process to have a stationary distribution, you need $\sum jp_j<\infty$, and in order to prove that, you use that the process has a stationary distribution.

When the list $(p_1,p_2,\dots)$ has too fat a tail, the process will never settle, and instead become more diffuse as time goes on.

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  • $\begingroup$ Thank you. That is what I had forgotten. $\endgroup$ – saulspatz Dec 5 '18 at 17:57
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The statement is false. Put $$ p_j=\frac{6}{\pi^2}\frac{1}{j^2}\quad (j\geq 1) $$ where the constant is for normalization. Let $N$ be distributed according to this pmf. Then $$ \sum_{j=1}^\infty jp_j=EN=\frac{6}{\pi^2}\sum_{j=1}^\infty\frac{1}{j}=\infty $$

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  • $\begingroup$ Can you find the error in the linked question? $\endgroup$ – saulspatz Dec 5 '18 at 16:28
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We have the series $\displaystyle\sum_{j=1}^\infty \frac{1}{j(j+1)} = 1$, but $\displaystyle\sum_{j=1}^\infty \frac{1}{j+1}$ diverges, so your affirmation is false.

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  • $\begingroup$ Can you find the error in the linked question? $\endgroup$ – saulspatz Dec 5 '18 at 16:29
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Another counterexample can be derived from the St. Petersburg paradox. Suppose that $p_j=\frac1j$ if $j$ is a power of $2$, and $0$ otherwise. Then $\sum p_j = \sum 2^{-k}=1$, but $jp_j=1$ whenever $j$ is a power of $2$, and thus $\sum jp_j$ diverges.

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It is known that $$\sum_{j=1}^\infty \dfrac{1}{j^2}=\dfrac{\pi^2}{6}.$$

So if you take $p_j=\frac{6}{(\pi j)^2}$, you have $\sum_{j=1}^\infty p_j=1$ yet $\sum_{j=1}^\infty jp_j=\frac{6}{\pi^2}\sum_{j=1}^\infty\frac{1}{j}=+\infty.$

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  • $\begingroup$ Can you find the error in the linked question? $\endgroup$ – saulspatz Dec 5 '18 at 16:28
  • $\begingroup$ @saulspatz Looked it up. Didn't find any mistake. Rather, for your answer to hold, you must show that $$\lim_{k\to +\infty}\sum_{j=k}^\infty(j-k)p_j=0,$$ which is not obvious to me. $\endgroup$ – Scientifica Dec 5 '18 at 17:10

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