Simple proof of a theorem on convergence of series

Question

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.

Answer 1

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.

Answer 2

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 $\sum_{j=1}^\infty jp_j=\frac{6}{\pi^2}\sum_{j=1}^\infty\frac{1}{j}=+\infty.$

Answer 3

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 $$ \sum_{j=1}^\infty jp_j=EN=\frac{6}{\pi^2}\sum_{j=1}^\infty\frac{1}{j}=\infty $$

Answer 4

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.

Answer 5

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.