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I'm doing some exercises and came across one that has two parts, as follows:

Given a transition matrix for a Markov Chain, $\mathbf{P}$, and a vector $\mathbf{f}$, $\mathbf{f}$ is harmonic if

$$ \mathbf{f} = \mathbf{P}\mathbf{f}$$

$(a)$ Show that if $\mathbf{f}$ is harmonic, then

$$ \mathbf{f}=\mathbf{P}^n\mathbf{f} $$

for all $n$

$(b)$ Using $(a)$, show that if $\mathbf{f}$ is harmonic,

$$ \mathbf{f} = \mathbf{P}^\infty \mathbf{f} $$

Am I incorrect in assuming that if $(a)$ holds, then $(b)$ holds by necessity? Are there any cases where proving that something holds for all $n$ does not prove that it holds as $n$ tends to infinity?

  • 3
    You need to refer back to the definition of $\textbf{P}^{\infty}$.2012-09-06
  • 1
    If something holds for all $n$, then it clearly holds for $n$ as $n$ tends to $+\infty$. But you don't seem to be asking what happens as $n$ tends to $+\infty$, but instead what happens **at** $+\infty$.2012-09-06
  • 0
    @Hurkyl "If something holds for all $n$, then it clearly holds for $n$ as $n$ tends to $+∞$." What about the counterexample below?2012-09-07
  • 0
    @Peter: It's an example of a statement that holds for every term in any sequence of natural numbers tending towards $+\infty$, but whose (suitably interpreted) **limit** fails.2012-09-07

2 Answers 2

13

Simple counterexample

$$\frac{1}{n}>0\text{ for all }n\in\Bbb N$$

6

The sum $$ \sum_{k = 1}^n \frac{1}{k} $$ is finite for all finite $n$, but $$ \sum_{k = 1}^\infty \frac{1}{k} $$

is infinite.

  • 0
    OK, let me be a nitpicker here, but what does it mean for a finite sum to be convergent?2012-12-03
  • 0
    @PeterTamaroff I agree it is an abuse of terminology. I meant only that the sum of finitely-many finite terms is finite. I've replaced my answer with something more concrete.2012-12-04