[ergodicity]fix_proof_ex2

[shlff]

Dear @jstac , thanks for the inspiring talk and really nice discussions with me.

As we discussed, for lecture ergodicity, this PR adds non-increasing condition on $(\lambda_k)$

• "Let $(\lambda_k)$ be a bounded sequence in $(0, \infty)$." -->> "Let $(\lambda_k)$ be a bounded non-increasing sequence in $(0, \infty)$."

and corrects the solution to exercise 2 as:

"Suppose to the contrary that $\phi \in \dD$ and $\phi Q = 0$.

Then, for any $j \geq 1$,

$$ (\phi Q)(j) = \sum_{i \geq 0} \phi(i) Q(i, j) = - \lambda_j \phi(j) + \lambda_{j-1} \phi(j-1) = 0$$

Since $(\lambda_k)$ is non-increasing, it follows that

$$ \frac{\phi(j)}{\phi(j-1)} = \frac{\lambda_{j-1}}{\lambda_j} \geq 1 $$

Therefore, for any $j\geq 1$, it must be either one of two cases:

1. $\phi(j)= \phi(j-1)$ or
2. $\phi(j) \geq \phi(j-1)$.

For case $1$, it follows that $\phi$ is constant on $\ZZ_+$.

For case $2$, it follows that $\phi$ is increasing on $\ZZ_+$.

But $\dD$ contains no constant or increasing functions when the state space is infinite.
(Why?)

Contradiction."

Look forward to your comments.

This PR fixes #82 .

[jstac]

Thanks @shlff , nice work.

Your "case 1" is a special case of "case 2" and hence, from a logical perspective, does not need to be handled separately.

Could you please drop case 1 and rewrite appropriately?

[shlff]

Dear @jstac , thanks for pointing this out. This point also bothered me yesterday.

I will update you very soon.

[shlff]

Dear @jstac , thanks for your patience.

Now the solution to Exercise 2 is modified as:

"Suppose to the contrary that $\phi \in \dD$ and $\phi Q = 0$.

Then, for any $j \geq 1$,

$$ (\phi Q)(j) = \sum_{i \geq 0} \phi(i) Q(i, j) = - \lambda_j \phi(j) + \lambda_{j-1} \phi(j-1) = 0$$

Since $(\lambda_k)$ is non-increasing, it follows that

$$ \frac{\phi(j)}{\phi(j-1)} = \frac{\lambda_{j-1}}{\lambda_j} \geq 1 $$

Therefore, for any $j\geq 1$, it must be:

$$\phi(j) \geq \phi(j-1)$$

It follows that $\phi$ is non-decreasing on $\ZZ_+$.

But $\dD$ contains no non-decreasing functions when the state space is infinite.
(Why?)

Contradiction."

I look forward to your comments.

[jstac]

This is perfect. Very nice job @shlff

[shlff]

This is perfect. Very nice job @shlff

Great thanks, @jstac .

I was very worried when you introduced this lecture series yesterday. I feel a bit relieved now.