[ergodicity] an issue in solution 2

[shlff]

Dear @jstac , as I mentioned in PR #81 , there is an issue in the solution 2's proof of lecture ergodicity, even after fixing.

In Exercise 2, we want to prove that the Markov semigroup $(P_t)$ with intensity $$Q=\begin{pmatrix} -\lambda_0 & \lambda_0 & 0 & 0 & \cdots \ 0 & -\lambda_1 & \lambda_1 & 0 & \cdots \ 0 & 0 & -\lambda_2 & \lambda_2 & \cdots\ \vdots & \vdots & \vdots & \vdots & \ddots \end{pmatrix}$$ has no stationary distribution.

After fixing, we have for any $j \geq 1$,

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

However, this brings us a new issue that the statement after this math expression, "It follows that $\phi$ is constant on $\mathbb Z_+$.", may not hold.

• This statement holds if we consider a special case of the pure birth process: when $\lambda_i=\lambda$ for all $i$.

It may require more discussion.

[jstac]

Thanks @shlff , this is very helpful. We will need some mild conditions on the lambda sequence.

(Are you able to propose something a bit weaker than the requirement that they are all constant?)

[shlff]

Thanks @jstac .

How about letting $\lambda_i = \phi(i)$ for all $i$?

[jstac]

Hi @shlff , this doesn't work because the point is to show that no stationary distribution exists for (P_t).

[shlff]

Thanks for your patient waiting, @jstac . I figured out that we can prove it by consider a Yule process $\lambda_i= \lambda i$ for all $i$, where $0<\lambda <\infty$.

Here is my updated proof:

• Suppose to the contrary that $\phi \in \mathcal D$ 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 a bounded sequence in $(0, \infty)$, $\lambda_j >0$ for any $j\geq1$.
• Therefore, it must be either one of the two cases: (i) $\phi(j)=0$ for all $j$ or, (ii) $ \lambda_j \phi(j) = \lambda_{j-1} \phi(j-1)$ for any $j\geq 1$.
• For case (i), since $\lambda_i = \lambda i$, we have $j \phi(j) = (j-1) \phi(j-1)$ or equivalently, $\phi(j) = \frac{j-1}{j} \phi(j-1)$ for any $j \geq 1$.
• From the last equation, we obtain $\phi(j) = \frac{j-1}{j} \frac{j-2}{j-1} \cdots \frac{0}{1} \phi(0)= 0$ for any $j \geq 1$.
• Because of $ \lambda_j \phi(j) = \lambda_{j-1} \phi(j-1)$ for any $j\geq 1$ and the last equation, we also have $\phi(0) = 0$
• For both cases, $\sum_{j\geq 0} \phi(j) = 0$, which contradicts with our assumption $\phi \in \mathcal D$.

Do you think it makes sense?

If we do not want to add that condition but consider a general pure birth process, then we need to turn to the relationship between its explosiveness feature and the existence of its invariant distribution.