# Semi-group property of branching processes?

See the edit below!

I do have a question about continuous-state branching processes while reading "Fluctuations of Lévy Processes with Applications" by Kyprianou:

Definition (Continuous-state branching process). A $$[0, \infty]$$-valued strong Markov process $$Y=\{Y_t : t \geq 0\}$$ with probabilities $$\{ P_x : x \geq 0 \}$$ is called a continuous-state branching process if it has càdlàg paths and satisfies the branching property.

Let $$Y$$ be a continuous-state branching process. Then it is shown that the semi-group property $$u_{t+s}(\theta) = u_t(u_s(\theta))$$ is true where $$u_r(\theta)$$ fulfills the following equation: $$E_x(e^{- \theta Y_r}) = e^{-xu_r(\theta)} \tag{\star} .$$ One step in the proof states $$E_x(E(e^{-\theta Y_{t+s}}|Y_t)) = E_x(e^{-Y_t u_s(\theta)}).$$ I don't understand this step. What I tried to do was the following: $$E(e^{-\theta Y_{t+s}}|Y_t) = E(e^{-\theta (Y_{t+s} - Y_t)} e^{-\theta Y_t}|Y_t) = E(e^{-\theta (Y_{t+s} - Y_t)}|Y_t) e^{-\theta Y_t} = E_{K}(e^{-\theta (\tilde{Y}_s - K)}|Y_t) e^{-\theta Y_t}$$ Here, I used in the second equality measurability. In the third equality, I want to use something of the form that $$Y_{t+s} - Y_t$$ is distributed like $$\tilde{Y}_s$$. But I don't want to start $$\tilde{Y}$$ at zero since such a branching process would then always be zero. As an alternative I thought of something like $$\tilde{Y}_s - K$$ where $$\tilde{Y}$$ is started at $$K = Y_t$$. However, I guess I am doing something wrong since I think we don't have any time homogeneity that I used and furthermore what I am doing with $$K$$ feels also weird but I don't know what else I should be doing since the process $$Y_s$$ is always zero if it is started at zero.

Assuming, however this is correct, I do can continue in the following way: $$E_{K}(e^{-\theta (\tilde{Y}_s - K)}|Y_t) e^{-\theta Y_t} = E_{Y_t}(e^{-\theta \tilde{Y}_s}|Y_t) e^{\theta K} e^{-\theta Y_t} = e^{-\tilde{Y}_t u_s(\theta)}$$ which shows the desired step.

So my question is how this step is done properly and maybe also what my misconceptions are? Thanks a lot for your help.

Edit:

I was told that from the Markov property we can follow the equation $$E(e^{-\theta Y_{t+s}}|Y_t) = E_{Y_t} (e^{-\theta Y_s})$$ and using $$(\star)$$ the desired result follows.

However, I don't understand this step with the Markov property formally. Intuitively, it makes sense that the equation is true since if we know $$Y_t$$ (or $$\mathcal{F}_t$$ by the Markov property), the process has only time s to evolve from this, so this describes the same situation when would start a process at $$Y_t$$ and let it again evolve for time s. However, this sounds to me like some time homogeneity is needed and I don't see how this follows formally from the (strong) Markov property, i.e.: $$P(Y_{t+s} \in B | \mathcal{F}_t) = P(Y_{t+s} \in B | Y_t).$$

So, I am wondering what I am missing/don't understand correctly in this situation. Thank you.

• The equation under Edit: just uses two properties, firstly independence of increments and secondly time homogeneity. Levy processes per definition satisfy that property. Feb 3, 2023 at 14:16
• Thanks for the comment. However, the process $Y$ is "only" a continuous-state branching process as far as I understood and hence we cannot use results for Levy processes or what am I missing? I also edited the question to make the definition of a continuous-state branching process more visible now.
– tor
Feb 3, 2023 at 14:42
• Alright, I am not quite sure what a branching process means, but from reading Wikipedia I see branching processes are random walks, i.e. sums of iid random variables (at least in discrete time they are). Such random walks are clearly Levy processes. Feb 3, 2023 at 15:18
• No, i don't think that is the case. The chapter, which this question is also about, concerns itself with a time transformation of continuous-state branching processes to get (spectrally positive) Levy processes. Hence, without this time transformation they shouldn't be Levy processes themself.
– tor
Feb 4, 2023 at 13:10

If $$(X_t)_{t \geq 0}$$ is any Markov process, then the very definition of the Markov property is that \begin{align} \tag{0} E_x[f(X_{t+s})|X_t] = E_{X_t}[f(X'_s)]. \end{align} where $$(X'_s)_{s \geq 0}$$ is a copy of the Markov process starting from $$X_t$$ under $$E_{X_t}$$.

We will use both the Markov property and the branching property (i.e. $$E_x[e^{-\theta Y_t} ] = e^{ - xu_t(\theta)}$$) to prove that $$u$$ satisfies the semigroup property.

As you observe, by the tower property we have $$$$\tag{1} E_x[e^{-\theta Y_{t+s}} ] = E_x[E_x[e^{-\theta Y_{t+s} } | Y_t]]$$$$

By the Markov property, $$$$\tag{2} E_x[e^{-\theta Y_{t+s} } | Y_t] = E_{Y_t}[e^{ - \theta Y'_s}],$$$$ where $$(Y'_s)_{s \geq 0}$$ is an independent copy of the CSBP. Both sides of (2) refer to $$\mathcal{F}_t$$-measurable random variables, where $$(\mathcal{F}_s)_{s \geq 0}$$ is the underlying filtration. ((2) is the special case of (0) for $$f(y) = e^{ - \theta y}$$.)

Using the definition of $$u_s(\theta)$$, we have $$E_{Y_t}[e^{ - \theta Y'_s}] = e^{ - Y_t u_s(\theta)}$$, so that plugging this fact into (2) we obtain $$$$\tag{3} E_x[e^{-\theta Y_{t+s} } | Y_t] = e^{-Y_tu_s(\theta)}.$$$$ Plugging (3) into (1) we have $$$$\tag{4} E_x[e^{-\theta Y_{t+s}} ] = E_x[e^{-Y_tu_s(\theta)}].$$$$ Writing $$\lambda := u_s(\theta)$$, the right-hand-side of (4) is $$E_x[e^{-\lambda Y_t}]$$, which is by definition of $$u_t(\lambda)$$ equal to $$e^{ - x u_t(\lambda)}$$. Substituting in $$\lambda := u_s(\theta)$$ we finally obtain $$$$\tag{4} E_x[e^{-\theta Y_{t+s}} ] = e^{ - xu_s(u_t(\theta))},$$$$ as required.

• Thanks for your answer. I guess I am just still confused regarding the definition of the Markov property. Wikipedia for example says that the Markov property says:\begin{align} E[f(X_{t+s})|\mathcal{F}_t] = E[f(X_{t+s})|X_t], \end{align} which doesn't match what you are saying I think. Or how are they equivalent?
– tor
Feb 5, 2023 at 17:27