Timeline for Construction of a Markov process with prescribed local behavior and state-dependent jump distribution
Current License: CC BY-SA 4.0
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| Aug 10, 2022 at 19:32 | comment | added | Nawaf Bou-Rabee | Write $III = III_1 + III_2 + III_3$ where $III_1 = (f (Y^0_{\tau_1}) - f(x)) 1_{t \ge \tau_1}$, $III_2 = (f(X_t) - f(Y_0^1)) 1_{ \tau_1 \le t \le \tau_2}$ and $III_3 = (f(X_t) - f(Y_0^1)) 1_{ t > \tau_2}$. Then $E[III_1] = E[ \int_0^{\tau_1} Q f(Y^0_s) ds ; t \ge \tau_1] \le t \| Q f \| (1-e^{-t}) = O(t^2)$, $E[III_2] = E[ \int_{\tau_1}^t Q f(Y_{s-\tau_1}^1) ds ; \tau_1 \le t < \tau_2] \le t^2 e^{-t} \| Q f \| = O(t^2)$ $E[III_3] \le \| f\| P( t \ge \tau_2) \le \| f \| (1-e^{-t} -t e^{-t}) = O(t^2)$. Hope that clarifies. | |
| Aug 2, 2022 at 19:51 | comment | added | Nawaf Bou-Rabee | Certainly. Have not forgotten about this. This week is difficult because I have quite a few reports due, but I'll try to add it asap. Thanks for the reminder. | |
| Aug 2, 2022 at 19:34 | comment | added | 0xbadf00d | Do you think you can edit the question and add how we can show $E(III)=O(t^2)$ soon? BTW, am I missing something, or does $$\tilde X_t:=\begin{cases}Y^0_t&\text{, if }t<\tau_1\\Y^1_{t-\tau_1}&\text{, if }\tau_1\le t\end{cases}$$ admit the same generator $X_t$? | |
| Jul 25, 2022 at 14:16 | history | bounty ended | 0xbadf00d | ||
| Jul 14, 2022 at 18:39 | vote | accept | 0xbadf00d | ||
| Jul 14, 2022 at 18:16 | history | undeleted | Nawaf Bou-Rabee | ||
| Jun 11, 2022 at 10:10 | history | deleted | Nawaf Bou-Rabee | via Vote | |
| Jun 10, 2022 at 11:23 | history | edited | Nawaf Bou-Rabee | CC BY-SA 4.0 |
added a remark at the end
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| Jun 9, 2022 at 23:24 | history | edited | Nawaf Bou-Rabee | CC BY-SA 4.0 |
deleted 6 characters in body
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| Jun 9, 2022 at 23:18 | history | edited | Nawaf Bou-Rabee | CC BY-SA 4.0 |
added a bit more detail based on the feedback of the OP
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| Jun 7, 2022 at 15:56 | comment | added | Nawaf Bou-Rabee | I think you are referring to $E[{\rm II}]$. To avoid confusion, maybe $P_x[X_{\tau_1} \in B \mid (X_t)_{0 \le t < \tau_1} ] = P_x[Y_0^1 \in B \mid Y^0_{\tau_1}] = \alpha(Y^0_{\tau_1}, B) = \alpha(X_{\tau_1-},B)$. In any case, I see no reason why the proof given in this answer does not work in the general homogeneous case with state-dependent jump rates, under suitable conditions on the jump rates. | |
| Jun 7, 2022 at 15:48 | comment | added | 0xbadf00d | I intended to write "we immediately obtain $\operatorname E[\text I]=\operatorname E_x\left[\int_0^t(Af)(Y^0_s)\:{\rm d}s\right]$ for all $x\in E$" in my comment above. Do you agree that we should impose the assumption $\operatorname P_x\left[Y^1_0\in B\mid \mathcal F^{Y^0}_{\tau_1}\right]=\alpha\left(Y^0_{\tau_1},B\right)$ for all $(x,B)\in E\times\mathcal E$ instead of the one you imposed before? | |
| Jun 7, 2022 at 14:03 | comment | added | Nawaf Bou-Rabee | Sorry, I misspoke: what you are asking about in the original question is a homogeneous MP with state-independent jump rates: I think the proof that is given applies to this case. However, what it seems you are interested in is the case of state-dependent jump rates. Both cases are not time-inhomogeneous MPs. In the case of state-dependent jump rates, a similar result does indeed hold, except $E[I] = E[ (f(Y^0_t) - f(x)) \exp(-\int_0^t \kappa(Y^0_s) ds)]$ and the additional factor contributes at only $O(t^2)$ provided that $\kappa$ is sufficiently regular, e.g. bounded. | |
| Jun 7, 2022 at 10:49 | comment | added | 0xbadf00d | As I wrote before, I'm mainly interested in the setting described in the paper and the $\tau_i$ defined therein. In that scenario, $Y^0$ is not independent of $\xi_0$. Or am I missing something? | |
| Jun 7, 2022 at 10:46 | comment | added | 0xbadf00d | I think we need to impose a different assumption. Instead of $\operatorname P_x\left[Y^1_0\in B\mid Y^0_{\tau_1}\right]=\alpha\left(Y^0_{\tau_1},B\right)$, we should need to impose the stronger assumption $\operatorname P_x\left[Y^1_0\in B\mid \mathcal F^{Y^0}_{\tau_1}\right]=\alpha\left(Y^0_{\tau_1},B\right)$ for all $(x,B)\in E\times\mathcal E$. With this assumption, we immediately obtain $\operatorname E_x\left[\int_0^t(Af)(Y^0_s)\:{\rm d}s\right]$ for all $x\in E$ without the need of assuming $\operatorname P_x$-independence of $Y^0$ and $\tau_1$. | |
| Jun 7, 2022 at 10:45 | comment | added | Nawaf Bou-Rabee | It is a strong assumption in the proof and I don’t know how to relax it, but it does work for some important special cases, including the time homogeneous cases (your original question) and time inhomogeneous piecewise deterministic Markov processes. | |
| Jun 7, 2022 at 9:20 | comment | added | 0xbadf00d | Well, okay, then it's hard to prove it :) However, I think this assumption is rather strong. In the paper $\tau_1=\inf\left\{t\ge0:\int_0^t\kappa(Y^0_s)\:{\rm d}s\ge\xi_0\right\}$ for some bounded $\mathcal E$-measurable $\kappa:E\to[0,\infty)$ and $\xi_0\sim\operatorname{Exp}(1)$ independent of $Y^0$. With this, $\tau_1$ is clearly not independent of $Y^0$. Can the claim still be proved? | |
| Jun 6, 2022 at 21:13 | comment | added | Nawaf Bou-Rabee | The claim that $Y_t^0$ and $\xi_0$ are independent is an assumption. | |
| Jun 6, 2022 at 21:11 | comment | added | 0xbadf00d | Which assumption is this? | |
| Jun 6, 2022 at 20:50 | comment | added | Nawaf Bou-Rabee | The inference is based on an assumption which really should be made explicit | |
| Jun 6, 2022 at 20:46 | comment | added | 0xbadf00d | For some reason, I've tried to prove independence of $Y^1_t$ from $\tau_1$ which is clearly not what you wrote ... Howerver, from what are you infering the independence of $Y^0_t$ and $\xi_0$? We only that $\xi_0$ is independent from $\xi_1,\xi_2,\ldots$ and $\mathcal L_x(Y^0_t)=\mathcal L_x(Y^0_0)\kappa^0_t$, where $\mathcal L_x(Y^0_t)$ is the distribution of $Y^0_t$ under $\operatorname P_x$ and $(\kappa^0_t)_{t\ge0}$ is the transition semigroup of $Y^0$. Now $\mathcal L_x(Y^0_0)=\delta_x$ ... | |
| Jun 6, 2022 at 20:28 | comment | added | Nawaf Bou-Rabee | Please note that I only said that $Y^0_t$ is independent of $\tau_1$, which is all that is needed to prove $E[I]=…$. | |
| Jun 6, 2022 at 20:24 | comment | added | 0xbadf00d | Okay, we are getting closer. First of all, note that we need that $(Y_t)_{t\ge0}$ is $(\mathcal F^Y_t)_{t\ge0}$-progressive in order to ensure that $Y_\tau$ is $\mathcal F^Y_\tau$-measurable for every $(\mathcal F^Y_t)_{t\ge0}$-stopping time $\tau$. Now I guess we need to assume that $\tau_i$ is an $(\mathcal F^{Y^{i-1}}_t)_{t\ge0}$-stopping time. However, for example, $\text P_x[Y^1_0\in B\mid Y^0_{\tau_1}]=\alpha(Y^0_{\tau_1},B)$ for all $(x,B)∈E×\mathcal E$. Since the right-hand side is clearly not independent of $τ_1$, I don't get how you infer independence of $Y^1_0$ and $τ_1=\xi_0$ ... | |
| Jun 6, 2022 at 19:52 | comment | added | Nawaf Bou-Rabee | Exactly, that is what i meant, and you indeed wrote it correctly/better by conditioning on the pre jump state of the process. | |
| Jun 6, 2022 at 19:52 | comment | added | 0xbadf00d | Don't you mean $\operatorname P_x\left[Y^i_0\in B\mid Y_{\tau_i - \tau_{i-1}}^{i-1}\right]=\alpha(Y_{\tau_i - \tau_{i-1}}^{i-1},B)$ as I suggested before? | |
| Jun 6, 2022 at 19:44 | comment | added | 0xbadf00d | I know that you're an experienced researcher and so I guess I'm just unfamiliar with your notation, but what is the expression on the right-hand side of $P[Y_0^i \in B] = \int_B \alpha(Y_{\tau_i-\tau_{i-1}}^{i-1}, dy)$? There is still a dependence on $\omega\in\Omega$ and if you integrate with respect to the second variable, your integral should simply be $\alpha(Y^{i-1}_{\tau_i-\tau_{i-1}},B)$. | |
| Jun 6, 2022 at 19:38 | comment | added | Nawaf Bou-Rabee | Yes, I’m being a bit lazy: what I mean is that $Law(Y_0^i)(dy)=\alpha(Y_{\tau_i-\tau_{i-1}}^{i-1},dy)$. In other words $P[Y_0^i \in B] = \int_B \alpha(Y_{\tau_i-\tau_{i-1}}^{i-1}, dy)$ — exactly as you wrote above in one of your questions. | |
| Jun 6, 2022 at 19:35 | comment | added | 0xbadf00d | Yes, but that doesn't (immediately) make sense. The law of $Y_0^i$ is a probability measure on $(E,\mathcal E)$ whereas $\alpha(Y_{\tau_i-\tau_{i-1}}^{i-1},\cdot)$ is a Markov kernel with source $(\Omega,\sigma(Y_{\tau_i-\tau_{i-1}}^{i-1}))$ and target $(E,\mathcal E)$. | |
| Jun 6, 2022 at 19:28 | comment | added | Nawaf Bou-Rabee | The law of $Y_0^i$ is $\alpha(Y_{\tau_i-\tau_{i-1}}^{i-1},\cdot)$. | |
| Jun 6, 2022 at 19:26 | comment | added | 0xbadf00d | Thank you for your explanations. Could you clarify what $Y^i_0 \sim \alpha(Y_{\tau_i - \tau_{i-1}}^{i-1}, \cdot)$ precisely means? | |
| Jun 6, 2022 at 19:25 | comment | added | Nawaf Bou-Rabee | (c) a key difference between the setup here and the one in arxiv.org/abs/1910.05037, is that here the jump distribution depends on the current state of the process whereas there it does not, but otherwise, the two constructions look the same. I’m not so confident about the thesis you shared. | |
| Jun 6, 2022 at 19:11 | comment | added | 0xbadf00d | $\operatorname P_x\left[Y^i_0\in B\mid Y_{\tau_i - \tau_{i-1}}^{i-1}\right]=\alpha(Y_{\tau_i - \tau_{i-1}}^{i-1},B)$ for all $x\in E$ and $B\in\mathcal E$? So, for example, $\operatorname P_x\left[Y^1_0\in B\mid Y^0_{\xi_1}\right]=\alpha(Y^0_{\xi_1},B)$ for all $x\in E$ and $B\in\mathcal E$. From this not even $Y^1_0$ should be independent from $\xi_1$ ... So, maybe you've meant that $Y^i$ is $\operatorname P_x$ independent from $\xi_{i-1}$ instead. However, I only see that $Y^i_0$ and $\xi_i$ are .$\operatorname P_x$. | |
| Jun 6, 2022 at 19:11 | comment | added | 0xbadf00d | (c) Hm, the paper you've linked to in an other post (arxiv.org/abs/1910.05037) is claiming that the process is the one described in section 11.1 ... What are we missing? (d) How do you obtain that $Y^i$ (the whole process) is $\operatorname P_x$-independent from $\xi$ for all $x\ni E$? It must stem from your assumption $Y^i_0 \sim \alpha(Y_{\tau_i - \tau_{i-1}}^{i-1}, \cdot)$ under $\operatorname P_x$ for all $x\in E$ and $i\in\mathbb N$. What precisely do you mean by this claim on the distribution? Does it mean | |
| Jun 6, 2022 at 16:31 | comment | added | 0xbadf00d | Thank you for your comments. I will consider the independence thing later. Note that is is chapter 11.1 (not 11.3) which is about the "concatenation of two processe". Chapter 11.3 is about the "concatenation of countably many processes". (a) Isn't the construction in Chapter 11.3 precisely the same as you gave in your answer? (b) If we define your $X_t$ in the same manner as it is done in chapter 11.1; i.e. $X_t=Y^1_t$ if $t<\tau_1$ and $X_t=Y^2_{t-\tau_1}$ if $\ge\tau_1$, shouldn't the resulting process still have the same generator? | |
| Jun 6, 2022 at 16:13 | comment | added | 0xbadf00d | The "transfer kernel" should be our $\alpha$, if I'm not missing something. However, I really struggle to understand why the rather complicated construction described in chapter 11.3 is necessary. If all $X^i$ are the same, doesn't this construction somehow mimic the construction in the proof of the Ionescu-Tulcea theorem from which we can infer the existence of independent processes? If you know something about this stuff, it would be great to hear what you can say. | |
| Jun 6, 2022 at 16:13 | comment | added | 0xbadf00d | BTW, the more general concatenation of Markov processes we have talked about in the other thread is also described in the following paper: google.com/…. It starts in Chapter 11 on p. 55. I'm not 100% sure, but the scenario considered here should be the special case described in chapter 13.1. | |
| Jun 6, 2022 at 16:06 | comment | added | 0xbadf00d | Yes, sorry, that's what I've meant. It also seems like that we need some kind of independence between $Y^i$ and $\tau_{i-1}$, since I don't understand how you obtain $E[\rm{I}] = e^{-t} ( \kappa_t f(x) - f(x) ) $ without that. | |
| Jun 6, 2022 at 16:05 | comment | added | Nawaf Bou-Rabee | Yes, except that I think you meant to say "the distribution of $Y^i_0$ under $P_x$ is $\alpha(Y^{i-1}_{\tau_i - \tau_{I-1}}, \cdot)$". | |
| Jun 6, 2022 at 15:46 | comment | added | 0xbadf00d | Thank you for your answer. Please clarify what you mean when you say that "$\{Y^i\}$ are independent realizations of $Y$ with $Y_0=x$". With respect to which probability measure are the $Y^i$ independent and what is $x$? Do you intend to introduce a family $\operatorname P_x$ of probability measures such that $\{Y^i:i\in\mathbb N_0\}$ is $\operatorname P_x$-indepndent, $\operatorname P[Y^0_0=x]=1$ and the distribution of $Y^i$ under $\operatorname P_x$ is $\alpha(Y_{\tau_i - \tau_{i-1}}^{i-1}, \cdot)$ for all $x\in E$? | |
| Jun 6, 2022 at 13:43 | history | answered | Nawaf Bou-Rabee | CC BY-SA 4.0 |