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Let $(\Omega,\mathcal A,\operatorname P)$ be a probability space, $(\mathcal F_t)_{t\ge0}$ be a filtration on $(\Omega,\mathcal A)$, $E$ be a $\mathbb R$-Banach space and $(X_t)_{t\ge0}$ be an $E$-valued process on $(\Omega,\mathcal A,\operatorname P)$. Remember that $X$ is called $\mathcal F$-Lévy if

  1. $X$ is $\mathcal F$-adapted;
  2. $X_0=0$;
  3. $X_{s+t}-X_s$ and $\mathcal F_s$ are independent for all $s,t\ge0$;
  4. $X_{s+t}-X_s\sim X_t$ for all $s,t\ge0$.

Assume $X$ is $\mathcal F$-Lévy and let $(Y_t)_{t\ge0}$ be another $E$-valued $\mathcal F$-Lévy process on $(\Omega,\mathcal A,\operatorname P)$ (or possibly on another probability space) with $X_t\sim Y_t$ for all $t\ge0$.

Are we able to show that $X$ and $Y$ have the same (finite-dimensional) distribution(s)?

EDIT: To be precise, the question is whether we can show that $$\operatorname P\left[X_{t_1}\in B_1,\ldots,X_{t_k}\in B_k\right]=\operatorname P\left[Y_{t_1}\in B_1,\ldots,Y_{t_k}\in B_k\right]\tag1$$ for all $B_1,\ldots,B_k\in\mathcal B(E)$, $k\in\mathbb N$ and $0\le t_1<\cdots<t_k$ or even $$\operatorname P\left[X\in B\right]=\operatorname P\left[Y\in B\right]\;\;\;\text{for all }B\in\mathcal B\left(E^{[0,\:\infty)}\right)\tag2.$$

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  • $\begingroup$ What is the question? You mean whether it is true that $(X_{t_1},\dots,X_{t_n})\sim (Y_{t_1},\dots,Y_{t_n})$ for arbitrary $t_1,\dots,t_n$, $n\in\mathbb{N}$? Or whether $X\sim Y$ as a random variable in the space of functions $[0,T]\to E$? I don't know if it helps but, for example, in general $X_t\sim Y_t$ for all $t$ does not imply that $\int_0^1 X_tdt \sim \int_0^1 Y_tdt$, so no, $X\sim Y$ (as random variables) is not necessarily true if $X_t\sim Y_t$ for all $t$. Not sure if that was your question. As for the former, you may be able to prove it using independent increments. $\endgroup$
    – Martingalo
    Nov 3, 2020 at 8:10
  • $\begingroup$ @Martingalo I've edited the question. Is it clear now? And couldn't we infer $(2)$ from $(1)$? $\endgroup$
    – 0xbadf00d
    Nov 3, 2020 at 8:16
  • $\begingroup$ Aha thanks, exactly, then I think I answered your question below $\endgroup$
    – Martingalo
    Nov 3, 2020 at 8:21

2 Answers 2

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Not sure what the question is, but in general, $X_t \sim Y_t$ for all $t$ does not imply that $X\sim Y$ as random variables in the (inifnite dimensional) space of functions $[0,T]\to E$. For example, $X_t\sim Y_t$ for all $t$ does not necessarily imply that $\int_0^1 X_tdt \sim \int_0^1 Y_tdt$, for the latter to hold you need the stronger condition that $X\sim Y$ as random variables.

This said, if your question is whether $(X_{t_1},\dots,X_{t_n})\sim (Y_{t_1},\dots,Y_{t_n})$ for arbitrary $t_1,\dots,t_n$, $n\in \mathbb{N}$ (equality of finite dimensional distributions) the answer is yes. Let us just look at $n=2$ for the sake of simpler notation: Take $0\leq s\leq t<\infty$ then,

$$\varphi_{(X_t,X_s)}(u,v) = \mathbb{E}[e^{iuX_t + iv X_s }] = \mathbb{E}[e^{iu(X_t-X_s+X_s) + iv X_s }] = \mathbb{E}[e^{iu(X_t-X_s)}] \mathbb{E}[e^{i(u+v)(X_s)}] = \mathbb{E}[e^{iu(X_{t-s})}] \mathbb{E}[e^{i(u+v)(X_s)}] =\varphi_{X_{t-s}}(u) \varphi_{X_{s}}(u+v),$$ where we used first that $X_t-X_s$ is independent of $X_s$ and then that $X_t-X_s\sim X_s$. Now, you can finally use that $X_t\sim Y_t$ and hence $\varphi_{X_t}(u)=\varphi_{Y_t}(u)$ and go backwards. As a result, $$(X_{t_1},\dots,X_{t_n})\sim (Y_{t_1},\dots,Y_{t_n}),$$ which, as said, is a consequence of the independent and stationary increments.

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  • $\begingroup$ I see a problem in my answer now! I realised that your question regards Banach-valued random variables... in that case the issue is trickier! You can not prove it using characteristic functions... but I assume that the idea is the same, start with Borel sets and I guess that using the same trick you may arrive at the same conclusion. $\endgroup$
    – Martingalo
    Nov 3, 2020 at 8:28
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    $\begingroup$ You can still write $\operatorname P\left[X_{t_1}\in B_1,X_{t_2}\in B_2\mid X_{t_1}\right]=1_{B_1}\left(X_{t_1}\right)\left.\operatorname P\left[X_{t_2-t_1}\in B_2-x\right]\right|_{x\:=\:X_{t_1}}$ a.s. for all $t_2>t_1\ge0$ and $B_1,B_2\in\mathcal B(E)$. $\endgroup$
    – 0xbadf00d
    Nov 3, 2020 at 9:08
  • $\begingroup$ You are right! I was trying to figure out something using indicators as well. Independent increments is stronger than Markov, so there was the trick :) As for (2) I don't think it's true. There are examples where $X_t\sim Y_t$ and $\int_0^1 X_tdt =0$ while $\int_0^1 Y_tdt \neq 0$ but I can't recall one right now. For the latter to also be equal you need $X\sim Y$, so there's the counterexample if you find such a choice of $X$ and $Y$. $\endgroup$
    – Martingalo
    Nov 3, 2020 at 10:31
  • $\begingroup$ We should be able to conclude $(2)$ from $(1)$ using Proposition 3.2 in the book of Kallenberg. $\endgroup$
    – 0xbadf00d
    Nov 3, 2020 at 10:47
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I think the answer whether $(1)$ holds is rather trivial by observing that $X$ is a time-homogeneous Markov process with transition semigroup $$\kappa_t(x,B):=\operatorname P\left[X_t\in B-x\right]\;\;\;\text{for }(x,B)\in E\times\mathcal B(E)\text{ and }t\ge0.$$ From this we know that, for all $k\in\mathbb N$ and $0\le t_0<\cdots<t_k$, $$\left(X_{t_0},\ldots,X_{t_k}\right)\sim\mathcal L\left(X_{t_0}\right)\otimes\bigotimes_{i=1}^k\kappa_{t_i-t_{i-1}}\tag3.$$ So, unless I'm missing something, it's only the question left whether $(2)$ holds as well. But this should be a general fact for any processes over an arbitrary index set.

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