# If $X$ and $Y$ are Lévy processes with $X_t\sim Y_t$ for all $t$, can we infer that $X\sim Y$?

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 $$0\le t_1<\cdots or even $$\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.$$

• 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. Nov 3 '20 at 8:10
• @Martingalo I've edited the question. Is it clear now? And couldn't we infer $(2)$ from $(1)$? Nov 3 '20 at 8:16
• Aha thanks, exactly, then I think I answered your question below Nov 3 '20 at 8:21

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.
• 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 $B_1,B_2\in\mathcal B(E)$. Nov 3 '20 at 9:08
• 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 $Y$. Nov 3 '20 at 10:31
• We should be able to conclude $(2)$ from $(1)$ using Proposition 3.2 in the book of Kallenberg. Nov 3 '20 at 10:47
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 $$0\le t_0<\cdots, $$\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.