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Proposition: Let us consider a Brownian motion $W(t)$, $t\geq0$. For fixed $t_0\geq0$, the stochastic process $\widetilde{W}(t)=W(t+t_0)-W(t_0)$ is also a Brownian Motion.

Proof: Let us take properties $1.$, $3.$ and $4.$ in here for granted as to stochastic process $\widetilde{W}(t)$ and focus on the proof of property $2.$ for $\widetilde{W}(t)$. First, consider that for any $s<t$: $$\widetilde{W}(t)-\widetilde{W}(s)=W(t+t_0)-W(s+t_0)\tag{1}$$ To check property $2.$, we may assume that $t_0>0$.
Then, for any $0\leq t_1<t_2<\ldots<t_n$, we have $0<t_0\leq t_1+t_0<\ldots<t_n+t_0$. By property $2.$ for $W(t)$, $W(t_k+t_0)-W(t_{k-1}+t_0)$, $k=1,2,\ldots,n$ are independent random variables. Thus, by $(1)$, the random variables $\widetilde{W}(t_k)-\widetilde{W}(t_{k-1})$, $k=1,2,\ldots,n$ are independent and so $\widetilde{W}(t)$ satisfies property $2.$.


Starting from $0<t_0\leq t_1+t_0<\ldots<t_n+t_0$, I would say that property $2.$ does not necessary apply for $W_t$ in correspondence of $t=1$, since, given the above assumptions, I cannot be sure that, considering $W(t_1+t_0)-W(t_0+t_0)$, $(t_1+t_0)>(t_0+t_0)$, that is, in other words, that $W(t_1+t_0)-W(t_0+t_0)$ can be actually considered as an increment in time.



So, I would say that there is a missing assumption so as that $(1)$ holds true for all $k\geq1$ and not just for $k>1$, that is $$t_1>t_0\tag{2}$$ Would you agree with me as to the fact that assumption $(2)$ is necessary so as that $(1)$ holds true for $k=1$? If not, why am I mistaken?

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  • $\begingroup$ We know, by assumption, that $t_0>0$. Indeed, suppose (for instance) $t_1=0$ and $t_0=1.2$. With such values, you would have that $(t_1+t_0)<(t_0+t_0)$, so $W(t_1+t_0)-W(t_0+t_0)$ would not be an increment in time of the Brownian motion $W$ @TSF $\endgroup$ Sep 25, 2020 at 13:12
  • $\begingroup$ Sorry, I cannot get it. Could you please explicit why, with $k=1$, could I consider $W(t_1+t_0)-W(t_0+t_0)$ as an increment in time? @TSF $\endgroup$ Sep 25, 2020 at 14:14
  • $\begingroup$ I just realized there is probably an error in your definition after all. Where did you find this "proof"? $\endgroup$ Sep 25, 2020 at 16:26
  • $\begingroup$ It is from "Introduction to Stochastic Integration" by Kuo (2006). Are you referring to the same error I am pointing at? @TSF $\endgroup$ Sep 25, 2020 at 18:36
  • $\begingroup$ It's more than that. Look at how increments are defined on the wikipedia page. $\endgroup$ Sep 26, 2020 at 19:02

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I think you should just rename $t_0$ to be $\tau$ and see if it doesn't fix things. You seem to be mixing the translating in time with the increments in time themselves.

Proof: Let us take properties $1.$, $3.$ and $4.$ in here for granted as to stochastic process $\widetilde{W}(t)$ and focus on the proof of property $2.$ for $\widetilde{W}(t)$. First, consider that for any $s<t$: $$\widetilde{W}(t)-\widetilde{W}(s)=W(t+\tau)-W(s+\tau)\tag{1}$$ To check property $2.$, we may assume that $\tau>0$.
Then, for any $0\leq t_0<t_1<\ldots<t_n$, we have $0<\tau\leq t_0+\tau<t_1 + \tau<\ldots<t_n+\tau$. By property $2.$ for $W(t)$, $W(t_k+\tau)-W(t_{k-1}+\tau)$, $k=1,2,\ldots,n$ are independent random variables. Thus, by $(1)$, the random variables $\widetilde{W}(t_k)-\widetilde{W}(t_{k-1})$, $k=1,2,\ldots,n$ are independent and so $\widetilde{W}(t)$ satisfies property $2.$.

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  • $\begingroup$ Hence, in my reference, according to your restating, there is "just" the omission of the red part below? $$0<t_0\color{red}{\leq t_0+t_0}\leq t_1+t_0<\ldots<t_n+t_0$$ @TSF $\endgroup$ Sep 27, 2020 at 14:31
  • $\begingroup$ No, it's not the same as that, I already answered you in the other comment. That condition doesn't make sense. The issue is calling the translation time, $\tau$, by the name $t_0$. This makes it seem like the first increment is $t_0$ when it's really $t_1$. $t_0$ is not an increment, it is the starting place. Do you see $\tau+\tau$ anywhere in my answer? $\endgroup$ Sep 27, 2020 at 14:33
  • $\begingroup$ So, if you are just replacing $t_0$ with $\tau$, I would just expect $0<\tau<t_1+\tau<\ldots<t_n+\tau$ and not $0<\tau\color{red}{\leq t_0+\tau}\leq t_1+\tau\leq\ldots<t_n+\tau$ @TSF $\endgroup$ Sep 27, 2020 at 14:38
  • $\begingroup$ I included a $t_0$ since it's how it was used in the definition. If you want to start from $t_1$ that's fine too but then you need to consider only $k=2,3,\ldots,n$. Do you understand what the mistake was now? Labelling both the translation and the increments as $t$ with a subscript confuses what conditions have to hold. $\endgroup$ Sep 27, 2020 at 14:40
  • $\begingroup$ Maybe I am on the way. Sorry for my confusion. So, If I refer the the original reference, could we say that $\color{red}{\text{red}}$ part and $\color{blue}{\text{blue}}$ part below are two different "objects"? $$\widetilde{W}(t)-\widetilde{W}(s)=W(t+\color{red}{t_0})-W(s+\color{red}{t_0})\tag{1}$$ $$0<\color{green}{t_0}\leq t_1+\color{blue}{t_0}<\ldots<t_n+\color{blue}{t_0}\tag{2}$$ $\endgroup$ Sep 27, 2020 at 14:59

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