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Let $\{B_1(t):t\geq0\}$ be a one dimensional Brownian motion. Then, $\{B^a_1(t):t\geq0\}$ is also a Brownian motion where $B^a_1(t):=a^{-1}B_1(a^2t)$ for $a\in\mathbb{R}$.

Let $\{B_2(t):t\geq0\}$ be a one dimensional Brownian motion independent of $\{B_1(t):t\geq0\}$. For $b\in\mathbb{R}$, $T:=\inf\{t\geq0:B_2(t)=b\}$ and $T^a:=\inf\{t\geq0:B^a_2(t)=b\}$ have same distribution where $B^a_2(t):=a^{-1}B_2(a^2t)$ for $a\in\mathbb{R}$.

I already know these facts is right.

Question

$B_1(T)$ and $B^a_1(T^a)$ have same distribution?

Thank you for your cooperation.

sate
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1 Answers1

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$B^a(T^a)=B(T)=b$ by definition of your stopping time (note that $T$ and $T^a$ are a.s. finite), so yes.

WoolierThanThou
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  • thank you for answer. sorry, my question was mistaken. – sate Jan 21 '20 at 16:03
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    The answer to your new question is yes because $(B_1, B_2, T)$ has the same distribution as $(B_1^a,B_2^a, T^a)$. – WoolierThanThou Jan 21 '20 at 16:07
  • Thanks, but I don't know much. Well, to prove $B_1(T),B^a_1(T^a)$ have the same distribution, should I show $P(B^a_1(T^a)\in A)=P(B_1(T)\in A)$ ? This follows what you said ? – sate Jan 22 '20 at 02:34
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    I think you're confusing yourself with the various realisations of the Brownian Motion you've created. You're really looking at the measurable space $X=C([0,\infty))\times C([0,\infty))\times [0,\infty)$ and the measurable map $(f,g,t)\mapsto f(t)$. So what's your measure on $X$? It's $W\otimes (W,\tau_b)$ where $W$ denotes the Brownian Motion measure and $(W,\tau_b)$ is the joint distribution of a Brownian Motion and its first hitting time of $b$. But that's the case whether or not you've added superscript $a$. – WoolierThanThou Jan 22 '20 at 06:48
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    Once the measure on $X$ are the same, then the push-forward measure under the above map (i.e., the distribution of $f(t)$) is the same - this is immediate. – WoolierThanThou Jan 22 '20 at 06:48
  • Thank you for your reply. Yes, I was very confused. Now, I get it thanks to your accurate comment ! – sate Jan 22 '20 at 18:33