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Given two same brownian motion with no drift and different variances:

$$dG_1= \sigma_1 G_1 dW $$ $$dG_2= \sigma_2 G_2 dW $$

and two barriers $P_1 > P_2$ enter image description here

assuming that $ \sigma_1 > \sigma_2 $ and given $τ_1=\inf [t:G_1(t)≥P_1] $; $τ_2= \inf [t:G_2(t)≥P_2]$

is it possible to find $P(τ_1<τ_2)$?

(the probability that a certain time $\tau\ $ "G1 will hit the barrier P1 earlier than G2 hits P2"?)

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  • $\begingroup$ Can you clarify your last sentence? I don't understand the role of $\tau$. And when you say "hit the barrier faster" do you just mean "earlier"? $\endgroup$ – Nate Eldredge May 27 '15 at 15:53
  • $\begingroup$ Yes, earlier is better. For $ \tau $ that's just to identify the value of time at which G1 hit the barrier P1 $\endgroup$ – Clemente Cortile May 27 '15 at 16:19
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    $\begingroup$ Ok, so if we let $\tau_1 = \inf\{t : G_1(t) \ge P_1\}$ and $\tau_2 = \inf\{t : G_2(t) \ge P_2\}$, your question is "What is $\mathbb{P}(\tau_1 < \tau_2)$?" Is that right? $\endgroup$ – Nate Eldredge May 27 '15 at 16:23
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    $\begingroup$ I'm also confused by your definition of the processes $G_1, G_2$. Is $G_1$ supposed to satisfy the SDE $dG_1 = \sigma_1 G_1 dW$ for $W$ a Brownian motion? If so, then $G_1$ is not itself a Brownian motion. Why does $W$ have a subscript $g$? And are both processes $G_1, G_2$ supposed to be driven by the same Brownian motion $W$? $\endgroup$ – Nate Eldredge May 27 '15 at 16:26
  • $\begingroup$ Yes they are. Thanks for the pointers I'll edit the question. $\endgroup$ – Clemente Cortile May 27 '15 at 16:51

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