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Hello Quant Finance StackExchange,

Is there a simple way to find the crossing probabilities of a moving barrier, namely a barrier written in the form $U(t)=\alpha_1t^2 + \beta_1t + \gamma_1$ and $L(t)=\alpha_2t^2 + \beta_2t + \gamma_2$. (or if no solution, reduce the complexity to linear polynomial $\alpha t + \beta$.

I understand that there's a simple solution if the barrier is fixed, $U(t) = U$ and $L(t) = L$ and that $P(U$before$L)$, written here as $H(x)$, for a process $X_t$is given by


where $x$, $L\leq x \leq U$ is the starting position and $s(x)$ is the scaling function such that $Y_t = s(X_t)$ is a martingale. Example, for GBM with $2\mu \neq\sigma^2$,


Now, if I were to apply optional stopping for a moving barrier as I did a fixed barrier, it would seem that I'll get something like,


which would make it a random variable as the first exit time $\tau$ is random. Would getting the crossing probability as simple as taking the expectation, namely


Is this method reasonably logical?

(My rusty notation comes from me being two years removed from doing stochastic calculus at college. A rough explanation would do. Yes, I know there's tons of mathematical intricacies I conveniently overlooked.)

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Just some notational questions: So $\tau$ is a stopping time taking values in $[0,T]$? If so, do you assume a discrete time model, i.e. $\tau$ takes values in $\{0,\dots,T\}$. Then your formula seems (mathematically) correct. But for a continuous setting, the math is more advanced and you should first check, if there is a density $f_\tau$ for $\tau$. Then $E[H(x,\tau)]=\int f_\tau(w) \frac{s(x)-s(L(w))}{s(U(w))-s(L(w))}dw$. You would get the expectation of your r.v. $H(x,\tau)$, i.e. the expected values of your crossing probabilities. –  user8 Jul 1 '13 at 9:53
Please note, I'm not very familiar with barrier, so I just commented on the math part. Where I assmued your formula for $H(x,\tau)$ is correct. –  user8 Jul 1 '13 at 10:02
Yup, discrete time model and $\tau$ takes values in ${0,...,T}$. –  Donny Lee Jul 1 '13 at 10:11
So assuming your fomula for $H(x,\tau)$ is correct, the formula for $E[H(x,\tau)]$ is also correct. But since $H(x,\tau)$ is a r.v. this gives really just the expectation of your crossing probability and not the crossing probability itself. This is given by the r.v. If your are interested in $H(x,\tau)$ you should try to figure out the distribution or try to approximate it numerically. –  user8 Jul 1 '13 at 10:17

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