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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

$H(x)=\frac{s(x)-s(L)}{s(U)-s(L)}$

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 $2\mu \neq\sigma^2$,

$s(x)=\frac{x^{1-2\mu/\sigma^2}}{1-2\mu/\sigma^2}$.

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,

$H(x,\tau)=\frac{s(x)-s(L(\tau))}{s(U(\tau))-s(L(\tau))}$

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

$H(x)=\text{E}_\tau(H(x,\tau))=\sum_{t=0}^T\left[H(x,t)P(\tau=t)\right]$

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
2  
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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