Assume the price follows a lognormal process. We can convert it, by taking the natural logarithm of the price, into a problem of finding the probability of a standard Brownian motion particle starting from and hitting before time , or its first passage time being less than . This can be derived through the reflection principle. The paths crossing are exactly paired up by the segment post crossing through mirror reflection about .

Case 1) No drift.

By the strong Markov property, at the moment a path first touches , the probabilities of the particle taking on either of two path mirror reflecting about the line are the same, therefore the total probability of touching is twice of that of particle reaching above $$P(\tau_x<t) = \frac{2}{\sqrt{2\pi t}}\int_x^\infty e^{-\frac{y^2}{2t}} {\rm d}y=\sqrt{\frac{2}{\pi}}\int_{\frac{x}{\sqrt t}}^\infty e^{-\frac{y^2}{2}} dy={\rm erfc}\Big(\frac{x}{\sqrt{2t}}\Big).$$

Case 2) The drift is , where is a constant.

The probability measure is \begin{align} dP(y) &= \frac{1}{\sqrt{2\pi}}\exp\Big(-\frac{(y-vt)^2}{2t}\Big)\frac{{\rm d}y}{\sqrt t} \\ &= \frac{1}{\sqrt{2\pi}}\exp\Big(vy-\frac{1}{2}v^2t\Big)\exp\Big(-\frac{y^2}{2t}\Big)\frac{{\rm d}y}{\sqrt t}. \end{align}

The set of paths crossing can be partitioned into two disjoint subsets, one ends at above and the other ends below. The probability of the first set is obtained directly using the first expression above \begin{align} P_1 &= \frac{1}{\sqrt{2\pi}}\int_x^{\infty}\exp\Big(-\frac{(u-vt)^2}{2t}\Big)\frac{{\rm d}u}{\sqrt t} \\ &=\frac{1}{\sqrt{2\pi}}\int_{\frac{x}{\sqrt t}-v\sqrt t}^\infty e^{-\frac{y^2}{2}} dy \\ &= \frac{1}{2}{\rm erfc}\Big(\frac{1}{\sqrt 2}\Big(\frac{x}{\sqrt t}-v\sqrt t\Big)\Big). \end{align}

In the second set, the paths that end in is a subset of the set of all the paths that start from and end in . That former set one-to-one corresponds to the set of paths starting from ending in . The second expression for the probability measure indicates we can treat the first exponential as a factor or a random variable dependent only on the final time and the second a new probability measure. The first factor can also be interpreted as a Radon-Nikodym derivative or Jacobian between two probability measures. This new measure makes any two path sets reflectively symmetric about have the same measure just as in the driftless case, which allows us to compute the probability of second set using that of the first set when driftless. So the probability of the second set is \begin{align} P_2 &= \frac{1}{\sqrt{2\pi}}\int_{-\infty}^x\exp\Big(vu-\frac{1}{2}v^2t\Big)\exp\Big(-\frac{(2x-u)^2}{2t}\Big)\frac{{\rm d}u}{\sqrt t} \\ &=\frac{e^{2vx}}{\sqrt{2\pi}}\int_{\frac{x}{\sqrt{t}}+v\sqrt t}^\infty e^{-\frac{y^2}{2}} {\rm d}y \\ &= \frac{e^{2vx}}{2}{\rm erfc}\Big(\frac{1}{\sqrt 2}\Big(\frac{x}{\sqrt t}+v\sqrt t\Big)\Big). \end{align} Therefore the probability of the particle passing or the first passage time of less than is the sum of the above two probability