For a vanilla option, I know that the probability of the option expiring in the money is simply the delta of the option... but how would I calculate the probability, without doing monte carlo, of the underlying touching the strike at some time at or before maturity?

  • $\begingroup$ It seems like for a vanilla there should be a non-simulation way to calculate this. $\endgroup$ – glyphard Feb 7 '11 at 22:51
  • $\begingroup$ This question is entitled the "probability of touching", but the OP was asking for the probability of an option expiring in the money. The two are not the same. Note how many folks mentioned "stopping time" and "barriers" in their answers. $\endgroup$ – wsw Nov 11 '12 at 5:52
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    $\begingroup$ @WilliamS.Wong - the OP is about probability of touching, before expiration. Read it again. $\endgroup$ – glyphard Nov 11 '12 at 19:12

10 Answers 10


There is a simple solution if there is no drift, as the probability $p(x,t)$ obeys a simple diffusion equation: $\mathrm{d}(p)/\mathrm{d}t = \frac{1}{2} \sigma^2 \frac{\mathrm{d}(\mathrm{d}(p))}{\mathrm{d}x^2}$, here $x$ is the price difference $\text{price}(t) - \text{price}(t=0)$. Of course there is a simple solution to the diffusion equation (using scaling as a method to solve the PDE):
$$ p(x,t) = (4\pi \frac{\sigma^2}{2} t)^{-\frac{1}{2}} \text{e}^{(-x^2/(4 \frac{\sigma^2}{2} t) )} $$ to find the probability of hiting a barrier $x$ on or before $T$ simply ( :} ) integrate, $$ \text{prob of hitting ($t \le T$)} = \int\limits_{t=0}^{T} p(x,t)\mathrm{d}t $$

  • $\begingroup$ that's what I was looking for, $\endgroup$ – glyphard Feb 10 '11 at 21:45
  • $\begingroup$ I am trying to use this in Sage to approximate the probability of touching on a vanilla option. However, I don't think I am getting the correct results. I am using the volatility of the underlying (for example .30 for 30%) for sigma (not sigma squared) and then using the price of strike minus the current price for x. Is that the correct interpretation? For example, inputting the current price of 123.97, strike of 130, volatility of .30, and 30 days .. I am getting a .00107 which is much much less than the 50+% I get on online calculators that do the same thing. $\endgroup$ – user1576 Oct 24 '11 at 18:57
  • $\begingroup$ I think something is wrong with the answer, I'm surprised so many people upvoted it. The original answer (before the edit) doesn't have the correct number of brackets so I'm not sure how the Editor has rewritten it into Latex. Would appreciate if someone could clarify / check this $\endgroup$ – mchangun Jul 24 '13 at 5:56
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    $\begingroup$ This formula is definitely wrong. @mchangun, your instinct is right. See my solution below. $\endgroup$ – Hans Jun 25 '14 at 17:00
  • $\begingroup$ @glyphard: This answer is wrong. You should unaccept this answer and instead accept the correct answer by Hans below. $\endgroup$ – Hans Apr 22 '16 at 6:18

Assume the price follows a lognormal process. We can convert it into a problem of finding the probability of a standard Brownian motion particle starting from $0$ and hitting $x$ before time $t$, or its first passage time $\tau_x$ being less than $t$. This can be derived through the reflection principle. The paths crossing $x$ are exactly paired up by the segment post crossing through mirror reflection about $x$. We divide the set of path crossing $x$ into two groups. The first

Case 1) No drift.

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

Case 2) The drift is $vt$, where $v$ 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 $x$ can be partitioned into two disjoint subsets, one ends at $t$ above $x$ and the other ends below. The probability $P_1$ 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 $[y-dy,y],\, y<x$ is a subset of the set of all the paths that start from $0$ and end in $[y-dy,y],\, y<x$. That former set one-to-one corresponds to the set of paths starting from $0$ ending in $[2x-y,2x-y+dy]$. 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 $t$ 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 $x$ 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 $P_2$ 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 $x$ or the first passage time $\tau_x$ of $x$ less than $t$ is the sum of the above two probability $$P(\tau_x<t)=P_1+P_2.$$

  • $\begingroup$ Sounds positive. Can anyone else verify the solutions? $\endgroup$ – SmallChess Jun 30 '14 at 4:34
  • $\begingroup$ An interesting reference using similar arguments seems to be "How Likely is it to Hit a Barrier? Thoretical and Emperical Estimates" by Lujing Su and Marc Oliver Rieger. $\endgroup$ – philippe Sep 24 '14 at 22:46
  • $\begingroup$ @philippe: Thank you for the reference. Their theoretical argument is indeed the same as mine. The empirical work is interesting. $\endgroup$ – Hans Oct 20 '15 at 18:34

Allow me to disagree with Jaydles' proposal ; his methodology is valid only if the events of touching the barrier on each were independent.

If you are working within the standard Black-Scholes framework, you're looking for the probability of a drifted Brownian motion hitting a fixed level before a fixed time ; this probability is derived in most stochastic calculus texts, see for example Karatzas-Shreve or Chesney-Jeanblanc-Yor.

Another way of seeing it : you're trying to price a knock-in digital option with 0 interest rate, or knock-in zero bond. You can find formulae for these in Peter Carr's work on barrier options.

  • $\begingroup$ the stocastic calculus approach is called 'first hitting time' or 'stopping time'. Shreve does not talk about actually calculating it. He just says, 'use monte carlo' There is a non-monte carlo method of calculating it, and that is what I am looking for. $\endgroup$ – glyphard Feb 8 '11 at 16:19
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    $\begingroup$ OK. An approach without tedious computations is as follows : first find a nonzero real number $\gamma$ such that $X_t=S_t^{\gamma}$ is a martingale. This process $X$ now satisfies a "multiplicative reflection principle" : for any stopping time $T$, $X_{T+s}$ has the same law as $X_T^2/X_{T+s}$. Use this at $T_H$ (first hitting time of $H$) and mimic the classic reasoning for standard Brownian motion to find an expression of $P(T_H < t)$ as a function of $P(X_t > H)$, and finally, go back to $S$. $\endgroup$ – egoroff Feb 8 '11 at 16:52
  • $\begingroup$ Shreve (in "Stochastic calculus for finance") does talk about calculating the distribution of the stopping time for Brownian motion. See chapter 4 or 5. $\endgroup$ – quant_dev Feb 11 '11 at 7:55
  • $\begingroup$ With all that discussion and references, why don't you write out the derivation and the final expression? Is your diffusion process more general than the Brownian motion? It is not that complicated and "tedious". See my answer below for standard Brownian motion. $\endgroup$ – Hans Jun 26 '14 at 12:23

First of all, the delta is not the probability of the option in the black scholes model, it is instead the closely related N(d2) (binary probability)

Secondly, the black scholes model gives risk neutral probabilities - for a binary event this is ok, but it gives no correct measure of, say, how far you would be through

Thirdly, the options you are interested in are traded in the market - they are called binary no touch or one touch options ... there are several mechanisms to price, depending on your model for volatility ... black scholes pricers for them are available online, eg here http://www.volopta.com/Matlab.html


This surely isn't the most efficient way, but if you want something quick and dirty:

You could run a vanilla model that calcs delta for each expiration date between now and expiration, and grab the delta for each. That would give you the likelihood that it's in the money at the close on any day.

From that, you can pretty easily calculate the odds that it's not in the money each day (just subtract the delta from one), multiply them all together, and subract the product from one to determine the likelihood that it closes above the strike between now and expiration.

This does require running the formula to calc delta many times, and it ignores the risk of an intra-day touch, but it doesn't require writing something to calc the exotic you're describing.


The required probability is equivalent to asking: what is the probability that the geometric brownian motion of the underlying touches the strike for the first time before the given time $T$?

A strategy for solving this related to Brownian motion - first passage time.

After transforming from geometric brownian motion to brownian motion via the log, the full distribution of the first passage time of a random walker (with a given drift and vol) to a boundary (which should be $log(K/S_0)$ where $K$ is the strike and $S_0$ is the initial stock price), is given by the Inverse Gaussian Distribution.

The explicit form for the distribution is given here http://goo.gl/lzGZ9Y. Using the notation in that link, the parameters of the IG distribution would be $\mu=log(K/S_0)/d$ and $\lambda=(log(K/S_0))^2/\sigma^2$ where $d, \sigma$ are the drift and volatility of the geometric brownian motion.

So, you will want to compute the probability that this random variable, the first passage time of ordinary brownian motion with drift $d$ and volatility $\sigma$ to a fixed boundary at $log(K/S_0)$, is $<= T$, by straightforward integration.

I believe the CDF is known in closed form (see e.g. http://goo.gl/DUK9fh)

  • $\begingroup$ This is wrong. The first passage time density function is NOT inverse Gaussian. The two look similar, but are not the same. The probability is shown in my answer above. The time derivative of that probability is the first passage time probability density, specifically, $p(\tau_x=t) = \frac{x}{\sqrt{2\pi t^3}}\exp\big(-\frac{x^2}{2t}\big).$ $\endgroup$ – Hans Jun 26 '14 at 5:14
  • $\begingroup$ No I think the density function of the first passage time of brownian motion to fixed barrier is well known to be the Inverse Gaussian. There are certainly more authoritative references in existence than Wikipedia (including the Springer link I linked to), but here is Wikipedia on the question: en.wikipedia.org/wiki/… $\endgroup$ – user915 Jun 27 '14 at 17:34
  • $\begingroup$ You are right. I realized that and was just about to edit my previous comment. But my main point still remains that the derivation of the first passage time density comes from the derivative of the integral which is what is sought after. The integral can be easily derived from the reflection principle. Now your approach is essentially to time differentiate the integral then time integrate it again. It is a bit absurd. Rather than taking all that time listing all the references, why don't you just derive the "closed form" explicitly yourself? It is not that hard. $\endgroup$ – Hans Jun 27 '14 at 22:09
  • $\begingroup$ Your comment is puzzling (if not actually absurd itself). The point of this site is to provide correct answers. I happened to know the answer to this question and tried to supply the poster with the answer. My "approach" as you put it is to integrate a function, nothing more. I agree the derivation is probably not too difficult, though it seems it was a bit of a challenge for you. $\endgroup$ – user915 Jun 27 '14 at 22:23
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    $\begingroup$ I have accepted your "a bit of a challenge", as you put it, and edited my answer above to include the derivation for drift. Now I pose my "bit of challenge" to you to show the inverse Gaussian distribution is indeed the right density function for the first passage time. If it comes from time differentiating the final probability in my answer above, then the integration process is redundant. That was what I meant by "absurd", albeit not a good choice of word. I would be very intrigued to see your derivation, were it not derived via time differentiation. $\endgroup$ – Hans Jun 29 '14 at 17:12

The answer requires reflected Brownin motion

$w=(\ln (p/b)+rt+ \frac{\sigma^2}{2} t)/(\sqrt{t}\sigma) $

$z=(\ln (p/b)-rt- \frac{\sigma^2}{2} t)/(\sqrt{t}\sigma) $


probability of hitting= $N(w)+(b/p)^gN(z)$

for $b>p$

$N$ cumulative normal distribution


I believe you can modify my https://github.com/barrycarter/bcapps/blob/master/box-option-value.m to do this.

You're effectively looking for the distribution of the maximum (or minimum) of the price for a given period of time.


I understand that a useful approximation for prob of touch is 2 * delta, though I don't recall which authority that is from.


An asymptotic answer is

$$ p(t > \tau) \sim \mathrm{e}^{(-t/\tau)} \qquad \text{where $\tau = x^2/(\frac{\sigma^2}{2} \pi^2)$}; $$

However, as in much of statistics, asymptotic answers = "In the long-run" are not usually helpful to traders, or even investors. Example: say initial price $x_0=\$20$, and you wish the probability of reaching say, $x= \$25$ before $x = \$15$ with an annual $\sigma$ of say $0.2$. Convert $X$ to fractions, $X=(\$25-\$15)/\$15$, and calculate $\tau = 1.126$ years. That is without some drift and for reasonable values of the volatility, you are going to have to wait a long time.

  • $\begingroup$ This is wrong. By asymptotics, do you mean, when $t$ becomes very large, or the strike becomes large? It is really vague. I think you want the former. If that is the case, the probability of reaching the strike or any price becomes certainty, or approaches $1$. $\endgroup$ – Hans Jun 25 '14 at 16:58

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