Calculating Barrier-Touch Probabilities for Options

# Probability of touching


# Probability of touching












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?

## Answer by Paul H. Lasky (score 14, accepted)

https://quant.stackexchange.com/a/403

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, $$ \text{prob of hitting ($t \le T$)} = \int\limits_{t=0}^{T} p(x,t)\mathrm{d}t $$

## Answer by Hans (score 19)

https://quant.stackexchange.com/a/12797

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 $0$ and hitting $x\ge0$ 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 $x$.

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 $$P(\tau_x<t)=P_1+P_2.$$

## Answer by egoroff (score 7)

https://quant.stackexchange.com/a/263

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.

## Answer by ffsffs (score 5)

https://quant.stackexchange.com/a/1829

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

## Answer by Yanyi Yuan (score 3)

https://quant.stackexchange.com/a/47223

For in-the-money options the probability of ever getting in-the-money (hitting the strike) before maturity naturally equals unity. The risk neutral probability for an out-of-the-money option ever getting in-the-money is equal to the barrier hit probability used for computing the value of a rebate, developed by Reiner/Rubinstein (1991):

$$p_c = (X/S)^{\mu+\lambda}N(-z) + (X/S)^{\mu-\lambda}N(-z + 2\lambda \sigma\sqrt{T})$$

$$p_p = (X/S)^{\mu+\lambda}N(z) + (X/S)^{\mu-\lambda}N(z - 2\lambda \sigma\sqrt{T})$$

with

$$z=\frac{ln(X/S)}{\sigma\sqrt{T}} + \lambda\sigma\sqrt{T}$$ $$\mu = \frac{b-\sigma^2/2}{\sigma^2}$$ $$\lambda=\sqrt{\mu^2+\frac{2r}{\sigma^2}}$$

Reiner, E., & Rubinstein, M., (1991), Breaking Down the Barriers, Risk Magazine, 4(8).

## Answer by Yanyi Yuan (score 3)

https://quant.stackexchange.com/a/47253

Shreve's book chapter exotic option, the probability of maximum Brownian Motion does not exceed a threshold is as follows.

The probability of touch, i.e. the probability of an option moving from OTM to ITM any time before expiry, is based on the above. The derivation is based on change of measure under Girsanov's Theorem.

For an out-of-the-money call option with strike K, the probability of the option falling in-the-money any time before expiry is given by the above formula as 1-P ̂{M ̂(T)≤m} where

r is the risk free rate.

For an out-of-the-money put option with strike K, the probability of the opton falling in the money any time before expiry is

## Answer by Jaydles (score 2)

https://quant.stackexchange.com/a/238

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.

## Answer by user915 (score 1)

https://quant.stackexchange.com/a/12804

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 https://meteor.springer.com/springerreference.jsf. 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. https://books.google.ca/books?id=Iy5f5vjfjfYC&pg=PA5&lpg=PA5&dq=cumulative+distribution+of+the+inverse+gaussian&source=bl&ots=m0o3Wxvea1&sig=DFPZ4Z-98SUOwO2i918WvfMkfPU&hl=en&sa=X&ei=n02rU9nVDY6YqAbggIHoAQ&redir_esc=y)

## Answer by Thomas Baert (score 1)

https://quant.stackexchange.com/a/31785

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) $

$g=2(\frac{\sigma^2}{2}+r)/\sigma^2$

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

for $b>p$

$N$ cumulative normal distribution

## Answer by Chuck (score 0)

https://quant.stackexchange.com/a/34032

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

## Answer by Paul H. Lasky (score -2)

https://quant.stackexchange.com/a/412

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.