# Estimating the knockout probability of a discretely observed autocall note

For simplicity, let's suppose the underlier follows a Geometric Brownian Motion $$S_t\sim\text{GBM}(\mu, \sigma), t\ge 0$$ with $$S_0=1$$. A discretely-observed binary autocall note is a derivative structure with observation dates $$t_1 and pays the investor a high coupon $$c$$ on the first observation date on which the underlier $$S_t$$ exceeds some preset barrier $$K$$. In mathematical notations, define the knockout date $$\tau=\inf\{t_i,i=1,\cdots,n\mid S_{t_i}>K\}$$ We are given the below tasks:

1. Estimate the distribution of $$\tau$$, i.e. calculate $$P(\tau=t_i)$$ and $$P(\tau=\infty)$$
2. Or at least calculate $$P(\tau < \infty)$$, if the first task proves too challenging.

Practically, MC simulation is clearly the way to go. But let's say we're more interested on the mathematical side of this problem and focus not so much on pragmatism. Are there any exact or approximate solutions that permit a simple numerical implementation? Thanks.

• The problem is definitely tractable when $\mu=0$ and $t_1=0$. For then $p(\tau > t_i) = 1 - 2p(S_t_i>K)$ by the reflection theorem. [sorry for format- can't seem to fix the mathjax]
– dm63
Commented Apr 1, 2023 at 14:43
• @dm63 we can assume $\mu=0$ but not $t_1=0$. In fact for a one-year long contract observations often start in the third or even sixth month i.e. $t_1=1/4$ or $1/2$.
– Vim
Commented Apr 2, 2023 at 8:02
• @dm63 you seem to have forgot to put the sub of $S$ between braces: S_{t_i} is the recognisable format. So your formula renders as $p(\tau > t_i) = 1 - 2p(S_{t_i}>K)$
– Vim
Commented Apr 2, 2023 at 8:07

There is a closed-form formula for the probability $$\mathbb{P}(\tau = t_i)$$.

First, we remind that $$S_t=S_0\cdot \exp\left(\left(\mu-\frac{1}{2}\sigma^2 \right)t+\sigma W_t \right)$$ For $$i=1$$, it's easy that \begin{align} \mathbb{P}(\tau = t_1) &= \mathbb{P}(S_{t_1}> K ) \\ &=\mathbb{P}\left(W_{t_1}> \frac{\ln \left(\frac{K}{S_0}\right) -\left(\mu-\frac{1}{2}\sigma^2 \right)t_1}{\sigma} \right)\\ &=\color{red}{\Phi\left( \frac{-\ln \left(\frac{K}{S_0}\right) +\left(\mu-\frac{1}{2}\sigma^2 \right)t_1}{\sigma \sqrt{t_1}} \right)} \end{align} where $$\Phi(\cdot)$$ the probability distribution function of the univariate standard normal distribution $$\mathcal{N} (0,1)$$.

For $$i \ge 2$$, we have \begin{align} \mathbb{P}(\tau = t_i) &= \mathbb{P}(\bigcap_{0 \leq k \leq i-1} \{S_{t_k}\le K \} \cap \{S_{t_i}> K \} ) \\ &= \mathbb{P}\left(\bigcap_{0 \leq k \leq i-1} \left\{W_{t_k} \le \frac{\ln \left(\frac{K}{S_0}\right) -\left(\mu-\frac{1}{2}\sigma^2 \right)t_k}{\sigma} \right\} \cap \left\{W_{t_i}> \frac{\ln \left(\frac{K}{S_0}\right) -\left(\mu-\frac{1}{2}\sigma^2 \right)t_i}{\sigma} \right\} \right) \tag{1}\\ \end{align} We notice that the vector $$(W_{t_1}, W_{t_2},...,W_{t_i})$$ is a $$i$$-variate normal distribution with zero mean and the covariance matrix $$\mathbf{\Sigma} \in \mathbb{R}^{i\times i}$$ defined by $$\Sigma_{hk} = Cov (W_{t_h},W_{t_k}) = \min \{t_h,t_k\} \qquad \text{for }1\le h,k\le i \tag{2}$$

By denoting $$\Phi_i(\mathbf{L},\mathbf{U};\mathbf{0}_i,\mathbf{\Sigma} )$$ the probability distribution function of the $$i$$-variate normal distribution $$\mathcal{N}_i (\mathbf{0}_i,\mathbf{\Sigma})$$ with

• zero mean $$\mathbf{0}_i$$,
• covariance matrix $$\mathbf{\Sigma}$$ defined by $$(2)$$
• from the lower bound $$\mathbf{L}$$ to the upper bound $$\mathbf{U}$$ with $$\mathbf{L}, \mathbf{U} \in \mathbb{R}^{i}$$ $$L_k=\begin{cases} -\infty & \text{if 0\le k\le i-1 }\\ \frac{\ln \left(\frac{K}{S_0}\right) -\left(\mu-\frac{1}{2}\sigma^2 \right)t_k}{\sigma} & \text{if k = i }\\ \end{cases}$$ $$U_k=\begin{cases} \frac{\ln \left(\frac{K}{S_0}\right) -\left(\mu-\frac{1}{2}\sigma^2 \right)t_k}{\sigma} & \text{if 0\le k\le i-1 }\\ +\infty & \text{if k = i }\\ \end{cases}$$

Then, from $$(1)$$, we have

$$\mathbb{P}(\tau = t_i) = \color{red}{\Phi_i(\mathbf{L},\mathbf{U};\mathbf{0}_i,\mathbf{\Sigma} ) }$$

• Thanks this is a neat representation. For what it's worth, a simple scipy implementation can be found here.
– Vim
Commented Apr 2, 2023 at 8:20
• I think $$\Sigma_{hk} = Cov (W_{t_h},W_{t_k}) = \color{red}{\min} \{t_h,t_k\} \qquad \text{for }1\le h,k\le i$$ though, rather than $\max$. Is this a typo?
– Vim
Commented Apr 2, 2023 at 8:51
• Applying your formula to the scenario where $\mu=5\%, \sigma=25\%,t_k=\frac{6+(k-1)}{12},k=1,\cdots,6$ gives the following probabilities in percentage terms: $$[45.45, 6.29, 4.1, 3.07, 2.43, 1.99]$$ which I believe is sound.
– Vim
Commented Apr 2, 2023 at 8:58
• @Vim yes, it’s a typo. I corrected it
– NN2
Commented Apr 2, 2023 at 10:19
• Excellent solution !
– dm63
Commented Apr 2, 2023 at 13:36