Estimating the knockout probability of a discretely observed autocall note

Question

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 $t_1<t_2<\cdots<t_n$ 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.

Answer 1

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

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