Applying Feynman–Kac to Down-and-Out Barrier Options

# Feynman Kac Formula for path-dependent options


# Feynman Kac Formula for path-dependent options












Consier geometric Brownian motion: $dS_t/S_t=\mu dt+\sigma dW_t$

Feynman Kac theorem tells us that

the conditional expectation $v(t,x)=E[ e^{-rT}\Psi(S_T) | S_t=x]$ can be computed by solving the following PDE:

$$ v_t(t,x)+1/2\sigma^2 x^2 V_{xx}(t,x)+rxV_{x}(t,x)-rV(t,x)=0 $$ with a boundary condition $V(T,x)=\Psi(x)$.

This is for path-independent European claim with the payoff $\Psi$.

My question is how to apply Feynman Kac theorem for path-depedent option like a barrier option.

the value of a down and out barrier option is $v(t,x)=E[ e^{-rT}\Psi(S_T)1_{m_{t,T}^S>L} | S_t=x]$ where $m_{t,T}^S=\inf_{\{t\leq s \leq T\}} S_s$.

However, Feynman Kac theorem is not available for $E[ e^{-rT}\Psi(S_T)1_{m_{t,T}^S>L} | S_t=x]$ due to path-dependent term $1_{m_{t,T}^S>L}$. However, we still have same PDE for those of vanilla option except that additional boundary condition $v(t,L)=0$.

So, I came to think that we can still utilize Fenyman Kac thoerem for a barrier option. Am I right??

## Answer by tatami goza (score 7, accepted)

https://quant.stackexchange.com/a/19530

In short answer, Yes: the backward PDE solution with $v(t,L)=0$ and the expectation coincides under the Black-Scholes market.

In the one dimensional case, this topic is mathematically treated in the theory of the scale function and the spead measure. See Revez-Yor 3rd.ed. Ch.VII.3 for details.

I don't know whether there are some rigorous theories on the general case. But I present an intuitive sketch here.

Assuming a no arbitrage market and let the underlying process is $X_t$. Then if we decide a numeraire $N_t$, we have the equivalent martingale measure $Q$ that makes any discounted financial trade price $U_t$ a martingale.

$$M_t = \frac{U_t}{N_t} = E^Q\left[\frac{U_T}{N_T} \Bigg| \mathbb{F_t}\right]$$

In the case of the Black Scholes market,

$$dX_t=X_t(\mu dt + \sigma dW_t), \qquad N_t=\exp(rt)$$

For simplicity, let us assume that $N_t$ can be described as $N_t = n(t, X_t)$.

Further assume that $X_t$ is a Markov process. And assume that if we have an expression $U_t = u(t,X_t)$, then the martingale $M_t$ follows the SDE with the infinitesimal generator $\mathcal{A}$ and a martingale $Y$ such that,

$$dM_t = \frac{\partial}{\partial t}g(t, X_t)dt + \mathcal{A}_t g(t, X_t) dt + dY_t, \qquad g(t,x) = \frac{u(t,x)}{n(t,x)}$$

In the case of the Black Scholes model, applying Ito formula yields,

$$\mathcal{A}_t = \frac{1}{2}\sigma^2x^2\frac{\partial^2}{\partial x^2} + \mu x \frac{\partial}{\partial x}, \qquad dY_t=\sigma X_t g(t,X_t) dW_t$$

Roughly speaking, what the Feynman-Kac theorem says is that if $g(t, x)$ satisfies,

$$\frac{\partial}{\partial t}g(t,x) + \mathcal{A}_t g(t,x) = 0$$

then it admits the stochastic representation:

$$g(t, x) = E^Q[g(T, X_T)|\mathbb{F}_t] = E^Q\left[\frac{U_T}{N_T} \Bigg| \mathbb{F}_t \right]$$

And the Feynman-Kac theorem is from the simple fact that $M_t$ becomes a martingale.

$$ \begin{align} E^Q[M_T | \mathbb{F}_t] &= M_t + E^Q\left[ \int_t^T \left( \frac{\partial}{\partial s}g(s, X_s) + \mathcal{A}_s g(s, X_s) \right) ds \Bigg| \mathbb{F}_t \right] + E^Q\left[\int_t^T dY_s \Bigg| \mathbb{F}_t \right] \\ &= M_t \end{align} $$

Let us consider to construct the same argument in the case of knock out options. Let $\tau$ be a hitting time $\tau=\inf\{t | X_t \leq L\}$ and let $J_t$ be the discounted price of the down and out option $V_t$ with the payoff function $f$.

$$J_t = \frac{V_t}{N_t} = E^Q\left[\frac{V_T}{N_T} \Bigg| \mathbb{F}_t \right] = E^Q\left[1_{\tau > T} \frac{f(X_T)}{N_T} \Bigg| \mathbb{F}_t \right]$$

Let us assume that $V_t$ can be expressed as $V_t=v(t, X_t)$.

Under the assumption, $J_t$ also follows the SDE on with the corresponding martingale part $Z$.

$$dJ_t = \frac{\partial}{\partial t}h(t, X_t)dt + \mathcal{A}_t h(t, X_t) dt + dZ_t, \qquad h(t,x) = \frac{v(t,x)}{n(t,x)}$$

Note that the price of the knock out option satisfies the following, since $J_{\tau}=J_T = 0$ on $\{\tau \leq T\}$.

$$E^Q[J_T| \mathbb{F}_t] = E^Q[J_{\tau \wedge T}| \mathbb{F}_t], \qquad \tau \wedge t := \min(\tau, t)$$

Using this equation,

$$ \begin{align} E^Q[J_T | \mathbb{F}_t]=E^Q[J_{\tau \wedge T} | \mathbb{F}_t] &= J_t + E^Q\left[\int_t^{\tau \wedge T} \left( \frac{\partial}{\partial s}h(s, X_s) + \mathcal{A}_s h(s, X_s) \right) ds \Bigg| \mathbb{F}_t \right] \\ & \qquad + E^Q\left[\int_t^{\tau \wedge T} dZ_s \Bigg| \mathbb{F}_t \right] \\ &=J_t + E^Q\left[1_{\tau > T} \int_t^T \left( \frac{\partial}{\partial s}h(s, X_s) + \mathcal{A}_s h(s, X_s) \right) ds \Bigg| \mathbb{F}_t \right] \\ & \qquad + E^Q\left[1_{t \leq \tau \leq T} \int_t^{\tau} \left( \frac{\partial}{\partial s}h(s, X_s) + \mathcal{A}_s h(s, X_s) \right) ds \Bigg| \mathbb{F}_t \right] \end{align} $$

If $h(t,x)$ satisfies the PDE,

$$ \frac{\partial}{\partial t} h(t,x) + \mathcal{A}_t h(t,x) = 0 \qquad x > L \\ h(t, L) = 0 $$

the second term is equal to zero.

$$E^Q\left[1_{\tau > T} \int_t^T \left( \frac{\partial}{\partial s}h(s, X_s) + \mathcal{A}_s h(s, X_s) \right) ds \Bigg| \mathbb{F}_t \right]=0$$

The last term is somewhat difficult to treat rigorously. Approximately, if we discretize the time interval $[t,T]$ with $\{t_0, t_1, \cdots, t_K\}$, $\Delta t=t_{i+1}-t_i$, we have the following.

$$ \begin{align} & E^Q\left[1_{t \leq \tau \leq T} \int_t^{\tau} \left( \frac{\partial}{\partial s}h(s, X_s) + \mathcal{A}_s h(s, X_s) \right) ds \Bigg| \mathbb{F}_t \right] \\ &\approx \sum_{i=0}^K E^Q\left[1_{t_i \leq \tau \leq t_{i+1}} \int_{t_i}^{t_{i+1}} \left( \frac{\partial}{\partial s}h(s, X_s) + \mathcal{A}_s h(s, X_s) \right) ds \Bigg| \mathbb{F}_t \right] \\ &\approx \sum_{i=0}^K E^Q\left[1_{t_i \leq \tau \leq t_{i+1}} \int_{t_i}^{t_{i+1}} \left( \frac{\partial}{\partial s}h(s, L) + \mathcal{A}_s h(s, L) \right) ds \Bigg| \mathbb{F}_t \right] \\ &\approx \sum_{i=0}^K E^Q\left[1_{t_i \leq \tau \leq t_{i+1}} \int_{t_i}^{t_{i+1}} \frac{\partial}{\partial s}h(s, L) ds \Bigg| \mathbb{F}_t \right] \\ &\approx \sum_{i=0}^K E^Q\left[1_{t_i \leq \tau \leq t_{i+1}} h(t_i, L) \Bigg| \mathbb{F}_t \right] \Delta t = 0 \end{align} $$

In fact, this approximation assumes a lot of things, especially on the regularity of the solution $h(t,x)$.

The above argument is just for showing the idea and is not at all rigorous mathematically. It assumes the existence and the regularity of the solution of the SDE and the PDE, and never mentions about the uniqueness.

If someone knows rigorous treatment, please put an answer or a comment to let me know.