Monte Carlo Payoffs for Knock-Out Options

Summary

The note corrects a proposed Monte Carlo payoff for a barrier knock-out call. Multiplying asset prices observed at maturity and at an intermediate time does not represent the stated contract. The payoff should be the positive part of the maturity asset price minus the strike, provided the asset has never reached or crossed the upper barrier during the option’s life; otherwise, the payoff is zero. The answer expresses this condition using the first barrier-hitting time.

For simulation, each path must be checked for a barrier breach, and a breached path receives no payoff; a surviving path receives its ordinary call payoff at maturity. The note warns that discrete simulation steps can miss crossings between observation times, so fine time slices may be needed. It also states that an analytical valuation exists under Black–Scholes assumptions. It does not give a convergence analysis or compare simulation methods for handling missed crossings.

Key ideas

• A knock-out call pays only if the asset avoids the barrier throughout the option’s life.
• Its surviving-path payoff is the positive part of the maturity price minus the strike.
• A product of asset prices at two dates does not encode the specified barrier payoff.
• Monte Carlo paths must be checked for barrier breaches before assigning payoffs.
• Coarse time steps can miss barrier crossings, while an analytical Black–Scholes valuation is available under the stated setting.

Tags

# Is this formula correct to estimate a knock out option price using monte-carlo?


# Is this formula correct to estimate a knock out option price using monte-carlo?












I have a knock-out option with barrier $L>0$ and strike $K$ that pays at maturity $(S-K)_+$. So, positive payoff occurs only in case the price stays below the barrier over life of the option.

I am analyzing estimation of the price of this option using Monte Carlo simulation and came across with the below definition of the payoff:

$$\Pi_0=e^{-rT} E^Q [S_T * S_{T/2}]= e^{-rT} \frac{ \Sigma \ S_T(\omega_i) S_{T/2}(\omega_i)}{N}$$

Can this be correct? If yes, could you please explain the logic behind the multiplication $S_T S_{T/2}$?

## Answer by Gordon (score 1, accepted)

https://quant.stackexchange.com/a/26121

Your valuation is NOT for the knock-out option that you have specified. Let \begin{align*} \tau = \inf\{t \mid 0 \le t \le T, S_t \ge L\}. \end{align*} Here, we set the infimum of an empty set to $\infty$. Then, the payoff of the knock-out option is of the form \begin{align*} (S_T-K)^+ 1_{\tau = \infty}. \end{align*} Under the Black-Scholes setting, this option can be valued analytically.

Monte Carlo simulation is not always good for such type of payoff. For example, you may need to have a particularly fine time slices, and set the payoff to $0$ on each path once the simulated asset value is at or above the barrier level $L$; otherwise, simulate until the maturity $T$, and set the payoff to $(S_T(\omega)-K)^+$, where we use $\omega$ to denote a path.

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This summary was written by Stratmill's research agent from the original; it is not a copy of the source.