Pricing Multi-Barrier Knock-In and Knock-Out Call Payoffs

# How to price "knock-in, knock-out" options having a payoff at $T$


# How to price "knock-in, knock-out" options having a payoff at $T$












I'm trying to discover every secrets of pricing "knock-in, knock-out" options. However, I'm a bit confused with some scenarios.

Those 3 scenarios are :

- The payoff is equal to $0$, if $H_1 = 140$ was not hit at any time between today and $T$

- The payoff is equal to $2\max(S_T - 120;0)$, if $H_1$ was hit but $H_2 = 160$ is not hit in $(t_0,T)$

- The payoff is equal to $\max(S_T - 120;0)$, if $H_2$ is hit

Let's use the following "up-and-out" option prices table for different barrier levels for the sake of the example $K = 120$ and maturity equal $T$ (same as the maturity above that we have to price the options)

```
    **Barrier level**     **Price**
          140              0.3986
          150              1.0607
          160              1.8897
          170              2.7140
          180              3.4286
        Infinity           5.3686
```

So please, explain to me how to price the "knock-in, knock-out" option in the three scenarios I have mentioned.

PS: it's the first time I post on stackexchange, so please have some indulgence :)

## Answer by Nico bellic (score 2)

https://quant.stackexchange.com/a/53640

You're going to have to use Monte Carlo to correctly account for the knock-in / knock-out effects. Stochastic models such as Black-Scholes are inherently memoryless and won't be able to account for that. Because of that I don't think a vanilla barrier options table will help you.

You can use binomial tree pricing models to create your payoff function, decompose your time period $T$ into $n$ steps of $\Delta_t = \frac{T}{n}$ in length, accordingly adjust the volatility for each $\Delta_t$ and run it a few times. For sufficiently large $n$, the results from each run should be fairly close.

## Answer by Harry (score 2)

https://quant.stackexchange.com/a/53660

In practice, we use stochastic normal vol and simulate the asset at each time step. However, its a pain. My personal approximation is to price it as a product of 2 (possibly windowed) barriers using closed form approximates.This way, you can structure your knock ins and knock outs as ridiculous payoffs without really stopping to think. Eg a down and in call paired with an up and out call. Or vice versa. Haug has closed form approx for window barriers. This will get you within the B/O provided you have the right level of vol/fwd vol

## Answer by Ivan (score 0)

https://quant.stackexchange.com/a/53667

Let $S^* = max(S_t)_{t\in[0...T]}$, let $C_K = Max(S_T-K,0)$ and let $\mathrm{1_{A}}$ mean $1$ if event $A$ occurs, and $0$ otherwise.

Your payoff is then $\phi_T=C_{120}\mathrm{1_{S^*>H1}} + C_{120}\mathrm{1_{S^*>H1}}\mathrm{1_{S^*<H2}} = 2.C_{120}\mathrm{1_{S^*>H1}} - C_{120}\mathrm{1_{S^*>H1}}\mathrm{1_{S^*>H2}}$

Now because $H2>H1$ this is actually $\phi_T = 2.C_{120}\mathrm{1_{S^*>H1}} - C_{120}\mathrm{1_{S^*>H2}}$

So what you have is $2$ calls strike $K=120$ up-and-in at $H1=140$ minus $1$ call strike $K=120$ up-and-in at $H2=160$.

Now you also know that $UIC(K,H) = C_K - UOC(K,H)$.

So $\phi_T = 2.(C_{120}-UOC(120,140)) - (C_{120}-UOC(120,160))$

And your table gives you the prices of all the necessary $UOC(K=120,H=140...160)$ as well as the vanilla call $C_{120}=UOC(K=120,H=\infty)$.