Monte Carlo Valuation of a Three-Year Autocallable Equity Payoff

# Price of a simple autocall - Sebastien Bossu Advanced Equity derivatives


# Price of a simple autocall - Sebastien Bossu Advanced Equity derivatives












I am reading Advanced Equity Derivates by Sebastien Bossu and trying to do the exercises. In chapter 1 we have the following question :

Consider an exotic option expiring in one, two, or three years on an underlying asset S with the following payoff mechanism:

- If after one year S1 > S0 the option pays off 1 + C and terminates;

- Else if after two years S2 > S0 the option pays off 1 + 2C and terminates;

- Else if after three years S3 > 0.7 × S0 the option pays off max(1 + 3C, S3∕S0);

- Otherwise, the option pays off S3/S0.

Assuming S0 = $100, zero interest and dividend rates, and 25% volatility, estimate the level of C so that the option is worth 1 using Monte Carlo simulations.

In the solution manual it gives C ≈ 12% without details. I executed the following code ( apologies as it is not clean or efficient, i am just trying to get the answer).

```
import numpy as np

sigma = 0.25
S0 = 100
N = 100000

simuls = []
for _ in range(N):
    S1 = S0 * np.exp(-0.5 * sigma**2 + sigma * np.random.normal())
    S2 = S1 * np.exp(-0.5 * sigma**2 + sigma * np.random.normal())
    S3 = S2 * np.exp(-0.5 * sigma**2 + sigma * np.random.normal())
    simuls.append([S1,S2,S3])
    
coupon = 0.0765
payoff = []
for simul in simuls:
    if simul[0]>S0:
        payoff.append(1+coupon)
    elif simul[1]>S0:
        payoff.append(1+2*coupon)
    elif simul[2]>0.7*S0:
        payoff.append(max(1+3*coupon, simul[2]/S0))
    else:
        payoff.append(simul[2]/S0)
        
print(np.mean(payoff))
```

I find that a coupon of around 7.65% makes this product worth 1 ( did not solve properly with optimisation). As I am struggling to see where I went wrong I wanted to know if you found 12% and if yes how ? Thank you very much in advance

## Answer by siou0107 (score 1)

https://quant.stackexchange.com/a/73862

I had about the same level as you when I tried: $0.076\%$.

Code is below: feel free to get inspired by it, as well as giving feedback. That's the best way to collaboratively learn!

```
import numpy as np
import numpy.random as npr
from statistics import mean

def payoff(path, coupon):
    if path[1] > path[0]:
        return 1 + coupon
    elif path[2] > path[0]:
        return 1 + 2 * coupon
    elif path[3] > 0.7 * path[0]:
        return max(1 + 3 * coupon, path[3]/path[0])
    else:
        return path[3]/path[0]
testpath = [100, 101, 120, 60]
print(payoff(testpath, 0.1))

spot, volatility, coupon = 100, 0.25, 0.076
nPaths = 500000
path = spot * np.ones((nPaths, 4))
epsilon = npr.randn(nPaths, 3)
for i in range(nPaths):
    for j in range(3):
        path[[i], [j + 1]] = path[[i], [j]] * np.exp(-0.5 *
    volatility**2 + volatility * epsilon[i, j])

print(mean([payoff(path[i], coupon) for i in range(nPaths)]))
```