Monte Carlo Pricing of a Leveraged Down-and-Out Call
Summary
The document describes a proposed Monte Carlo approach to valuing a leveraged down-and-out call under geometric Brownian motion. The contract sketch has the buyer contribute part of the asset price while financing the remainder, pay interest on that financing, and lose the position if the asset crosses a barrier that rises with financing costs. At the barrier, proceeds are the greater of zero and the asset value less the accumulated financing. The author asks how to model the barrier and leverage consistently.
The Python example simulates a terminal asset price and discounts a vanilla call payoff when a barrier condition is met. It is presented as an uncertain attempt rather than a validated pricing method, and gives no benchmark prices or evidence of accuracy. In particular, a single terminal draw cannot establish whether a continuously monitored barrier was crossed; the barrier adjustment and financing payoff also need a clearly specified time convention and contract definition. The option is described as potentially open-ended, while the code uses a fixed maturity, so the example does not resolve that modeling choice.
Key ideas
- The contract combines financing costs with a down-and-out barrier condition.
- The proposed simulation uses geometric Brownian motion to generate terminal asset prices.
- A barrier option generally requires modeling whether the asset crossed the barrier during the contract period.
- The document does not establish a correct barrier adjustment or provide validated pricing results.
- The contract's maturity and financing payoff conventions need to be specified for valuation.
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Full text
# How to price a down-and-out leveraged barrier call option using Brownian motion?
# How to price a down-and-out leveraged barrier call option using Brownian motion?
I am trying to price a type of leveraged down-and-out (LDAO) barrier call option, using geometric Brownian motion.
My python script is below. I am not sure how to correctly model the increasing barrier `B` and leverage factor that multiplies the payoff when the stock price goes up.
The characteristics of this option are as follows.
- The "leveraged" part op the LDAO looks like this. When you buy the call, you only pay part of the spot price `S0` of the underlying asset. The seller provides financing `F0` to buy the rest. In other words, the buying price `P` of the option is: `P = S -/- F`.
- As a buyer, you pay interest `i` on the financing level `F`.
- The "down-and-out" part is structured as follows. When the price `S` of the underlying drops below the barrier `B`, the option is cancelled and the underlying asset is sold at the active market price. When the spot price `S1` is lower than the financing `F`, your end value is zero. If the spot price is higher than the financing, your payout is `S1 - F1` (F1 includes the interest). Your payout cannot be negative.
- The barrier B is increased daily by the amount of the interest on the financing level `F`. So `B1 = B0 + F*(1 + i)^t`
- The barrier B is always below the price of the underlying asset, but above the financing level: `S0 > B > F`
I have tried to implement this in the Python 3 script below. Any suggestions on how to get this right?
```
import numpy as np
from math import log, e
P = 30 # This is what you pay (S -/- F)
S = 360 # spot price of the stock
K = 340 Exercise price is equal to the stop-loss barrier (as far as I understand it)
B = 340 # the stop-loss barrier at which the option is cancelled
F = 330 # The financing level
T = 1 # Time to maturity. But in priciple, the option runs indefintely as long as S > B
i = 0.02 # Annualised interest rate on the financing F
r = 0.00135 # Risk-free rate of return
impliedVolatility = 0.3
num_reps = 100
def barrier_option(option_type, s0, strike, B, F, maturity, i, r, sigma, num_reps):
payoff_sum = 0
for j in range(num_reps):
st = S
st = st*e**((r-0.5*sigma**2)*maturity + sigma*np.sqrt(maturity)*np.random.normal(0, 1))
B_shift = B + F*(1+i)*np.sqrt(maturity) # Here the interest I on F gets adjusted by increasing B
non_touch = (np.min(st) > B_shift)*1
if option_type == 'c':
payoff = max(0,st-strike)
elif option_type == 'p':
payoff = max(0,strike-st)
payoff_sum += non_touch * payoff
premium = (payoff_sum/float(num_reps))*e**(-r*maturity)
return premium
Bar = barrier_option('c', S, K, B, F, (T*252)/365, i, r, impliedVolatility, num_reps)```
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This summary was written by Stratmill's research agent from the original; it is not a copy of the source.