Question Regarding B-S Option Pricing

[Hob]

Hi all,

I have a question regarding the B-S method for pricing an option.

I have some actual data for a stock (shown in the image as the blue line). On the assumption from the B-S that dS changes as: ds = (u dt + o- dW) * S, I have simulated a predicted stock market change for 3 years (red line).

Given that it is random walk, there is a large variation in the predicted stock price at the end of 3 years.
Assuming someone wanted to price an option at the point indicated on the graph (Stock Price is $15), would you use a Strike price of $27.2 (the predicted one ?).

If you do, you have the variables (S = $15, K = $27.2, r = 0.03 (assumed), T = 3 (3 years to expiry) and Volatility = 4.52 (estimated)).

This gives a value to the Call option of $13.78 and the Put option at $24.84.

But given that you have dW (and I am assuming that K is generated from the end of the walk?), the Call/Put valuation would change every time you calculated it as the walk is random?


many thanks,

Hob

 

[aaronhotchner]

I'm not sure where to start.

K is not some assumed, calculated, or random value. It's defined by the contract. You don't get K from anything other than the parameters of the option. It is a constant, and it is decided by the people trading the contract. No generation. Nothing special about it.

Your one simulated stock price path doesn't mean anything, either. It has virtually no bearing on valuing the option, or at least it shouldn't -- you should have (hundreds of) millions of paths, and then you calculate the option value for each path and average them.

However, you're asking a question about Black-Scholes, so I suppose the valuation process is even easier: you don't do any simulations at all. Given the current stock price, strike price, interest rate, time to maturity, and (constant) volatility, the call and put have particular values defined by BS, because it is just a function.

 

[Hob]

I'm not sure where to start.

K is not some assumed, calculated, or random value. It's defined by the contract. You don't get K from anything other than the parameters of the option. It is a constant, and it is decided by the people trading the contract. No generation. Nothing special about it.

Ah ok, I see now.

Your one simulated stock price path doesn't mean anything, either. It has virtually no bearing on valuing the option, or at least it shouldn't -- you should have (hundreds of) millions of paths, and then you calculate the option value for each path and average them.

So a MC simulation would converge to the options price given from the BS pricing?

 

[aaronhotchner]

Yes, Monte Carlo will converge to Black-Scholes price as long as the simulations of the stock price are consistent with Black-Scholes (u*dt + sigma*dW with u, sigma being constant and a sufficiently small dt).