Why Risk-Neutral Pricing Uses the Risk-Free Drift

# Intuitive Reasoning for Using Risk-Neutral Measure


# Intuitive Reasoning for Using Risk-Neutral Measure












Although we thoroughly covered risk-neutral pricing in university I never fully understood it in the context of continuous-time processes.

But first of all, lets consider a discrete time example:

Here we want to evaluate the call option price $C_0$ with strike $K=100$. If the interest rate (until the option expiry) is $r=2\%$, then we need to solve $$\Delta\cdot S_u + \phi\cdot(1+r)=C_u$$ $$\Delta\cdot S_d + \phi\cdot(1+r)=C_d$$ for $\Delta,\phi$, which then gives $C_0 = \Delta\cdot S_0 + \phi \approx 10.6952$.

Here I can see how the real-world probaility $p$ $-$ and ultimately the real-world drift $-$ do not matter per se as we exactly replicate the option with the stock itself and some cash account.

But on the continuous side, things are not that simple. And I am just not sure why we would always use the risk-free $r$ as drift instead of the real drift $\mu$. For example, say we have 2 stocks that are exactly the same (same current price & volatility) but differ only in terms of their drift parameters $\mu_1,\mu_2$. Then a call option on stock 1 will have exactly the same price as a call option on stock 2 (given that strike and maturity are the same), because both would use $r$ as the "drift" for pricing. But if $\mu_1>\mu_2$ then everybody would want buy the call option on stock 1.

Any advice would be greatly appreciated.

## Answer by Mark Joshi (score 11, accepted)

https://quant.stackexchange.com/a/17759

this is probably the most asked question in quantitative finance... There are many answers. One nice example to consider is what if the calls were struck at zero. The call then pays the stock price at time $T$ and so it's value today must the stock price today since we can replicate by holding one unit of stock. This will be true regardless of the drift of the stock.

Another point is that put-call parity forces puts and calls to have the same value at the money (actually at the forward.) Drift arguments that send call prices up, send put prices down, but they have to be equal to each other.

(A large part of my book Concepts etc is devoted to this question.)

## Answer by Yawning Lion (score 1)

https://quant.stackexchange.com/a/17737

You may bet on stock 1 by buying a call option on stock 1, and drive up the option price. But some arbitrageurs will immediately short the option and hedge with stock 1, pocketing the profit. These arbitrages will force the call option back to normal.

Discrete or continuous-time, the logic is the same.