I am reading a paper which tries to convey the intuition behind the Black-Scholes pricing formula. In that paper, the author states the following two things without proof, and I would like to know why they are true.

It turns out that one can adjust the probability distribution of the stock price in such a way that the current value of any stock-price contingent claim equals the expected future payoff to the claim, computed using the adjusted probabilities, discounted at the riskless rate.

The author also states the practical aspect of how such risk adjustment is performed under Black-Scholes:

Risk adjustment of the probabilities in this model consists in replacing $\mu$ by $r$, the riskless interest rate. The risk-adjusted probability distribution is such that $S_t$ is still lognormally distributed, but the mean and variance of the normally distributed variable $logS_t$ are now $logS + (r − \sigma^2/2)t$ and $\sigma^2t$, respectively.

I think this is a loaded question, and so even reference material will be very much appreciated.


1 Answer 1


This holds due to a change of measure. There is the real-world $\mathbb{P}$ and the risk-neutral world $\mathbb{Q}$. (I am going to assume constant interest rate $r$)

The first fundamental theorem of asset pricing states that if there are no arbitrage strategies in a market, then there exists at least one probability measure $\mathbb{Q}\sim\mathbb{P}$ such that discounted stock prices $(S_te^{-rt})$ are $\mathbb{Q}$-martingales, i.e. for any $T\geq t$, we have $$\mathbb{E}^\mathbb{Q}[S_Te^{-rT}\mid\mathcal{F}_t]=S_te^{-rt}.$$ We can then show that the value process of any admissible claim is alslo a $\mathbb{Q}$-martingale. The value process of a derivative with terminal payoff $V_T$ is simply the discounted payoff, i.e. $V_t^\mathbb{Q}=\mathbb{E}^\mathbb{Q}[V_T\mid\mathcal{F}_t]$ (some authors define the value process directly as conditional expectation of the discounted payoff). This means $$V_t^\mathbb{Q}= e^{-r(T-t)} \mathbb{E}^\mathbb{Q}[V_T].$$

For example, consider a vanilla call option and set $t=0$. You obtain $$\mathrm{Call} =e^{-rT}\cdot \mathbb{E}^\mathbb{Q}[\max\{S_T-K,0\}].$$

So, in order to price the call option, you only need to compute this expectation in the risk-neutral world.

The second question is about the distribution of $(S_t)$ under $\mathbb{P}$ and $\mathbb{Q}$. You need the latter in order to compute the expectation above. Black and Scholes (1973) assume that the stock price follows a geometric Brownian motion $$\mathrm{d}S_t=\mu S_t \mathrm{d}t+\sigma S_t\mathrm{d}W_t.$$ To switch from $\mathbb{P}$ to $\mathbb{Q}$, you need to change the drift $\mu$. The volatility $\sigma$ remains the same. This follows from Girsanov's Theorem. It turns out that you change $\mu$ (under $\mathbb{P}$) to $r$ (under $\mathbb{Q}$). Remember that all stocks have return $r$ since no one pays you a premium for risk. Thus, in the risk-neutral world, the price follows $$\mathrm{d}S_t=r S_t \mathrm{d}t+\sigma S_t\mathrm{d}W_t.$$ You can solve this SDE with Ito's Lemma to get a geometric Brownian motion. This process is for each time point log-normally distributed. Using this distribution, you can compute the above expectation and you will arrive at the solution from Black and Scholes (1973).

Risk-neutral pricing, i.e. finding derivatives prices as expectation of discouned payoffs in the risk-neutral world is a super powerful tool :)

  • 1
    $\begingroup$ This is a very thorough answer. I absolutely loved it. Many thanks! $\endgroup$ Aug 29, 2019 at 6:09

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.