Why Discounted Option Prices Are Martingales Under Risk-Neutral Pricing

# Why is $C(t,S_t)/B_t$ a martingale?


# Why is $C(t,S_t)/B_t$ a martingale?












In the derivation of the Black-Scholes formula given by Joshi (extract below), he says $C(t,S_t)/B_t$ is a martingale. Why?

I understand this can be deduced from the Black-Scholes PDE since the drift term is equal to zero. But how can he deduce $C(t,S_t)/B_t$ is a martingale before we have derived the Black-Scholes PDE.

## Answer by pbr142 (score 8, accepted)

https://quant.stackexchange.com/a/18131

It is important to note that he says: "In the risk-neutral world, $\frac{C(t,S_t)}{B_t}$ is a martingale." That is true by definition of what the risk-neutral measure is, also called martingale measure for exactly that reason.

A risk-neutral measure is defined such that asset prices deflated by the numeraire (unit with which prices are measured) are martingales. In your example, the standard numeraire is used: a bank account which is continuously reinvested at the risk-free rate. There are plenty of sources that discusses the (almost) equivalence of the existence of a risk-neutral measure and the absence of arbitrage (or more precisely: No free lunch with vanishing risk). The most comprehensive book on these issues is by Delbaen and Schachermeyer, but it's mathematically very demanding. I personally like the book by Duffie, but opinions about the books vary. A good compromise between mathematical completeness and financial intuition is the book by Björk.

In Joshi's Book "The Concepts and Practice of Mathematical Finance", the whole chapter 6 is devoted to this topic.

## Answer by Gordon (score 3)

https://quant.stackexchange.com/a/18136

In the Black-Scholes world, it is assumed that the option value $C(t, S_t)$ is replicable by an admissible self-financing trading strategy $\phi$, where $\phi_t=(\alpha_t, \beta_t)$. That is, \begin{align*} C(t, S_t) = \alpha_t B_t + \beta_t S_t, \end{align*} and \begin{align*} dC(t, S_t) = \alpha_t dB_t + \beta_t dS_t. \end{align*} Since $dB_t = rB_t dt$, and $dS_t = S_t(rdt + \sigma dW_t)$, then \begin{align*} d\bigg(\frac{C(t, S_t)}{B_t} \bigg) &= \frac{dC(t, S_t)}{B_t}-\frac{C(t, S_t)}{B_t^2}dB_t\\ &=\beta_t \frac{dS_t}{B_t} - \beta_t \frac{S_t}{B_t^2}dB_t\\ &=\beta_t \frac{S_t(rdt + \sigma dW_t)}{B_t} - \beta_t \frac{S_t}{B_t}rdt\\ &=\sigma\beta_t\frac{S_t}{B_t}dW_t. \end{align*} Therefore, $\{C(t, S_t)/B_t \mid 0\leq t \leq T\}$ is a martingale.

## Answer by Ulysses (score 0)

https://quant.stackexchange.com/a/18130

Not sure whether author explains that, but when talking about replicating portfolios, the discounted portfolio must be a martingale for non-arbitrage conditions. That's a very important fact in the theory of risk-neutral pricing. More details are in Shiryaev's "Essentials of Stochastic Finance", and maybe in some probability theory oriented books on the topic, e.g. Musiela and Rutkowski.

## Answer by SmallChess (score 0)

https://quant.stackexchange.com/a/18146

Let's go to Chapter 6 in the book where he talks about risk-neutrality. In the chapter he proved the existence of a risk-neutral measure. Now, let's stop and think what this means.

It means our hedging portfolio for the option can't beat the risk-free rate. Although not exactly mathematically correct, you can think C/B as a ratio of the option price to the zero-coupon-bond. If this ratio has a drift (i.e: a trend), the option (i.e: heading portfolio) performs better than the zero-coupon-bond. Obviously, it'll create an arbitrage opportunity. Mathematically, this is simply a martingale under the risk-neutral measure.

Now, we know the option price must be a martingale under the risk-neutral measure and we also know that the drift is zero, it's not hard to see why Mark Joshi derived the way he did in the book.