Why discounted derivative price is a martingale?

Question

Usually after showing that discounted stock price process is martingale under the risk-neutral measure, most authors say that this implies that the discounted derivative price process is a martingale as well. But I have difficulties to see how the former implies the latter. Generally a function of a martingale is not a martingale. Could anyone put some more light on this?

Answer 1

Under a Black-Scholes framework, the dynamics of the stock price under the risk-neutral measure $\mathbb{Q}$ are given by ...

$$ S_t = r S_tdt +\sigma S_tdW^{\mathbb{Q}}_t $$

... and those of the risk-free bond by:

$$ \begin{align} dB_t = rB_tdt \end{align} $$

Let us define the derivative value as $V(t,S_t)$, which only depends on the time $t$ and the stock price $S_t$. For notational clarity we will also write $V_t$.

Because the value of the derivative only depends on time and stock price, we can form a portfolio made up of $w_S(t)$ shares of stocks and $w_B(t)$ bonds which replicates the derivative's payoff. Because the portfolio replicates the payoff, by no arbitrage both the derivative and the portfolio must have the same value for all $t$ between $0$ and the derivative's maturity $T$:

$$ V_t = w_S(t)S_t+w_B(t)B_t$$

This portfolio needs to be self-financing, meaning that the net impact of changes in the allocation $w(t)=(w_S(t),w_B(t))$ must be equal to $0$ $-$ i.e. no cash injections into or withdrawals from the portfolio:

$$ dV_t = w_S(t)dS_t+w_B(t)dB_t \quad (1)$$

To ensure property $(1)$ is verified, a trivial strategy is to choose:

$$ w_B(t) = \frac{V_t - w_S(t)S_t}{B_t} $$

Indeed, at each time step you rebalance your portfolio by buying (selling) $w_S(t+dt)-w_S(t)$ shares of stock and selling (buying) bonds in a such a quantity that makes the derivative value and the portfolio value match.

We end up for the following dynamics for the derivative's value:

$$ dV_t = w_S(t)dS_t + r(V_t-w_S(t)S_t)dt \quad (2)$$

Now, consider the dynamics of the discounted stock price. By Ito's lemma:

$$ \begin{align} d\left(e^{-rt}S_t\right) & = -re^{-rt}S_tdt + e^{-rt}dS_t \\[9pt] & = e^{-rt}\sigma S_tdW_t^{\mathbb{Q}} \end{align}$$

Hence, as stated in your question, the discounted stock price is a martingale$^1$ under $\mathbb{Q}$. Let us now derive the dynamics of the discounted derivative value process with Ito's lemma again:

$$ \begin{align} d\left(e^{-rt}V_t\right) = -re^{-rt}V_tdt + e^{-rt}dV_t \quad (3) \end{align} $$

Now, combining $(2)$ and $(3)$:

$$ \begin{align} d\left(e^{-rt}V_t\right) & = -re^{-rt}V_tdt + w_S(t)e^{-rt}dS_t + r(V_t-w_S(t)S_t)e^{-rt}dt \\[9pt] & = w_S(t)e^{-rt}dS_t - w_S(t)e^{-rt}rS_tdt \\[9pt] & = w_S(t)e^{-rt}rS_tdt + w_S(t)e^{-rt}\sigma S_tdW_t^{\mathbb{Q}} - w_S(t)e^{-rt}rS_tdt \\[9pt] & = w_S(t)e^{-rt}\sigma S_tdW_t^{\mathbb{Q}} \\[9pt] & = w_S(t) \, d\left(e^{-rt}S_t\right)\end{align} $$

The dynamics of the discounted derivative value are drift-less $-$ i.e. we only have a term in $dW_t^{\mathbb{Q}}$ left $-$ hence the discounted derivative price is a martingale under the risk-neutral measure$^1$.

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Technical point $1$: an Ito process $X_t$ with $0$ drift is strictly speaking a local martingale. A further technical condition is required to ensure that the process is also a martingale: one such possible condition is that the expected quadratic variation of the process must be finite. In general, the local martingales we work with in financial engineering verify this type of condition so we do not bother proving that the local martingale is also a martingale.

Here, in the case of the discounted stock price:

$$ \mathbb{E}\left[[S,S]_t\right] = \mathbb{E}\left[\int_0^t\sigma^2e^{-2ru}S_u^2du\right] < \infty$$

We have:

$$ \begin{align} \mathbb{E}\left[\int_0^t\sigma^2e^{-2ru}S_u^2du\right] & = \sigma^2S_0^2 \mathbb{E}\left[\int_0^te^{-2ru}e^{2\left((r-\frac{\sigma^2}{2})u + \sigma W_u\right)}du\right] \\[12pt] & = \sigma^2S_0^2 \int_0^te^{-\sigma^2u} \mathbb{E}\left[e^{2\sigma W_u}\right]du \\[12pt] & = \sigma^2S_0^2 \int_0^te^{-\sigma^2u} e^{2\sigma^2 u}du \\[12pt] & = \sigma^2S_0^2 \int_0^te^{\sigma^2u}du \\[12pt] & = S_0^2 \left(e^{\sigma^2t}-1 \right) \end{align}$$

Which is finite for all $t$. For the derivative:

$$ \mathbb{E}\left[\int_0^t\left(w_S(u)\sigma e^{-ru}S_u\right)^2du\right] < \infty$$

Conditions over $w_S(t)$ are needed to ensure the expectation is finite. In the case of a European call option, we would have:

$$ \begin{align} & w_S(t) = -\frac{\partial V}{\partial S}(t,S_t) = -\mathcal{N}(d_1) < 1 \\[6pt] & \Rightarrow 0< w_S(t)^2 < 1 \end{align} $$

Thus the European call price is also a martingale:

$$ \mathbb{E}\left[\int_0^t\left(w_S(u)\sigma e^{-ru}S_u\right)^2du\right] < \mathbb{E}\left[\int_0^t\left(\sigma e^{-ru}S_u\right)^2du\right] < \infty$$

Answer 2

We decree that $D_t$ has a certain process which makes it a martingale. In particular, we let $$ D_t = \mathbb{E} ( D_T \, | \, \mathcal{F}_t) $$ This is trivially a martingale by the tower law. Since the discounted stock price and discounted bond price are martingales, we have made everything a martingale and that ensures that there is no arbitrage in the martingale measure. However, the real world measure is equivalent to it so there are none there either.

Answer 3

This has nothing to do with Black Scholes or any other stock evolution model. Although they are interesting examples.

The fact that prices from european derivatives are martingales comes from the market completeness, a more general assumption in asset pricing.

Market completeness implies (by the second theorem of asset pricing) that any european derivative can be replicated/attained. The value of the replication strategy is a martingale and the price, which is equal to the value, is therefore a martingale.

Any function of a Martingale is not a martingale. The heart of the question can be seen in a discrete model easily. Let $H$ be the payoff of a european derivative, $(\hat X_t)_{t\in\{0,..T\}}$ the replication strategy, $\pi_t$ the vector of the replicating portfolio, and $S_t$ the vector of all assets.

The following equality holds: $$\hat X_t=\hat X_0+\sum_{\tau=1}^T\pi_t\Delta \hat S_t\\ $$ Where $\Delta \hat S_t= \hat S_t-\hat S_{t-1}$

$\hat X_t$ is a martingale because $\hat S_t$ is a martingale and $\pi_t$ is a previsible process.