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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?

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    $\begingroup$ Functions of a martingale are not generally martingales. But expectations of functions of martingales are always martingales by the tower property. $\endgroup$ – user9403 Jun 15 '17 at 14:13
  • $\begingroup$ @user9403 Take a simple random walk on integers - a martingale. Consider its square (a function). It is not a martingale. Is the expectation of the square a martingale? (perhaps I'm missing something obvious) $\endgroup$ – Dan Getz Jun 23 '17 at 13:27
  • $\begingroup$ @DanGetz in our case the derivative price is $V_t = \mathbb{E}[e^{-r(T-t)}P_T|\mathcal{F}_t]$ where $P_T$ is the terminal payoff. Letting $s<t$ our presumed martingale is $M_t=e^{-r(t-s)}V_t$. Thus by the tower property: $\mathbb{E}[e^{-r(t-s)}V_t|\mathcal{F}_s] = \mathbb{E}[e^{-r(t-s)}e^{-r(T-t)}P_T|\mathcal{F}_s] = \mathbb{E}[e^{-r(T-s)}P_T|\mathcal{F}_s]=V_s$ by definition. $\endgroup$ – Daneel Olivaw Jun 23 '17 at 16:13
  • $\begingroup$ @DaneelOlivaw Thanks. My comment was regarding the generality of "expectations of functions of martingales are always martingales by the tower property", which seems not to be the case (unless I am still mistaken, but see the example given in my comment). $\endgroup$ – Dan Getz Jun 23 '17 at 16:44
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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.

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    $\begingroup$ Suppose prices are equal to expectation under some measure: $P_{t0}=E_{t0}[g(S_T)]$, $P_{t1}=E_{t1}[g(S_T)]$, etc. What can we say about the evolution of prices in this setup? Let us compute $E_{t0}[P_{t1}]=E_{t0}[E_{t1}[g(S_T)]]=E_{t0}[g(S_T)]=P_{t0}$ which shows that P is a Martingale. QED. Tedious perhaps but explicitly shows how the Tower Property of Expectation is used (to eliminate nested exectations). $\endgroup$ – Alex C Jun 16 '17 at 2:50
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    $\begingroup$ @Mark Joshi Agreed with the answer. However, you state that $D_t$ is a martingale by decree and the line below you say that $D_t$ is a martingale by the tower law. It is either one or the other, not both. $\endgroup$ – JejeBelfort Jun 16 '17 at 9:35
  • $\begingroup$ I do agree with @JejeBelfort, the first part of the answer is misleading, specially because just below it is mathematically explained why the derivative price process is a martingale. $\endgroup$ – Daneel Olivaw Jun 16 '17 at 10:02
  • $\begingroup$ @DaneelOlivaw Thanks, but it seems not everyone agrees as my edit to the post was reverted... $\endgroup$ – JejeBelfort Jun 16 '17 at 10:28
  • $\begingroup$ i have clarified. $\endgroup$ – Mark Joshi Jun 17 '17 at 0:38
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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$^1$.


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:

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

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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.

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