How would I prove that a Black-Scholes Model is not a Martingale if it has drift. In many cases it is just stated as a fact (without proof). For instance if Im looking at: $$dS_{t} = \mu S_{t} + \sigma S_{t} dB_{t}$$ $$S_{0} = 1, \beta _{t} = e^{et}, \tilde{\beta}=B_{t}+((\mu-2r)/ \sigma)t $$

From this I got to: $dS_{t} = \mu S_{t} + \sigma S_{t} d (\tilde{\beta}-((\mu-2r)/ \sigma)t)$ Which when expanded leads to: $$dS_{t} = \sigma S_{t}d \tilde{\beta}_{t} + 2rS_{t}dt$$ Is there a way to prove this is not a Martingale with something more substantial rather than "has drift term". Im assuming it would have to lead back to Solving the SDE. Starting with it being under P $$Z(t)=S(t)e^{-rt}= S(0)*e^{(\mu -r-1/2 * \sigma^{2})t +\sigma B(t)}$$ Then changing it to being under Q. $$Z(t)=S(0)*e^{(\sigma^{2})t +\sigma W(t)}$$ Any help on how to actually prove no drift is a martingale (hence with drift it isnt) would be most appreciated.

  • $\begingroup$ No drift is a martingale does not mean that it's not a martingale if it has a drift. Those are completely different assertions. $\endgroup$
    – Arshdeep
    May 30 at 13:02

That no drift is a martingale:

That ito integrals are martingales requires a simple but algebraically cubersome proof. You can refer to Shreve (continuous time) for the proof. You can also intuitively observe it as Brownian increments that are multiplied with their respective integrands are allocated independently of the integrand value. Thus, when all terms are added, the sum isn't biased upward or downward. This is exactly the martingale property.

That martingales have no drift:

The martingale representation theorem (MRT) asserts that (loosely speaking) a martingale can be represented as a stochastic integral. This necessitates that any other representation of the martingale as an Ito process:

$dX/X = a(t,X)dt+b(t,X)dW(t)$

must have $a(t,X):=0$. If not, we have:

$a(t,X)dt+b(t,X)dW(t)=c(t,X)dW(t)$ for all $t$, for some function $c$ due to MRT.

$a(t,X)dt=c(t,X)dW(t)-b(t,X)dW(t)$ for all $t$

As LHS has no quadratic variation, so must the RHS and thus $c(t,X)=b(t,X)$, leaving $a(t,X)$ to be identically 0.

Hence, nothing that is a martingale can have drift.

Thus martingales are equivalent to no drift.


I see a bit of confusion here. I try to clarify a bit. First of all, it is NOT the Black-Scholes model to be a martingale. It can be that a stock is a Martingale.

Let me recall here what does it mean to be a martingale. A stochastic process (which have "nice properties") $X_t$ is said to be a martingale (wrt a certain filtration $\mathcal{F}_t$) if: $$\mathbb{E}\left[ \ X_t \ | \ \mathcal{F}_s \right] = X_s$$ Let me move to Ito's processes. I define an Ito's process as: $$dX_t = \mu(t, X_t)dt + \sigma(t, X_t)dB_t$$ where $B_t$ is a SBM. I recall that the above is just a mere short-hand notation for: $$X_t - X_0 = \int_0^t \mu(s, X_s)ds + \int_0^t \sigma(s, X_s)dB_s$$ Consider (it suffices) the simple case in which $\sigma(s, X_s) = \sigma$ and $\mu(s, X_s) = \mu$. Then we have: $$X_t - X_0 = \mu t + \sigma B_t$$ this follows since constants can be brought outside the integral and that $\int_0^t dB_s = B_t$.

Now recall that $\mathbb{E}[B_t | \mathcal{F}_s ] = B_s$. So we have: $$\mathbb{E}\left[ \ X_t | \mathcal{F}_s \right] = X_0 + \mu t + \sigma B_s$$ recalling that: $$X_s = X_0 + \mu s + \sigma B_s$$ we see the two differ. Hence, $$\mathbb{E}\left[ \ X_t | \mathcal{F}_s \right] \neq X_s$$ i.e. $X_t$ is not a martingale.

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    $\begingroup$ Everything you write is correct, but keep in mind that your process, $X_t$, is an arithmetic Brownian motion. The Black Scholes model considers a geometric Brownian motion (with exponentials). Needless to say, your analysis easily applies to that case too! $\endgroup$
    – Kevin
    May 30 at 8:47
  • $\begingroup$ That is true, just wanted to save me some time :D $\endgroup$ May 30 at 8:52

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