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My math finance professor once said someting that I can't make sense of. Hope you can answer:

For a foward process the non-discounted price for a European call option under Bachelier is $$C_t = \left(f_t-K^*\right)\Phi\left(\frac{f_t-K^*}{v(t)}\right) +v(t)\,\phi\left(\frac{f_t-K^*}{v(t)}\right)$$

$C_t$ is a martingale

How come $C_t$ is a Martingale? I have been through most of Bjork's book Arbitrage Theory and I know that the fair valur of a derivative $X$ is $$E^Q_t\left[\frac{X}{B_t}\right]$$which is a Martingale, right?

How come the non-discounted Bachelier call option price is martingale?

However this is from my personal notes, so I might have it wrong.

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marked as duplicate by Daneel Olivaw, skoestlmeier, amdopt, byouness, Attack68 Jul 25 at 19:25

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  • $\begingroup$ The question here is, martingale under which measure? $\endgroup$ – byouness Jul 14 at 21:09
  • $\begingroup$ I am not sure what you mean. This under Bachelier model. So using the dynamics of $f_t$: $df_t$ and then compute $dC_t$ we should end up getting a process without a drift.. This is what I understand when I read these notes. $\endgroup$ – econmajorr Jul 14 at 21:35
  • $\begingroup$ $X_t/B_t$ is the non discounted option price. It’s a martingale in the measure of the zero coupon bond maturing at the expiration date. $\endgroup$ – dm63 Jul 15 at 0:04
  • $\begingroup$ What I mean is that when saying that some process is a martingale, you need to specify under which measure. For example, the discounted option price is a martingale under the risk-neutral measure, but it is not a martingale under other measures. When you change the measure the drift changes. $\endgroup$ – byouness Jul 15 at 7:53
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Let $P(t,T)$ denote the time $t$ price of a zero-coupon bond maturing at time $T$ and $\mathbb{Q}_T$ be the associated equivalent martingale measure which uses $P(t,T)$ as numeraire. Then, for any $\mathcal{F}_T$-measurable payoff $\xi$, the time $t$ value of $\xi$ is given by $$V_t=P(t,T)\cdot\mathbb{E}^{\mathbb{Q}_T} [\xi\mid\mathcal{F}_t].$$ The undiscounted time $t$ price is given by $$\tilde{V}_t = \frac{V_t}{P(t,T)} = \mathbb{E}^{\mathbb{Q}_T} [\xi\mid\mathcal{F}_t].$$ And indeed, $(\tilde{V}_t)$ is a $\mathbb{Q}_T$-martingale. Assuming integrability and adaptness (trivial), we need to show the martingale property. To this end, let $0\leq s<t\leq T$. Then, by the tower law, \begin{align*} \mathbb{E}^{\mathbb{Q}_T}[\tilde{V}_t\mid\mathcal{F}_s] &= \mathbb{E}^{\mathbb{Q}_T}\left[\mathbb{E}^{\mathbb{Q}_T} [\xi\mid\mathcal{F}_t]\bigg|\mathcal{F}_s\right] \\ &= \mathbb{E}^{\mathbb{Q}_T}[\xi\mid\mathcal{F}_s] \\ &= \tilde{V}_s. \end{align*}

Please note the following:

  • This result is completely independent of the Bachelier model and equally applies to the Black-Scholes model, the Heston model and others.
  • If interest rates are deterministic, so are bond prices and back accounts. Thus, the forward measure $\mathbb{Q}_T$ coincides with the ``standard'' risk-neutral measure $\mathbb{Q}$ which uses a risk-free bank account $B_t=e^{\int_0^t r(s)\mathrm{d}s}$ as numeraire.
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  • $\begingroup$ That is the point. They do not need to be deterministic. Even if they are stochastic, the statement holds. Just the measure changes. If they are stochastic, we use the $T$ forward measure and if they are deterministic, we may use the "standard" risk-neutral measure. The theorem holds anyway. Well, provided there are equivalent martingale measures in the first place, i.e. no arbitrage. $\endgroup$ – KeSchn Jul 16 at 15:41
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    $\begingroup$ Just that kinda answer one could hope for. Thanks! $\endgroup$ – econmajorr Jul 17 at 11:51
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    $\begingroup$ Happy to help. :) I should have mentioned that the above argument does not only apply to call options but to any other claim whose price we can find as conditional expectation, as well. Finally, note the following well-known theorem from martingale theory: For any integrable random variable $X$, the process $(X_t)$ defined by $X_t=\mathbb{E}[X\mid\mathcal{F}_t]$ is a martingale which immediately follows from the tower law. $\endgroup$ – KeSchn Jul 17 at 11:55

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