2
$\begingroup$

Up to now, I was only presented with stochastic integral processes of the form $\left(\int_{0}^{t}\phi_s\mathrm{d}B_s\right)_{t\geq 0}$ and the general way to show that such a process is a martingale goes like this:

  1. Show that $\phi$ is progressively measurable, e.g. by showing that $\phi$ is (a.s.) continuous and adapted.
  2. Show that $\mathbb{E}\left[\int_0^\infty \phi_s^2 \mathrm{d}s\right]<\infty$.

Let $\lambda>0$ and take $\left(\int_{0}^{t}e^{-\lambda s}\mathrm{d}B_s\right)_{t\geq 0}$ for example.

Define $\phi = (\phi_t)_{t\geq 0}$ by $\phi_t := e^{-\lambda t}$.

  1. The process $\phi$ is deterministic and therefore adapted. We then have that $\phi$ is progressive since it is continuous and adapted.
  2. $$\mathbb{E}\left[\int_0^\infty \phi_s^2 \mathrm{d}s\right] = \left[-\frac{1}{2\lambda} e^{-2\lambda s}\right]_{s=0}^{s=\infty} = \frac{1}{2\lambda} < \infty$$

It follows that the stochastic integral progress $\left(\int_{0}^{t} \phi_s \mathrm{d} B_s\right)_{t\geq 0}$ is a martingale.


Now let $\lambda>0$ and consider $\left(\int_{0}^{t}e^{-\lambda\left(t-s\right)}\mathrm{d}B_s\right)_{t\geq 0}$. The main problem here is that the integrand depends on $t$. I therefore can't solve it like above. If I try to drag $t$ out of the integrand, e.g. like

$$\int_{0}^{t}e^{-\lambda\left(t-s\right)}\mathrm{d}B_s = e^{-\lambda t}\int_{0}^{t}e^{\lambda s}\mathrm{d}B_s$$

then $\mathbb{E}\left[\int_0^\infty \phi_s^2 \mathrm{d}s\right]<\infty$ would no longer hold. I am confident that $\left(\int_{0}^{t}e^{-\lambda\left(t-s\right)}\mathrm{d}B_s\right)_{t\geq 0}$ is a martingale, but how can I show it?

$\endgroup$

2 Answers 2

5
$\begingroup$

The criterion you mentioned is certainly a sufficient condition, but not a necessary condition. So, let's go back to the basic.

Let $M_t = \int_{0}^{t} e^{-\lambda (t-s)} \, \mathrm{d}B_s$. Then for $0 \leq u \leq t$ and with the natural filtration $\mathcal{F}_t = \sigma(B_s : s \leq t)$,

\begin{align*} \mathbf{E}[M_t \mid \mathcal{F}_u] &= \mathbf{E}\biggl[ \int_{0}^{t} e^{-\lambda (t-s)} \, \mathrm{d}B_s \,\biggm|\, \mathcal{F}_u \biggr] \\ &= e^{-\lambda t} \mathbf{E}\biggl[ \int_{0}^{t} e^{\lambda s} \, \mathrm{d}B_s \,\biggm|\, \mathcal{F}_u \biggr] \\ &= e^{-\lambda t} \int_{0}^{u} e^{\lambda s} \, \mathrm{d}B_s \\ &= e^{-\lambda (t-u)} \int_{0}^{u} e^{-\lambda(u-s)} \, \mathrm{d}B_s \\ &= e^{-\lambda (t-u)} M_u. \end{align*}

In the third step, we utilized the fact that $t\mapsto \int_{0}^{t} e^{\lambda s} \, \mathrm{d}B_s$ is a martingale.

Since $M_u$ is not identically zero (Why?), this implies that $(M_t)$ is not a martingale.

$\endgroup$
1
  • $\begingroup$ Can you please show that $\left(\int_0^t e^{\lambda s}\mathrm{d}B_s\right)_{t\geq 0}$ is a martingale? $\endgroup$ Apr 7 at 13:32
2
$\begingroup$

Define $A_t := \int_0^t e^{\lambda s} dB_s$. You are asking if $X_t := e^{-\lambda t} A_t$ is a martingale. By Ito's formula, we have \begin{align*} dX_t &= e^{-\lambda t} dA_t -\lambda e^{-\lambda t} A_t dt \\ &= e^{-\lambda t} e^{\lambda t} dB_t - \lambda e^{-\lambda t} A_t dt \\ &= dB_t - \lambda e^{-\lambda t} A_t dt \end{align*} Since the drift term $- \lambda e^{-\lambda t} A_t$ is not $0$, $X_t$ is not a local martingale, and hence also not a martingale.

$\endgroup$
1
  • 1
    $\begingroup$ Sorry, that should be $e^{-\lambda t} e^{\lambda t} dB_t = dB_t$. It follows because $dA_t = e^{\lambda t} dB_t$ by definition. $\endgroup$ Apr 7 at 17:07

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .