# Is the stochastic integral process $\left(\int_{0}^{t}e^{-\lambda\left(t-s\right)}\mathrm{d}B_s\right)_{t\geq 0}$ a martingale?

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?

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.

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

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.

• 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. Apr 7 at 17:07