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Let $B$ be a brownian motion and consider the stochastic integral $I_t:=\int_0^t (1-B_s-s)e^{-B_s-\frac{s}{2}} dB_s$, where $s\leq t$. I want to show that this is a martingale.

At the moment I know that this is a local martingale and I know that if $$\Bbb{E}(\langle I\rangle_t)=\Bbb{E}\left(\int_0^t (1-B_s-s)^2e^{-2B_s-s} ds\right)<\infty$$we have a true martingale. There is also another statement telling that if $(I_t)$ is a local martingale such that $\sup_{0\leq s\leq t} |I_s|\in L^1$ for all $t$ then $(I_t)$ is a martingale.

I tried it in both ways but I somehow cannot prove it. I also thought about that $s\mapsto (1-B_s-s)e^{-B_s-\frac{s}{2}}$ is a continuous function on a compact interval so it has a maximum, but I don't know if this works.

Can someone help me?

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1 Answer 1

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Use Fubini's Theorem to compute the expectation first.

$$\Bbb{E}\left(\int_0^t (1-B_s-s)^2e^{-2B_s-s} ds\right)=\int_{0}^{t}\Bbb{E}((1-B_s-s)^2e^{-2B_s-s})\,ds$$

Firstly, just bound $e^{-s}$ by $1$.

Simplify it a bit by noting that $B_{s}/\sqrt{s}$ has $N(0,1)$ distribution.

$$\int_{0}^{t}\Bbb{E}\bigg( s(1/\sqrt{s}-B_{s}/\sqrt{s}-\sqrt{s})^{2}e^{\sqrt{s}\cdot2B_{s}/\sqrt{s}}\bigg)\,ds$$

$$=\int_{0}^{t}\Bbb{E}(s(\frac{1}{\sqrt{s}}-X-\sqrt{s})^{2}e^{-\sqrt{s}X}\bigg)\,ds$$

where $X$ has $N(0,1)$ distribution.

Now expand it out, $$\int_{0}^{t}\Bbb{E}\bigg(\bigg(s\frac{1}{s}+sX^{2}+s^{2}-2\sqrt{s}X-2s+2s\sqrt{s}X\bigg)e^{-\sqrt{s}X}\bigg)\,ds$$

I'll leave it for you to check that $\Bbb{E}(p(X)e^{-tX})$ is finite for a standard normal variate $X$ and a polynomial $p(X)$ in $X$ . Infact the above expression can be calculated explicitly but why bother at all.

So you'll end up with terms of the form $\int_{0}^{t}p(\sqrt{s})e^{q(s)}\,ds$ w for some polynomials $p$ and $q$ which will be finite obviously as $[0,t]$ is a compact interval and $\sqrt{s}$ is a continuous function. (Actually, q will be atmost of order $2$) by basic properties of the moment generating function.

And thus the integrand has finite $L^{2}(\lambda|_{[0,t]}\otimes \Bbb{P})$ norm and thus the stochastic integral makes sense.

The maximum of a continuous function in a compact interval is finite logic "will" work but it will give you a potentially more complicated expresion. You see that the maximum is not uniformly bounded. Instead it has a distribution. So even if you use Fubini. You'll have to contend with $M_{s}=\sup_{r\in [0,s]} B_{r}$ has the same distribution as $|B_{s}|$ to work things out.

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  • $\begingroup$ Thanks a lot, but do you really mean $\int_{0}^{t}\Bbb{E}\bigg(s\frac{1}{s}+sX^{2}+s^{2}-2\sqrt{s}X-2s+2s\sqrt{s}X\bigg)e^{-\sqrt{s}X}\,ds$ and not $\int_{0}^{t}\Bbb{E}\bigg(\bigg(s\frac{1}{s}+sX^{2}+s^{2}-2\sqrt{s}X-2s+2s\sqrt{s}X\bigg)e^{-\sqrt{s}X}\bigg)\,ds$? $\endgroup$
    – user123234
    Commented Jan 8 at 10:23
  • $\begingroup$ Yeah sorry. I'll edit it along with a comment about how you "cannot" use the maximum of a continuous function logic. @user123234 $\endgroup$ Commented Jan 8 at 10:24
  • $\begingroup$ Thanks a lot. And for the part that $\Bbb{E}(p(X)e^{-tX})$ is finite can I say $\Bbb{E}(p(X)e^{-tX})=\int_\Bbb{R} p(x)e^{-tx} f_X(x)dx$ where $f_X$ is the density of $X$ but now how can I bound this from above? $\endgroup$
    – user123234
    Commented Jan 8 at 10:30
  • $\begingroup$ @user123234 I mean that's pretty basic. you have $e^{-x^{2}/2}$ . So you can just complete the square. It's a very well known fact that the Gaussian distribution will have that moments of the form $E(p(X)e^{ax})$ will exist easily. $\endgroup$ Commented Jan 8 at 10:35
  • $\begingroup$ I see perfect thanks! $\endgroup$
    – user123234
    Commented Jan 8 at 10:36

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