0
$\begingroup$

Let $W_1(t)$ and $W_2(t)$ be independent Brownian motions starting at positive points (not necessarily at the same point). Let $X_t=\log(W_1^2+W_2^2)$ and show that it is a local martingale but not a martingale.

I don't know how to construct a localizing sequence for the process.

Also the expectation is finite, so the martingale property that fails in this case is the conditional expectation, but again I don't seem to be able to compute the conditional expectation. I tried to write the conditional expectation $E_s$ for $s<t$ as $$ E_s\left( \log \frac{W_1^2(t)+W_2^2(t)}{W_1^2(s)+W_2^2(s)} \right) $$ and somehow tried to use concavity of log to show that conditional expectation is not zero, but I didn't succeed. Thank you

$\endgroup$
  • $\begingroup$ For any continuous local martingale $X_t$, the stopping times $\tau_n = \inf\{t \ge 0 : |X_t| \ge n\}$ are a localizing sequence. $\endgroup$ – Nate Eldredge Nov 10 '15 at 22:41
2
$\begingroup$

The process $X$ in this case is a squared Bessel process of dimension two, which means that it is a (weak) solution to the SDE $$ \text{d}X_t=2\,\text{d}t+2\sqrt{X_t}\,\text{d}B_t, $$ for all $t\geq 0$, where $B$ is some Brownian motion. An application of Ito's formula now yields $$ \text{d}\ln X_t=\frac{2}{\sqrt{X_t}}\,\text{d}B_t, $$ for all $t\geq 0$. In particular, $\ln X$ is a local martingale. Finally, the transition density of a squared Bessel process of dimension two is given by $$ q(t,x,y):=\frac{1}{2t}\text{e}^{-\frac{x+y}{2t}}I_0\Bigl(\frac{\sqrt{xy}}{t}\Bigr), $$ for all $t>0$ and all $x,y\in[0,\infty)$, where $I_0(\cdot)$ denotes the modified Bessel function of the first kind, of degree zero (see e.g. Revuz and Yor, 1999). Hence, $$ \textsf{E}_x(\ln X_t)=\int_0^\infty\ln y\,q(t,x,y)\,\text{d}y =\ln x-\text{Ei}\Bigl(-\frac{x}{2t}\Bigr), $$ for all $t>0$ and all $x>0$, where $\text{Ei}(\cdot)$ denotes the exponential integral function. In particular, $\textsf{E}_x(\ln X_t)\neq\ln x$ implies that $\ln X$ is not a martingale.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.