# Logarithm of Brownian motion which is a local martingale but not a martingale

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

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

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.