# Martingale processes

I've been finding some difficulties in solving this exercise;

Let $$\{B_t\}_{t\geq 0}$$ a standard Brownian motion with respect to the natural filtration $$\{\mathcal{F}_t\}_{t\geq 0}$$ and define the process $$Y_t=\ln(1+\alpha B_t^2)$$. Find the real values $$\alpha$$ for which $$Y_t$$ is martingale with respect to the filtration $$\{\mathcal{F}_t\}_{t\geq 0}$$

MY ATTEMPT

By Jensen inequality: $$$$E(\ln(1+\alpha B_t^2)\lvert\mathcal{F}_s)\leq\ln(E(1+\alpha B_t^2\lvert\mathcal{F}_s))=\ln(1+\alpha E(B_t^2\lvert\mathcal{F}_s))$$$$ Now rewriting $$B_t^2=(B_t-B_s)^2+2B_s(B_t-B_s)+B_s^2$$ and using the properties of the brownian motion we find: $$$$\ln(1+\alpha E(B_t^2\lvert\mathcal{F}_s))=\ln(1+\alpha E((B_t-B_s)^2+2B_s(B_t-B_s)+B_s^2\lvert\mathcal{F}_s))\\ =\ln(1+\alpha(t-s)+\alpha B_s^2)$$$$ since the logarithm is increasing I find immediately that if $$\alpha<0$$ we have that putting all together: $$$$E(Y_t\lvert\mathcal{F}_s)=E(\ln(1+\alpha B_t^2)\lvert\mathcal{F}_s)\leq\ln(1+\alpha(t-s)+\alpha B_s^2)<\ln(1+\alpha B_s^2)=Y_s$$$$ And this means that the process is not a martingale (it is a supermartingale).

If $$\alpha=0$$ the process is trivially a martingale because $$Y_t=0$$ for all $$t\geq 0$$.

It remains to prove what happens if $$\alpha >0$$ but in this case I don't have any ideas.

Using the hint

We want to apply ito formula to the process $$Y_t=\ln(1+\alpha B_t^2)$$, in particular we have that: $$$$dY_t=\dfrac{2\alpha B_t}{1+\alpha B_t^2}dB_t + \dfrac{1}{2}\dfrac{2\alpha(1+\alpha B_t^2)-4\alpha^2 B_t^2}{(1+\alpha B_t^2)^2}dt=\dfrac{2\alpha B_t}{1+\alpha B_t^2}dB_t + \dfrac{\alpha-2\alpha^2 B_t^2}{(1+\alpha B_t^2)^2}dt$$$$ Now in order to have a martingale it is necessary that $$\frac{\alpha-2\alpha^2 B_t^2}{(1+\alpha B_t^2)^2}=0$$ and hence: $$$$\alpha(1-2\alpha B_t^2)=0 \iff \alpha=0 \ \vee \ \alpha=\dfrac{1}{2B_t^2}$$$$ It seems that the last value of $$\alpha$$ does not make any sense, any hint on how to conclude?

• hint: use Ito's formula on $Y_t$ and find the $\alpha$ that nullifies the $dt$-term . Commented Jul 31, 2022 at 19:20
• @KurtG. I have updated some computation but I'm still not completely sure on it. Was this your idea? Commented Jul 31, 2022 at 19:37
• It makes sense ! If you want to allow only deterministic $\alpha$s you can throw that stochastic one ino the bin. Commented Jul 31, 2022 at 19:44
• Ok so with this fact the solution is just $\alpha=0$ isn't it? Commented Jul 31, 2022 at 19:58

If $$\alpha<0$$, then $$Y_t$$ is undefined with positive probability for each $$t>0$$.
If $$\alpha>0$$, then $$Y_t>0$$ a.s. for each $$t>0$$, so $$E(Y_t)>0=E(Y_0)$$, whence $$Y_t$$ is not a Martingale.
In fact, it is easy to verify that $$E(Y_t) \to \infty$$ as $$t \to \infty$$ when $$\alpha>0$$, since $$P(|B_t|. More precisely, this observation yields the lower bound for the asymptotic relation $$E(Y_t)=(1+o(1)) \log t$$ as $$t \to \infty$$. The upper bound can be inferred from Jensen's inequality.
Thus $$Y_t$$ is a Martingale only for $$\alpha=0$$.