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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: \begin{equation} 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)) \end{equation} 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: \begin{equation} \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) \end{equation} since the logarithm is increasing I find immediately that if $\alpha<0$ we have that putting all together: \begin{equation} 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 \end{equation} 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: \begin{equation} 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 \end{equation} 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: \begin{equation} \alpha(1-2\alpha B_t^2)=0 \iff \alpha=0 \ \vee \ \alpha=\dfrac{1}{2B_t^2} \end{equation} It seems that the last value of $\alpha$ does not make any sense, any hint on how to conclude?

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

1 Answer 1

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Ito's formula is a wonderful tool, but it is overkill for this question.

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 $P(|B_t|<t^{1/2}/(\ln t)) \to 0$. 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 $\alpha=0$.

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