# Show that $f(B_t)$ is a martingale iff $f(x)=a+bx$.

One question from Durrett's 5th Probability textbook is the following. Suppose that $$f\in C^2$$. Show that $$f(B_t)$$ is a martingale iff $$f(x)=a+bx$$.

For "$$\Leftarrow$$" direction, if $$f(x)=a+bx$$, then for s<t $$E[f(B_t)|\mathcal{F}_s]=E[a+bB_t|\mathcal{F}_s]=a+bB_s$$

For "$$\Rightarrow$$" direction, I want to use the Martingale representation theorem that if $$M_t$$ is a martingale and $$M_t\in L^2$$, then there exists a unique stochastic process $$g(s,\omega)$$ s.t. $$M_t=EM_0+\int_0^tg(s,\omega)dB(s)$$ is a martingale w.r.t. $$\mathcal{F}_t$$.

It suffices to take the derivative of $$f$$ that $$df(B_t)=f'(B_t)dB_t+\frac{1}{2}f''(B_t)dt$$ Then $$f(B_t)=f(0)+\int_0^tf'(B_s)dB_s+\frac{1}{2}\int_0^tf''(B_s)ds$$

Can we get $$f''(B_s)=0$$ from the martingale representation theorem? That implies $$f=a+bx$$?

• I don't see the need for martingale representation. Your assumption that $f(B_{t})$ is a martingale, together with the expression obtained from Ito in the last equation entail already that necessarily $f''=0$, which in 1d means $f$ is linear. Feb 13, 2022 at 9:17
• Why do we have $f''=0$ from the Ito equation? Feb 13, 2022 at 20:19
• 'cause $f(B_{t})$ is supposed to be a martingale, which I wrote already above. Feb 13, 2022 at 20:33
• @Tobsn Sorry, I am still confused. Hot to get $f''=0$ from $f(B_t)$ is a martingale? I am confused so I used the Martingale representation theorem, which implies there could not be a term like $\int_0^t f''(B_s)ds$. Feb 13, 2022 at 20:41

Your solution for the backwards direction is correct. For the forward direction, it's best to use Itô's formula instead of the martingale representation theorem.

To show the forward direction, assume $$X_t = f(B_t)$$ is a martingale and apply Itô's lemma to get: $$dX_t = f_x(B_t) dB_t + \frac{1}{2}f_{xx}(B_t)dt$$ Since $$X_t$$ is a martingale, it must be driftless (i.e. the $$dt$$ term must be zero), which implies that $$f_{xx}=0$$. Integrating twice yields that $$f(x) = a +bx$$, as desired.

• There seems to be a $1/2$ missing 🙂. Also, if it wasn’t proven in the book already, showing that „being martingale implies driftless“ seems to be the hardest part of the exercise. Feb 14, 2022 at 11:05
• @MaximilianJanisch woops, thanks for spotting the typo! And you're right: driftlessness requires a short proof, so thanks for adding that. Feb 14, 2022 at 12:42

We have by Itô's formula [1; (25.10)] $$X_t=X_0+\int_0^t f'(B_s)\,\mathrm dB_s+\frac 12\int_0^t f''(B_s)\,\mathrm ds.$$ By [1; Satz 25.18], since $$\mathsf E\left(\int_0^T f'(B_s)^2\,\mathrm ds\right)<\infty$$ for all $$T\in[0,\infty[$$ (exercise: prove this!), we have that $$t\mapsto\int_0^t f'(B_s)\,\mathrm dB_s$$ is a continuous martingale on $$[0,\infty[$$. Therefore, $$t\mapsto\int_0^t f''(B_s)\,\mathrm ds$$ must also be a continuous martingale on $$[0,\infty[$$ (exercise: prove this!).

Lemma. The stochastic process $$t\mapsto \int_0^t f''(B_s)\,\mathrm ds$$ has almost surely finite local variation.

Since a continuous martingale with almost surely finite local variation is almost surely constant ([1; Korollar 21.72]), we may thus conclude with the help of the Lemma that $$t\mapsto\int_0^t f''(B_s)\,\mathrm ds$$ is constant almost surely. This implies (exercise: state why) that $$f''(B_s)=0$$ for all $$s\in[0,\infty[$$ almost surely.

Now conclude that $$f''=0$$ using continuity of $$f''$$ and the fact that the support of $$B_s$$ is $$\mathbb R$$ for all $$s>0$$ and thus that $$f$$ must be an affine function.

Proof of Lemma. Let $$T>0$$ and $$n\in\mathbb N$$. For any partition $$0=t_0, we have $$\begin{equation*}\sum_{k=0}^{n-1} \left\lvert\int_0^{t_{k+1}} f''(B_s)\,\mathrm ds-\int_0^{t_k} f''(B_s)\,\mathrm ds\right\rvert\le\int_0^T\lvert f''(B_s)\rvert\,\mathrm ds\le T\sup_{s\in[0,T]} \lvert f''(B_s)\rvert,\end{equation*}$$ which is almost surely finite. (Furthermore, if $$t\mapsto B_t(\omega)$$ is continuous for all $$\omega\in\Omega$$, then the local variation is finite for all $$\omega\in\Omega$$.) $$\square$$

# Literature

[1] Achim Klenke, Wahrscheinlichkeitstheorie. 3. Auflage. Springer-Verlag Berlin/Heidelberg (2013).

Suppose $$f(B_t)$$ is a martingale. Let $$c<0 be real numbers. It's well known that if $$T:=\inf\{t:B_t\notin(c,d)\}$$ then $$P[B_T=c]={d\over d-c}.$$ Because $$f(B)$$ is a martingale, $$f(0)=E[f(B_0)]=E[f(B_T)]=f(c){d\over d-c}+f(d){-c\over d-c}.\qquad (1)$$ In particular (take $$c=-1$$) $$f(d) = f(0)+[f(0)-f(-1)]d,\qquad (2)$$ for $$d>0$$. Likewise $$f(c)=f(0)+[f(1)-f(0)]c,\qquad (3)$$ for $$c<0$$. Finally (take $$c=-1, d=1$$ in (1)): $$f(0)={f(-1)+f(1)\over 2},$$ implying that $$f(1)-f(0)=f(0)-f(-1)$$. Now (2) and (3) together imply that $$f(x)=f(0)+mx$$ for all $$x$$, where $$m:=[f(-1)+f(1)]/2$$.

This argument doesn't need $$f\in C^2$$ as a hypothesis. Indeed it works knowing just that $$f$$ is Lebesgue measurable and locally bounded.