Say we have a probability space $(\Omega, \mathscr{F}, \mathbb{P}$), and $\mathbb{F} = (\mathscr{F}_t)_{t \geq 0}$ is a filtration of sub-$\sigma$-algebras of $\mathscr{F}$ satisfying the standard conditions. I've been told to consider a one-dimensional $\mathbb{F}$-Wiener process $(W_t)_{t \geq 0}$ and also the process $(X_t)_{t \geq 0}$ given by: $$X_t = f(W_t) - \frac{1}{2} \int_{0}^{t} f''(W_s) ds$$ for every $t > 0$, where $f: \mathbb{R} \mapsto \mathbb{R}$ is a $C^2$ function. Furthermore, I've been told to assume that there exists a constant $C > 0$ such that $|f''(x)| \leq C$ for all $x \in \mathbb{R}$. Show that $(X_t)_{t \geq 0}$ is a martingale (with respect to $\mathbb{F}$).

I quote here my old question (Tricky Proof in Stochastic Processes/ Probability Theory). This question was answered by @user6247850 who uses Ito's formula to show that it is a local martingale. His answer below I will paste here:

Applying Ito's formula to $f$, we have \begin{align*} f(W_t) &= f(W_0) + \int_0^t f'(W_s)dW_s + \frac 12 \int_0^t f''(W_s)ds \end{align*} so \begin{align*} X_t = f(W_t)-\frac 12 \int_0^t f''(W_s)ds &= f(W_0) + \int_0^t f'(W_s)dW_s \end{align*} is a (local) martingale. To prove it is a true martingale, use the fact that $|f''(x)| \le C$ to show that \begin{align*} \mathbb{E}\left[\int_0^t |f'(W_s)|^2 ds \right] < \infty \end{align*} for all $t \ge 0$.

How can I go on to prove that it is a True martingale? I am new to this topic and these kind of questions so any demonstration of a solution would really help me. Many thanks.


1 Answer 1


By mean value theorem


for some $0\leq c \leq x$ when $x>0$and hence

$$-Cx+f'(0)\leq f'(x)\leq f'(0)+C x$$

$f'(0)$ exists because $f$ is $C^2$. When $x<0$

$$Cx+f'(0)\leq f'(x)\leq f'(0)-C x$$

Then $|f'(x)|^2 \leq \max((f'(0)+Cx)^2,(f'(0)-Cx)^2)$. Hence $|f'(x)|^2$ is bounded by a max of two integrable polynomials and is hence integrable. A result presented in any stochastic calculus textbook is that a stochastic integral has mean zero if the integrand function is square integrable. Hence


and $X_t$ is a true martingale.


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