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Assume $f\left(t,x\right)$ is $C^1\left(\left[0,T\right]\times\mathbb{R}, \mathbb{R}\right)$ and that the process $f\left(t, B_t\right)$ is a martingale w.r.t $\mathcal{F_t}=\sigma\left(B_s, s\leq\,t\right)$, with $B_t$ a standard Brownian motion.

Can we apply the Itô formula to get an expression for $d f\left(t,B_t\right)$ while we have only $C^1$ differentiability?

If yes, can we relax the conditions even further by assuming $f$ only $C^1$ differentiable w.r.t to $x$ since the derivatives $\frac{\partial}{\partial t}$ and $\frac{\partial^2}{\partial x^2}$ should in principle be unnecessary since $f\left(t,B_t\right)$ is a martingale?

BS.
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  • Do you have an example with a non-smooth $f$ where you know $f\left(t, B_t\right)$ is a martingale? –  Feb 05 '18 at 12:58
  • For example $f\left(t, x\right):=\mathbb{E}\left[u\left(B_T\right)\left|B_t=x\right.\right]$ for some Borel function $u$ such that $\mathbb{E}\left[\left|u\left(B_T\right)\right|\right]<\infty$. – BS. Feb 05 '18 at 13:04
  • That $f$ will be a very smooth function. –  Feb 05 '18 at 13:57
  • To the point of being $C^2\left(\left[0,T\right]\times\mathbb{R}, \mathbb{R}\right)$? But how if we don't impose any differentiability condition on $u$? – BS. Feb 05 '18 at 14:32
  • Think of a convolution with a heat kernel, as soon as $t<T$ (not for $t=T$, obviously). It's not exactly the same, your function won't satisfy the heat eqation, but it's a very similar effect. –  Feb 05 '18 at 17:08
  • @BS. No need to assume $u \in C^2$ to get differentiability of $f$. For instance if $u$ is bounded and Borel measurable, then $x \mapsto \mathbb{E}^x u(X_t)$ is smooth for any $t>0$. – saz Feb 05 '18 at 18:49

2 Answers2

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Of course it ought to be the case that $f(t,B_t) =\int_0^t f'_2(s,B_s)\,dB_s$, where $f'_2$ is the partial derivative of $f$ with respect to its second variable. And this is indeed true. First note that by localization one can reduce to the case in which $f$ and both of its first-order partials are bounded. Things being so, the continuity of $f$ and those partials allow one to use the argument in section IV.41 of volume 2 of Diffusions, Markov Processes, and Martingales by Rogers and Williams demonstrating Clark's Formula for the integrand in the stochastic integral representation of a Brownian functional. That argument shows that $$ f(T,B_T) =f(0,B_0)+\int_0^Tf'_2(s,B_s)\,dB_s $$ almost surely. Now take conditional expectations with respect to $\mathcal F_t$ in this identity to obtain $$ f(t,B_t) =\Bbb E[f(T,B_T)|\mathcal F_t] = f(0,B_0)+\int_0^t f'_2(s,B_s)\,dB_s, $$ a.s. for each fixed $t\in[0,T]$. As the extreme terms in this last display are a.s. continuous functions of $t$, it follows that $$ f(t,B_t) = f(0,B_0)+\int_0^t f'_2(s,B_s)\,dB_s,\qquad\forall t\in[0,T], $$ almost surely.

John Dawkins
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Since $f(t,B_t)$ is a martingale, it follows from well-known martingale representation theorems (see e.g. [1]) that we can write

$$f(t,B_t) = f(0,B_0) + \int_0^t X_s \, dB_s$$

for some progressively measurable process $X$. A very useful tool to identify $X$ is the Malliavin derivative. One of the key properties of the Malliavin derivative $D$ is that

$$D \left( \int_0^t X_s \, dB_s \right) = X_t \tag{1}$$

On the other hand, the Malliavin derivative also satisfies a chain rule and we get

$$Df(t,B_t) = f'(t,B_t).$$

Combining both equations we get $X_t = f'(t,B_t)$ almost surely. If you are interested in these kind of things, then take a look at [2,3]; of particular interest (for your question) is the Clark Ocone representation formula.

[1] R.S. Schilling, L. Partzsch: Brownian motion - An Introduction to Stochastic Processes. (De Gruyter)

[2] D. Nualart: The Malliavin Calculus and Related Topics (Springer)

[3] I. Nourdin, G. Peccati: From Stein's Method to Universality. (Cambridge)

Remark: Under slightly stronger assumptions we don't need all the fancy tools. Assume that we know that $f(s+t,x+B_t)_{t \geq 0}$ is a martingale for any fixed $s \geq 0$ and $x \in \mathbb{R}$, and assume that $f$ grows at most sub-exponentially. Then we have by the martingale property

$$f(s,x) = \mathbb{E}(f(s+t,x+B_t)).$$

If we denote by $p_t$ the density of $B_t$, then

$$f(s,x) = \int_{\mathbb{R}} f(s+t,x+y) p_t(y) \, dy = \int_{\mathbb{R}} f(s+t,z) p_t(z-x) \, dz.$$

Since the density $p_t$ is smooth, we can deduce easily that $x \mapsto f(s,x)$ is smooth and therefore we can apply Itô's formula to deduce that

$$f(t,B_t) = f(0,B_0) + \int_0^t f'(s,B_s) \, dB_s.$$

saz
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  • What implies that $X$ is square integrable? – John Dawkins Feb 06 '18 at 16:40
  • @JohnDawkins I was thinking about the particular case that $f$ is bounded; in this case $X$ is trivially integrable because $(f(t,B_t))_t$ is an $L^2$-bounded martingale. I didn't realize (until now) that the OP doesn't assume boundness of $f$. – saz Feb 06 '18 at 17:33