1

Let $W_t$ be the one-dimensional Wiener process.

Does the following integral $$\int_0^t f(W_s) dW_s,$$ where $f:\mathbb{R} \to \mathbb{R}$ continuous with second derivative, have zero mean for any fixed $t>0$?

The integral can be considered as a limit in probability of the Riemann sums $\sum_i f(W_{t_i}) (W_{t_{i+1}} - W_{t_i})$, whose means are zero. But dominated convergence theorem cannot be applied here.

newbie
  • 3,441

2 Answers2

1

Short answer: The stochastic integral, in Kunita-Watanabe sense, defines a martingale. Read P.195 of the article http://www-math.mit.edu/~dws/ito/ito7.pdf.

Long answer: We need to ask how is the stochastic integral defined. To answer such question, we need a lot of preparation (e.g. a chapter in a textbook). You may consult the famous book "Probabilities and Potential" by Claude Dellacherie and Paul-Andre Meyer.

Although it is true that the Riemann sum converges to the stochastic integral in probability, this fact is very hard to prove and there is a lot of technical difficulty. For example, imagine that we have not defined the stochastic integral yet, then: How can we tell the Riemann sums converge to some random variable in probability? In what sense? (It is not a sequence of random variables, but involving choices of $t_i$)

Danny Pak-Keung Chan
  • 12,250
  • Since $f\in \mathcal{L}_{loc}^2$, it is only a local martingale in general. – newbie Aug 28 '17 at 07:31
  • The trick is to apply a constant stopping time. Let $T$>0 be an arbitrary constant stopping time. Note that $f$ restricted on the compact set $[0,T]$ is bounded. Then $E [\int_0^T f^2(W_t) dt]<\infty$. Hence the integral defines a martingale for $t\in[0,T]$. – Danny Pak-Keung Chan Aug 28 '17 at 12:02
  • I am familiar with the analytical concepts, but not used to the probabilistic ideas. What's the motivation behind, to deal with the whole time horizon as $T = \infty$? – newbie Aug 28 '17 at 21:54
  • Even with $T<\infty$, $E[\int_0^T e^{W_t^4} dt]$ is not finite, but $f(x) = e^{x^4}$ is continuous. – newbie Aug 29 '17 at 21:46
  • I mixed-up something. We cannot restrict $f$ on $[0,T]$. Rather, we are considering $f\circ X$. Let me think about it once I have time. – Danny Pak-Keung Chan Aug 30 '17 at 11:55
  • You are right. For each $\omega$, $f(W_t(\omega))$ is bounded for $t\in[0,T]$, due to the continuity of sample paths. However, this bound depends on $\omega$, causing trouble in taking expectation. More precisely, for each $\omega$, there exists $M_\omega>0$ such that for each $t\in[0,T]$, $|f(W_t(\omega)|\leq M_\omega$. Now, I doubt that the random variable $\int_0^t f(W_s) dW_s$ is not even integrable in general. – Danny Pak-Keung Chan Aug 30 '17 at 12:10
1

This is not really rigorous (the other answer covers that) and might be trivial, but I wanted to add this answer from a more intuitive perspective. For me, $\int f(W_s)\,dW_s$ always makes me do a double take, since $dW_s$ is such an odd creature, i.e. not an independent variable in the sense one is used to seeing.

Consider the Brownian increment $\Delta W_{t_{i+1}} = W_{t_{i+1}} - W_{t_i}$. By the definition of a Wiener process, if $\Delta t = t_{i+1}-t_i$, then (1) $\Delta W_{t_{i+1}} \sim \mathcal{N}(0,\Delta t) $ and (2) $W_{t_i}$ is independent from $\Delta W_{t_{i+1}}$.

Thus, expanded as a sort of Ito-Riemann sum: \begin{align} \mathbb{E}\left[ \int_0^t f(W_s)\;dW_s \right] &\approx \mathbb{E}\left[ \sum_i f(W_{t_i})[W_{t_{i+1}} - W_{t_i}] \right] \\ &= \sum_i \mathbb{E}\left[ f(W_{t_i})\Delta W_{t_{i+1}} \right] \\ &= \sum_i \mathbb{E}\left[ f(W_{t_i})\right] \mathbb{E}\left[\Delta W_{t_{i+1}} \right]\;\;\;\;\;\;\;\;\;\;\;\text{ By (2)} \\ &= \sum_i \mathbb{E}\left[ f(W_{t_i})\right] 0 \;\;\;\;\;\;\;\;\;\;\;\;\,\;\;\;\;\;\;\;\;\;\;\;\text{ By (1)} \\ &=0 \end{align}

user3658307
  • 10,433