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How to prove that the integral $\int_{0}^{+\infty}\upsilon e^{-ru}S_{u}dW_{u}^{Q}$ is a martingale under Q where $S_{t}$ is a martingale under Q and $\mathbb{E}^{Q}[\int_{0}^{+\infty}|\upsilon e^{-ru}S_{u}|^{2}dt]<\infty$? After that, prove the expected value of the integral equals 0?

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What is the parameter? What are $v$ and $r$? – Siméon Jan 2 '13 at 12:58
v and r are all constants. – Tony Jan 2 '13 at 13:02
$\int_0^\infty v \cdot e^{-r \cdot u} \cdot S_u \, dW_u^Q$ is a random variable (so a trivial martingale). Probably you mean $\int_0^t v \cdot e^{-r \cdot u} \cdot S_u \, dW_u^Q$? – saz Jan 2 '13 at 13:16
Apparently, related question – Ilya Jan 2 '13 at 13:19
A general hint: You should add all* information you have, not only some of them! – saz Jan 2 '13 at 15:43
up vote 2 down vote accepted

My suggestion: Take a look at a book on stochastic processes/stochastic analysis. If you knew some basics of stochastic analysis this exercise would be (more or less) trivial. Since it's obviously not that easy for you, you should read more about this topic, because it's simply too much to explain it in detail.

Here are some basic definitions/theorems you should know before thinking about this exercise: Let $(W_t)_{t \geq 0}$ a Brownian motion and $(\mathcal{F}_t)_{t \geq 0}$ a admissible filtration. Let $T>0$.

Definition 1 The $\sigma$-algebra $\mathcal{P}$ defined as $$\mathcal{P} := \{\Gamma \subseteq [0,T] \times \Omega; \forall t \leq t: \Gamma \cap ([0,t] \times \Omega) \in \mathcal{B}[0,t] \times \mathcal{F}_t\}$$ is called progressive $\sigma$-algebra.

Definition 2 Let $f:[0,T] \times \Omega \to \mathbb{R}$ such that

  1. $[0,T] \times \Omega \ni (t,\omega) \mapsto f(t,\omega)$ is $\mathcal{P}$-measurable (so-called progressively measurable).
  2. $\mathbb{E} \left( \int_0^T |f(t,\cdot)|^2 \, dt \right) < \infty$

We denote the set of these functions by $L_T^2$.

Theorem Let $f \in L^2_T$ and define $$X_t := \int_0^t f(u) \, dW_u$$ Then $(X_t)_{t \leq T}$ is a (continuous) martingale. In particular $$\mathbb{E}X_t = \mathbb{E}\underbrace{X_0}_{0} = 0$$

Define $f(t,\omega) := v \cdot e^{-r \cdot t} \cdot S_t(\omega)$. You have to prove $f \in L_T^2$ (then you are done; just apply the above theorem). Some hints how to prove the conditions in definition 2:

  1. One can show that an adapted continuous process is progressively measurable. You can apply this statement since $S$ is a continuous martingale.
  2. Use the assumption $\mathbb{E} \left(\int_0^{\infty} |v \cdot e^{-r \cdot u} \cdot S_u|^2 \, du \right)< \infty$.

These definitions and the theorem are contained in

Brownian motion - An Introduction to Stochastic Processes - René L. Schilling/Lothar Partzsch; Chapter 14.

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Could I use the Fokker-Planck equation to prove that $\mathbb{E}^{\mathbb{Q}}[X_{t}]=0$ directly? $X_{t}$ is defined in theorem above. – Tony Jan 3 '13 at 3:14
@Tony Perhaps, I'm not familiar with this equation. But even if you could prove it using Fokker-Planck-equation I would say it's overkill. If you know that $(X_t)_t$ is a martingale, it's a monostitch to prove $\mathbb{E}X_t=0$ as you can see above (using $X_0=0$). – saz Jan 3 '13 at 7:53
What if $S_{t}$ is not defined as a martingale? – Tony Jan 3 '13 at 10:18
@Tony It depends. It suffices (for the argumentation above) that $S_t$ is $\mathcal{F}_t$-measurable for all $0 \leq t \leq T$. Otherwise it's getting more complicated because you have to think about how to define the stochastic integral in this case. – saz Jan 3 '13 at 14:28

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