1
$\begingroup$

Let

  • $(\Omega,\mathcal{A},\operatorname{P})$ be a probability space
  • $\mathbb{F}=(\mathcal{F}_t,t\ge 0)$ be a filtration on $(\Omega,\mathcal{A})$
  • $H=(H_t,t\ge 0)$ be a stochastic process on $(\Omega,\mathcal{A})$ with $$H_t(\omega)=\sum_{i=1}^nh_{i-1}(\omega)1_{(t_{i-1},t_i]}(t)$$ where $0=t_0<\ldots<t_n$ and $h_{i-1}$ is bounded and $\mathcal{F}_{t_{i-1}}$-measurable
  • $\tau$ be a bounded $\mathbb{F}$-stopping time
  • $(W_t,t\ge 0)$ be a Brownian motion on $(\Omega,\mathcal{A},\operatorname{P})$ and $$I^W_t(H):=\sum_{i=1}^nh_{i-1}\left(W_{t_i\wedge t}-W_{t_{i-1}\wedge t}\right)$$ where $x\wedge y:=\min(x,y)$ for all $x,y\in\mathbb{R}$

By definition, we've got $$\operatorname{E}\left[I_\tau^W(H)\right]=\sum_{i=1}^n\operatorname{E}\left[h_{i-1}\left(W_{t_i}^\tau-W_{t_{i-1}}^\tau\right)\right]\tag{1}$$ where $W_t^\tau:=W_{\tau\wedge t}$ is the so called stopped process. By definition of the conditional expectation, $$\operatorname{E}\left[W_{t_i}^\tau-W_{t_{i-1}}^\tau\right]=\operatorname{E}\left[\operatorname{E}\left[W_{t_i}^\tau-W_{t_{i-1}}^\tau\mid\mathcal{F}_{t_{i-1}}\right]\right]$$


However, I don't understand why we've got $$\sum_{i=1}^n\operatorname{E}\left[h_{i-1}\left(W_{t_i}^\tau-W_{t_{i-1}}^\tau\right)\right]=\sum_{i=1}^n\operatorname{E}\left[h_{i-1}\operatorname{E}\left[W_{t_i}^\tau-W_{t_{i-1}}^\tau\mid\mathcal{F}_{t_{i-1}}\right]\right]$$ That would be true iff $W_{t_i}^\tau-W_{t_{i-1}}^\tau$ and $h_{i-1}$ are uncorrelated; but why is that the case?

$\endgroup$
2
$\begingroup$

the answer to this is quite simple if you look carefully to the definition of H, you should notice that $h_{i-1}$ is an $\mathcal{F}_{t_{i-1}}$-measurable random variable. From this and the following elementary properties on conditional expectation we get the result :

If X is $\mathcal{F}$-measurable and for any bounded random variable $Y$, we have a.s. :

$$1-E[X.Y|\mathcal{F}]=X.E[Y|\mathcal{F}]$$

Property 1 allows you to get :

$$E\left[h_{i-1}.(W_{t_i}^\tau-W_{t_{i-1}}^\tau)\mid\mathcal{F}_{t_{i-1}}\right]= h_{i-1}E\left[(W_{t_i}^\tau-W_{t_{i-1}}^\tau)\mid\mathcal{F}_{t_{i-1}}\right]$$

Now let's remind the defining property of conditional expectation of a integrable random variable with respect to a $\sigma$-algebra $\mathcal{F}$ which is the only (modulo a.s. equivalence) $\mathcal{F}$-measurable random variable such that we have for any bounded $\mathcal{F}$-measurable random variable $Y$ the following property :
$$2-E[Y.X]=E[Y.E[X|\mathcal{F}]]$$

In our situation we are done because $h_{i-1}$ is bounded $\mathcal{F}_{t_{i-1}}$-meausarble and $(W_{t_i}^\tau-W_{t_{i-1}}^\tau)$ is an integrable random variable so we get by property 2 :

$$E[h_{i-1}E\left[(W_{t_i}^\tau-W_{t_{i-1}}^\tau)\mid\mathcal{F}_{t_{i-1}}\right]]=E[h_{i-1}.(W_{t_i}^\tau-W_{t_{i-1}}^\tau)]$$

Best regards

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.