1

How do you derive $\mathbb{E}\left[\left(\int_0^{t+\tau} e^{-\frac{t+\tau-s}{T_{\tilde{\chi}}}} \mathrm{d} \mathcal{W}_s\right) \left( \int_0^t e^{-\frac{t-s}{T_{\tilde{\chi}}}} \mathrm{d} \mathcal{W}_s\right)\right]$ ?

By decomposing from 0 to t and from t to t+$\tau$ and using Itô isometry, one has

$\mathbb{E}\left[\left(\int_0^{t+\tau} e^{-\frac{t+\tau-s}{T_{\tilde{\chi}}}} \mathrm{d} \mathcal{W}_s\right) \left( \int_0^t e^{-\frac{t-s}{T_{\tilde{\chi}}}} \mathrm{d} \mathcal{W}_s\right)\right] = \int_0^{t} e^{-\frac{2t+\tau-2s}{T_{\tilde{\chi}}}} \mathrm{d}s + \mathbb{E}\left[\left(\int_t^{t+\tau} e^{-\frac{t+\tau-s}{T_{\tilde{\chi}}}} \mathrm{d} \mathcal{W}_s\right) \left( \int_0^t e^{-\frac{t-s}{T_{\tilde{\chi}}}} \mathrm{d} \mathcal{W}_s\right)\right]$

but I can't see what to do with the second term...

Thanks for the help !

Romain
  • 153

1 Answers1

0

Following your result. Without loss of generality, take $T_{\tilde{\chi}}=1$.

Provided that $$ e^s{\rm d}W_s $$ is an Ito integral, for which $$ \int_0^te^s{\rm d}W_s $$ is a martingale. Thanks to this fact, $$ \int_0^te^s{\rm d}W_s\quad\text{and}\quad\int_t^{t+\tau}e^s{\rm d}W_s $$ are independent from each other. Therefore, $$ \mathbb{E}\left(C_1\int_t^{t+\tau}e^s{\rm d}W_s\right)\left(C_2\int_0^te^s{\rm d}W_s\right)=\mathbb{E}\left(C_1\int_t^{t+\tau}e^s{\rm d}W_s\right)\mathbb{E}\left(C_2\int_0^te^s{\rm d}W_s\right)=C_1C_2\mathbb{E}\left(\int_t^{t+\tau}e^s{\rm d}W_s\right)\mathbb{E}\left(\int_0^te^s{\rm d}W_s\right). $$ Finally, note that $$ \int_0^te^s{\rm d}W_s\quad\text{and}\quad\int_t^{t+\tau}e^s{\rm d}W_s $$ are both a sum of independent zero-mean normal random variables, for which their expectations are both zero, i.e., $$ \mathbb{E}\left(\int_t^{t+\tau}e^s{\rm d}W_s\right)=\mathbb{E}\left(\int_0^te^s{\rm d}W_s\right)=0. $$ Thus we conclude that $$ \mathbb{E}\left(C_1\int_t^{t+\tau}e^s{\rm d}W_s\right)\left(C_2\int_0^te^s{\rm d}W_s\right)=0, $$ meaning that your second term vanishes, and thus could be neglected.

hypernova
  • 6,072
  • Thank you for your detailed answer ! I understand globally the reasoning but I'm still having trouble understanding why the two integrals are independent regardless $t$ and $\tau$. Should not this two be correlated for $\tau$ not to large ? – Romain May 09 '18 at 08:37
  • 1
    @Romain: No problem :-) Well, since $\int_0^te^s{\rm d}W_s$ and $W_t=\int_0^t{\rm d}W_s$ are both martingales, let us take $W_t$ for instance. You will find that $W_{t+\tau}-W_t=\int_t^{t+\tau}{\rm d}W_s$ is independent from $W_t=\int_0^t{\rm d}W_s$, because a martingale yields an independent increment. Similar property also applies to $\int_0^te^s{\rm d}W_s$, because it is also a martingale. – hypernova May 09 '18 at 11:47
  • I guess the point I'm missing then is why $\int_0^t e^s dW$ is also a martingal because I though that the point of the OU process (and hence the $\int e^{s-t} dW$) was to generate a correlated signal (i.e. to color the spectrum) which I thought would mean that you necessarily don't have independent increments anymore. In other words you have independent increments in $W_t$ but not in $\int e^{s-t} dW$. Somehow I feel this is a basic misunderstanding of what independence and correlated means but I can't understand the issue. – Romain May 09 '18 at 13:03
  • Sorry if it's basic stuff...maybe you have a reference to suggest me ? – Romain May 09 '18 at 13:04
  • @Romain: Never mind. Here is a martingale property for Ito integrals. If $M_t$ is a martingale, and $X_t$ is an $M_t$-adaptive process with $\mathbb{E}\left|X_t\right|<\infty$ for each fixed $t$, then $\int_0^tH_s{\rm d}M_s$, the Ito integral of $H_t$ with respect to the martingale $M_t$ is also a martingale. You may check here for more information. – hypernova May 09 '18 at 16:20
  • @Romain: Also, the expression $\int_0^te^{s-t}{\rm d}W_s$ could have confused you. But this is no more than $e^{-t}\int_0^te^s{\rm d}W_s$, and the seemingly correlated part is actually outside the integration. Thus the key part of this integral is still $\int_0^te^s{\rm d}W_s$. Obviously, $e^t$ is bounded for each given $t$ (it is adaptive because it is even not stochastic). Thus the integral gives a martingale. – hypernova May 09 '18 at 16:23
  • @Romain: Generally, whenever you find ${\rm d}X_t=H_t{\rm d}M_t$ with $M_t$ being a martingale (most typically, $M_t=W_t$), in the majority of the cases, $X_t$ is a martingale. This is a really helpful result. Here is an example, showing that $\int_0^tsW_s{\rm d}s$ is a martingale, just because it could be written as an Ito integral with respect to $W_t$. – hypernova May 09 '18 at 16:28
  • @Romain: Please pardon these massive responses. I hope they could be helpful for you :-) – hypernova May 09 '18 at 16:28
  • They are indeed helpful, thanks ! The wikipedia page left me a bit hungry for more since it does not give proof of the Ito integral's property, but your example surely helps and your point on the correlated part saved the rest ! Thanks again for your help :-) Greatly appreciated. – Romain May 09 '18 at 16:44
  • @Romain: My pleasure. Good luck to you :-) – hypernova May 09 '18 at 16:47
  • Hi again ! I've used your answer to derive the variance of a filtered OU process. I'm not sure whether it is correct though...if you have few mintues, I'd love to know what you think ! https://math.stackexchange.com/questions/2785097/filtering-of-ou-process-need-validation-of-theoretical-developments – Romain May 19 '18 at 09:05
  • @Romain: Sure. I will have a look at it. – hypernova May 19 '18 at 12:20