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I do not understand the following result:

Suppose $dz_\chi$ and $ dz_\xi$ are correlated increments of standard Brownian motion with $dz_\chi dz_\xi=\rho dt$ you have the following expectation giving the following result:

$\mathbb{E} [\int_{t}^{T}e^{\kappa s} dz_\chi \int_{t}^{T}dz_\xi] = \rho \int_{t}^{T}e^{\kappa s} ds$

I really do not get it. This equation is justified by the Itô's isometry. In the book that I am reading the author just passes by saying that is easy to see, but if someone could show more explicit the steps that are taking to get this equation.

Carlos Lobato
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    What do YOU get when you apply Itô's lemma to the situation? And are you sure the correlation of the processes is written like that in your notes? – Did Jul 17 '14 at 22:51
  • Sorry the correct is $\rho dt$ I see that you can see this expected value as the covariance between two stochastic integrals, because the expected value of a stochastic integral with a deterministic integrand is zero, that justify the use of Itô's lemma. – Carlos Lobato Jul 17 '14 at 23:47
  • This is not Itô's lemma. – Did Jul 18 '14 at 06:42
  • @Did Why not? Perhaps it is not the most natural way to prove it (... I would prefer Itô's isometry), but it can be certainly proved using Itô's lemma. – saz Jul 18 '14 at 08:28
  • @saz My aim is that the OP gives much more precise statements than "justify the use of Itô's lemma" since such statements would lead right away to the result. The invocation of Itô's lemma would be convincing if one knew the process $X$ one applies it to, and if one saw a formula for $\mathrm dX$ (alternatively, one can appeal to Itô's isometry, as you mention). – Did Jul 18 '14 at 08:35
  • @saz Note that the OP seems determined to avoid providing any personal input and prefers to repost the same question rather than completing this one. – Did Jul 18 '14 at 08:41
  • @Did I see. Good luck ;). – saz Jul 18 '14 at 09:13
  • OP: I thought you (pretended to have) solved this question 14 months ago? – Did Jul 18 '14 at 09:18
  • @Did sorry for the misunderstending in the question above. The result is achieved with the use of the Ito's isometry as you said. The question that I am trying to make is that: The isometry stats that the variance of a Itô integral of a simple process $ \int_0^t \sigma_u dB_u$ name $I_t$ is $Var(I_t) = E( \int_0^t \sigma^{2}_u du)$. Ok, but in the case of the $\mathbb{covariance}$ why that $\rho$ emerge? In the duplicated (sorry about that) I used this as an example of my doubt which is: "Why in the covariance case the result has this $\rho$?" – Carlos Lobato Jul 18 '14 at 14:42
  • @Did about the question that I made 14 months ago. That was a different question. There I wanted to know why such correlation implicate that result. Here I am trying to understand why the covariance of two ito integrals with correlated increments (as in the old question) has the above result. thank you very much – Carlos Lobato Jul 18 '14 at 14:49

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