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I want to apply Ito for the following process:
\begin{equation*} X_t = tW_t + \int_0^t W_u du, \end{equation*} where $W$ is a Brownian motion. I have no trouble with the part $tW_t$ This can be written $tW_t = f(t,W_t)$ with $f(t,x) = tx$. However, which function is used for the integral? Can I write $X_t = g(t,W)$ but how is the integral a function of $t$ and $x$?

math
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  • You have to use Itô's formula for Itô processes. There is no $g$ such that $X_t = g(t,W_t)$. (What do you want to show/prove? Do you want to calculate the differential $dX_t$ or...?) – saz May 23 '15 at 08:14
  • @saz exactly I would like to have the differential $dX_t$. Just for the process $\int_0^t W_u du$ it must be equal $W_tdt$. This looks like normal calculus, but we have also the dependency of $W$ within the integral, or not? – math May 23 '15 at 08:22
  • Actually it is "normal" calculus since the mapping $t \mapsto \int_0^t W_u , du$ is differentiable. – saz May 23 '15 at 09:36
  • @saz thanks for your comment. I'm just struggeling with the following. The occurence of $W_t$ can be replaced by $x$, why not the occurence of $W_u$ within the integral? Why do I not have to take the partial derivative of $\int W_u du$ wrt to $x$. It seems $W_t$ and the $W_u$ under the integral sign are different things. – math May 23 '15 at 11:47
  • Itô's formula applies to functions of the form $f(t,W_t)$, i.e. a function which does depend on time and the value of Brownian motion at some (fixed) time $t$. The integral $\int_0^t W_u , du$ is something totally different since its value (at time $t$) does not only depend on the value of the Brownian motion at time $t$, but on the the whole path $(W_u)_{u \leq t}$. – saz May 23 '15 at 12:00
  • @saz brilliant! you helped me a lot. many thanks – math May 23 '15 at 12:01

1 Answers1

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You can solve this problem using Ito's Lemma two times. The first time is by the multidimensional version: if $A_t = B_t + C_t$ then $dA_t = dB_t + dC_t$ (Linearity).

In your example: $$d(X_t) = d(tW_t) + d\left(\int_0^tW_udu\right)$$ Next you apply Ito to the first component (Product Rule) $$d(tW_t) = W_tdt + tdW_t$$ and the second one is the definition of a time integral $$d\left(\int_0^tW_udu\right) = W_t dt$$ So the answer is $$dX_t = 2W_tdt+tdW_t$$

muaddib
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