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I am not quite sure how to solve this integral to be able to do numerical calculations with it. $\lambda$ is a constant, $u$ is time, and $Z_u$ is a wiener process. Can anyone provide some direction on how to solve it please?

$f = \int_{0}^t e^{\lambda u } dZ_u$

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2 Answers 2

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I'll try to use Ito's Lemma to come up with a solution. Ito's lemma states that:

$$F(Z_t,t)=\int_0^t\left(\frac{\partial F}{\partial u}+\frac{\partial F}{\partial Z}a+0.5\frac{\partial^2 F}{\partial Z^2}b^2\right)du+\int_0^t\frac{\partial F}{\partial Z}bdZ_u$$

We have $a=0$ and $b=1$ (because $Z_t=\int_0^t0du+\int_0^t1 dZ_u$).

My strategy is to find a function so that the integral $\int_0^te^{\lambda u}dZ_u$ appears in the expression of Ito's Lemma under $\int_0^t\frac{\partial F}{\partial Z}dZ_u$: therefore, I will try $F(Z_t,t):=Z_te^{\lambda t}$. Then we get:

$$Z_te^{\lambda t}=\\=\int_0^t\left(\frac{\partial F}{\partial u}+\frac{\partial F}{\partial Z}a+0.5\frac{\partial^2 F}{\partial Z_t^2}b^2\right)du+\int_0^t\frac{\partial F}{\partial Z}bdZ_u=\\=\int_0^tuZ_ue^{\lambda u}du+\int_0^te^{\lambda u}dZ_u$$

Now we can isolate the term of interest on the RHS and write:

$$Z_te^{\lambda t}-\int_0^tuZ_ue^{\lambda u}du=\int_0^te^{\lambda u}dZ_u$$.

As per @siou0107's answer, the above is a normally distributed random variable with:

$$\mathbb{E}\left[Z_te^{\lambda t}-\int_0^tuZ_ue^{\lambda u}du\right]=\\=\mathbb{E}\left[Z_t\right]_{=0}e^{\lambda t}-\int_0^tu\mathbb{E}\left[Z_u\right]_{=0}e^{\lambda u}du=0$$

$$Var\left(Z_te^{\lambda t}-\int_0^tuZ_ue^{\lambda u}du\right)=Var\left(\int_0^te^{\lambda u}dZ_u\right)=\\=\mathbb{E}\left[\left(\int_0^te^{\lambda u}dZ_u\right)^2\right]-\left(\mathbb{E}\left[\int_0^te^{\lambda u}dZ_u\right]_{=0}\right)^2$$

Using Ito Isometry, we get:

$$\mathbb{E}\left[\left(\int_0^te^{\lambda u}dZ_u\right)^2\right]=\mathbb{E}\left[\int_0^t\left(e^{\lambda u}\right)^2du\right]=\mathbb{E}\left[\int_0^te^{2\lambda u}du\right]=\int_0^te^{2\lambda u}du$$

PS: we can see that using Ito's lemma was not the smart way of solving the problem, because instead of simplifying, it made it more complicated. Directly taking the expectation and variance of the Ito integral, as per @siou0107's answer, is the best way to approach the problem.

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    $\begingroup$ When $Z$ is a Brownian motion, you do not have a first-order derivative in the space variable in Itō's formula. $df = \left(\partial_t f + \frac{1}{2} \partial_{xx}^2\right)f \, dt + \partial_x f \, dZ_t$ $\endgroup$
    – siou0107
    May 31 at 6:19
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    $\begingroup$ Yes, correct. The above should be correct now. $\endgroup$ May 31 at 7:16
  • $\begingroup$ Your last line seems very strange to me; $\text{Var}(A - B) \neq \text{Var}(A) - \text{Var}(B)$! Idem for variance of integral, which is not integral of variance. And the variance of $Z_t e^{\lambda t}$ is $e^{2 \lambda t} \text{Var} (Z_t) = e^{2 \lambda t} t$. $\endgroup$
    – siou0107
    Jun 7 at 18:24
  • $\begingroup$ @siou0107: yes, sorry man, I was in a rush. Your answer got deservingly more votes :) I made a correction. $\endgroup$ Jun 8 at 9:38
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This is just a Wiener integral (stochastic integral with respect to a Brownian motion and deterministic integrand), hence a centered Gaussian random variable with variance $$ \int_0^t{e^{2\lambda u}\mathrm{d}u} = \left[\frac{e^{2\lambda u}}{2\lambda}\right]_0^t = \frac{e^{2\lambda t} - 1}{2 \lambda} $$

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