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Typically in stochastic integrals, it is common to have integrals with respect to the Wiener increment $dW(t)$.

For example

$$ \int_{0}^{t}W(t)dW(t) = \frac{W^{2}(t)}{2}-\frac{t}{2} $$

by Ito's lemma. But, how does one go about computing

$$ \int_{0}^{t}W(t^{\prime})dt^{\prime} $$

? I know that $dW(t)^{2}\equiv dt$ but I am not sure how to proceed integrating

$$ \int_{0}^{t} W(t^{\prime})\sqrt{dW(t^{\prime})}. $$

What should be the approach here?

FShrike
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kowalski
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  • Hello. I've edited your post to add to the title another detail about your question that makes it more interesting at a glance - a user might have looked at the old title and thought you were talking about a normal stochastic integral. If you don't like this, feel free to change it. – FShrike Nov 08 '22 at 21:58
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    The ordinary Riemann integral $ \int_{0}^{t}W(t^{\prime})dt^{\prime} $ cannot be computed explicitly. It has nothing to do with stochastic integration. – geetha290krm Nov 08 '22 at 23:14
  • $X_t =\int_{0}^{t}W(t^{\prime})dt^{\prime}$ is a random variable, it doesn't have a closed-form expression as far as I know, but with Itô's lemma, you can rewrite it as the integral of a determinstic function against $dW_{t'}$ and check some of its properties : for instance you can show that it is a Gaussian process and compute its mean and variance easily. – Stratos supports the strike Nov 09 '22 at 06:52
  • @StratosFair Can you elaborate more on this? What are the statistical properties of $X_t$? Is $X_t \sim N(0,t)$ since $W_{t} \sim N(0,t)$? – kowalski Nov 09 '22 at 17:33

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As in the answer of your another question, you have $$\begin{align} \int_0^tW_t'dt' &= \int_0^t(t-u)dW_u \\ \end{align}$$ and then you can determine easily the law of this integral $$\int_0^tW_t'dt' \sim \mathcal{N}\left(0,\int_0^t(t-u)^2du \right)$$

NN2
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