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This question already has an answer here:

My question is about a stochastic integral of brownian motion w.r.t time.

Let $W(t)$ the Wiener process (or brownian motion). I want to calculate this: \begin{eqnarray} X(t)=\int_{0}^t dt' W(t'). \end{eqnarray} My strategy:

1) Itô's Fórmula: \begin{eqnarray} d(tW(t))=tdW(t)+W(t)dt \implies W(t)dt=d(tW(t))-tdW(t). \end{eqnarray} 2) Integrate: \begin{eqnarray} X(t)=\int_{0}^t dt'W(t')=\int_0^t d(t'W(t'))+\int_{0}^tdW(t')t'=tW(t)-\frac{t}{\sqrt{3}}W(t)=\left(1-\frac{1}{\sqrt{3}}\right)tW(t). \end{eqnarray} I used: \begin{eqnarray} \int_{0}^t dW(t')f(t')=\left(\frac{1}{t}\int_{0}^t dt'|f(t')|^2\right)^{1/2}W(t)\implies \int_0^t dW(t')t'=\frac{t}{\sqrt{3}}W(t), \end{eqnarray} because... \begin{eqnarray} \int_{0}^t dW(t')f(t')\sim \mathcal{N}\left(0,\int_0^t dt'|f(t')|^2\right), \hspace{0.5cm} W(t)\sim \mathcal{N}(0,t). \end{eqnarray}

The "problem" is: \begin{eqnarray} \sigma^2_X=\left(1-\frac{1}{\sqrt{3}}\right)^2 t^3 \end{eqnarray} But the correct is: \begin{eqnarray} \sigma^2_X=\frac{t^3}{3} \end{eqnarray} Can some illuminated mind tell me where the error is?

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marked as duplicate by LocalVolatility, JejeBelfort, skoestlmeier, amdopt, muffin1974 Sep 29 '18 at 19:35

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  • $\begingroup$ This is one of the most common introductory problems and has been answered multiple times before. Please spend a bit of time searching for existing questions before opening a new question. $\endgroup$ – LocalVolatility Sep 26 '18 at 9:41
  • $\begingroup$ My question is about a proposed solution to the problem and what is wrong with it. It is not a duplication. $\endgroup$ – Leonardo S. Vieira Sep 26 '18 at 16:47
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enter image description here

Let me know if any of the steps is not clear.

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  • $\begingroup$ Thank you very much. Quite clear. But I can not see what is wrong with my solution. If you can help me I'm very grateful. $\endgroup$ – Leonardo S. Vieira Sep 23 '18 at 19:13
  • $\begingroup$ Your approach ignores the covar between the components, I think $\endgroup$ – Magic is in the chain Sep 23 '18 at 20:24
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Re-comment and your second question as to why your transformation leads to a different result, it is because the transformation does not take into account the covariance between the two terms. As you can see below, the covaraince is distorted;

enter image description here

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  • $\begingroup$ Yes! $\int_0^t d(t'W(t'))-\int_0^t dW(t')t'\neq tW(t)-\frac{1}{\sqrt{3}}W(t)$. $\endgroup$ – Leonardo S. Vieira Sep 26 '18 at 16:59

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