# Integral of Brownian Motion w.r.t Time: what is wrong with this solution? [duplicate]

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?

## marked as duplicate by LocalVolatility, JejeBelfort, skoestlmeier, amdopt, muffin1974Sep 29 '18 at 19:35

• 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. – LocalVolatility Sep 26 '18 at 9:41
• My question is about a proposed solution to the problem and what is wrong with it. It is not a duplication. – Leonardo S. Vieira Sep 26 '18 at 16:47

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