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how can I solve these two stochastic integrals?

$$\int_0^T B_t\,dB_t$$ $$\int_0^T f(B_t)\,dB_t$$

where B_t is the BM.

Thank you very very much!

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Hi Chris, thank you for the link, yes I know that I should use the Ito lemma, but can you give me some hints about the two integrals? I do not know how to apply the Ito on this two problems. Thank you very much! – Jorko Jun 2 '13 at 0:51

Besides applying the Itô formula, there is also the possibility to calculate a stochastic integral using approximation by step functions. It works fine for the integral $\int_0^T B_t \, dB_t$:


$$f_n(t,\omega) := \sum_{j=1}^n 1_{[t_{j-1},t_{j})}(t) \cdot B_{t_{j-1}}(\omega)$$

where $\Pi_n$ is a partition of $[0,T]$ such that $$\max_{t_j \in \Pi_n} |t_j-t_{j+1}| \to 0 \qquad (n \to \infty)$$

Since the Brownian motion has continuous paths, it's not difficult to show that $(f_n)_{n \in \mathbb{N}}$ is an approximating sequence as required in the definition of the stochastic integral, therefore

$$\int_0^T f_n(t) \, dB_t \stackrel{L^2}{\to} \int_0^T B_t \, dB_t \qquad (n \to \infty)$$

By definition, it's easy to calculate the stochastic integral of step functions:

$$\int_0^T f_n(t) \, dB_t = \sum_{j=1}^n B_{t_{j-1}} \cdot (B_{t_{j}}-B_{t_{j-1}}) \tag{1} $$

On the other hand, we have

$$\begin{align} B_T^2 &= \left( \sum_{j=1}^n B_{t_j}-B_{t_{j-1}} \right) \cdot \left( \sum_{k=1}^n B_{t_k}-B_{t_{k-1}} \right) =\underbrace{\sum_{j=1}^n (B_{t_j}-B_{t_{j-1}})^2}_{\stackrel{L^2}{\to} T} + 2 \underbrace{\sum_{j=1}^n B_{t_{j-1}} \cdot (B_{t_j}-B_{t_{j-1}})}_{\stackrel{(1)}{\to} \int_0^T B_t \, dB_t}. \end{align}$$

Thus, $$B_T^2 = T + 2 \int_0^T B_t \, dB_t.$$

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Since by Ito's lemma you have for $\phi(t,x) \in \mathcal C ^{1,2}$

$$ \phi(t, X_t) = \phi(o,X_0) + \int _0 ^t (\partial_t +b\partial_x +\frac{\sigma^2}{2}\partial^2_{xx})\phi(s,X_s) ~ ds + \int_0^t\partial_x\phi(s,X_s) ~dB_s$$ if $X$ is a Ito's diffusion

$$ X_t = X_0 + \int _0 ^t b~ ds + \int_0^t\sigma ~dB_s$$

you must search for the simples $\phi$ such that $\partial_x \phi =x $ with $X =B$ ( so $b=0$ and $\sigma =1$) for the first integral and $\partial_x \phi =f $ for the second one also with $X =B$ .

Indeed, $$ \int_0^T B_s ~dB_s=\frac{1}{2}(B^2_T-B^2_0 -T)$$ and $$ \int_0^T f(B_s) ~dB_s=F(B_T)-F(B_0) -\frac{1}{2}\int_0^T\partial_x f (B_s) ~ds$$

where $F$ is a primitive of $f$.

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