How to solve this stochastic integrals?

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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 $\int_0^T B_t \, dB_t$:

Let

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

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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Updated on March 11, 2020

Comments

  • Jorko
    Jorko over 3 years

    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 Brownian Motion.

    Thank you very very much!

    • Chris Janjigian
      Chris Janjigian over 10 years
    • Jorko
      Jorko over 10 years
      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!