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## Hot answers tagged stochastic-integrals

3

I'm not too familiar with Brownian motions, but I think the problem is basically the same as in the case of individual random variables as in muaddib's answer: "Take the process $X$ where $X_0$, $X_1$ are iid positive random variables and $X_2$ is a normal random variable with variance $X_0+X_1$ and mean zero." Here $X_2$ could be said to have "random ...

2

Write the squared integral as $$\left( \int_0^t X_s \, ds \right)^2 = \left( \int_0^t X_s \, ds \right)\left( \int_0^t X_u \, du \right) = \int_0^t \int_0^t X_s X_u \, ds \, du,$$ where I have applied Fubini's theorem since the intervals are finite and the functions are integrable. Now use the definition of expectation: $$\mathbb{E}\left( \int_0^t X_s \, ... 2 There is no problem integrating a brownian motion pathwise with respect to a continuous (say) integrand. This is because a brownian motion has continuous almost everywhere sample paths. In the Ito integral the brownian motion is used as an integrand, i.e. it has the form \int_{[0,T]} f(s,\omega) dW(s,\omega). One can show that if every continuous ... 2 If you have an Ito process$$dX_t=\mu_tdt+\sigma_tdB_t$$then Ito's lemma states that$$df(X_t)=f'(X_t)dX_t+\frac{1}{2}f''(X_t)\sigma_t^2dtHere \mu_t=0.4S_t and \sigma_t=0.5S_t. We also have f'(x)=\frac{1}{x} and f''(x)=-\frac{1}{x^2}. So d(\log S_t)=\frac{dS_t}{S_t}-\frac{1}{2}\frac{(0.5S_t)^2}{S_t^2}dt=0.4dt+0.5dB_t-0.125dt 1 Note that \begin{align*} \int_t^T X_s\, ds \int_t^T Y_s \,ds = \int_t^T\int_t^T X_s Y_u \,ds \,du. \end{align*} Then, by assuming the Fubini, \begin{align*} E\left(\int_t^T X_s\, ds \int_t^T Y_s\, ds\right) = \int_t^T\int_t^T E\left( X_s Y_u\right)\, ds \,du, \end{align*} which is generally not equal to \int_t^T E( X_s Y_s) \,ds. 1 Ito's formula tells you thatf(t,X_t)=f(0,X_0)+\int_0^t \frac{\partial f}{\partial s}(s,X_s) ds + \int_0^t \frac{\partial f}{\partial x}(s,X_s) dX_s + \int_0^t \frac{1}{2} \frac{\partial^2 f}{\partial x^2}(s,X_s) (dX_s)^2$$where (dX_s)^2 is a notational shorthand for the quadratic variation of X_s. If X_t is a 1D diffusion process satisfying ... 1 Partial answer: We will derive the unconditional variance. By the Ito isometry, we get that  \mathbb{E}(\int_{0}^{r}\exp\{(r-s)\phi+\lambda(W_r-W_s)\}dB_s)^2 = \mathbb{E}\int_{0}^{r}\exp\{(r-s)\phi+\lambda(W_r-W_s)\}^2ds  Noting that \exp\{r\phi + \lambda W_r\} is a constant, we move that outside of the integral. ... 1 Following up on @saz's comment: In expanding Y:=\log X using Ito's formula you express Y in terms of \mu, \sigma, and W. This then leads to an explicit formula for (a putative solution) X=e^Y in terms of those same three ingredients. One of your jobs is then to check that this formula does indeed provide a solution. The first thing to notice is ... 1 Concerning the unboundedness of \alpha and \beta: you are absolutely right, there is some inaccuracy in Ikeda-Watanabe. In order for the argument to work, one needs to impose a stronger assumption e.g. the one from Remark 1.1, Chapter 4: For any T>0 and M>0$$ \sup_{t\in[0,T],\|w\|_T\le M} \big(\|\alpha(t,w)+ \|\beta(t,w)\|\big) ...

1

First, recall the following elementary result: Lemma: Let $f:[0,\infty) \to \mathbb{R}$ be a continuous function and set $$f_n(t) := \sum_{j=0}^{\infty} f(t_j) 1_{[t_j,t_{j+1})}(t)$$ where $t_j := j 2^{-n}$. Then $f_n(t) \to f(t)$ as $n \to \infty$ for all $t \geq 0$. Applying this result for each fixed $\omega$, we get that \$\phi_n(t,\omega) \to ...

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