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3

Let $X_t = \int_0^t e^s dB_s$. Then $$r_t = f(t, X_t) = .1 + .1e^{-t} + e^{-t}X_t$$ By Ito's Lemma: $$dr_t = \frac{\partial f}{\partial x}dX_t + (\frac{\partial f}{\partial t} + \frac{1}{2}\frac{\partial^2f}{dx^2}Var(X_t))dt$$ So: \begin{eqnarray*} \frac{\partial f}{\partial x}dX_t + (\frac{\partial f}{\partial t} + ...


3

EDITED to meet edit of question The first equation is (after the edit) true. Consider the twodimensional continuous semimartingale $\left( t,B_t\right)$, and function $f(x,y)=xy$ we get $$D_xf(x,y)=y\quad D_yf(x,y)=x\quad D_1D_1f=D_2D_2f=0\quad D_1D_2f=D_2D_1f=1$$ And therefore ITO's formula gives $$tB_t=0+\int_0^t s\; dB_s+\int_0^t B_s\; ds.$$ The 2nd ...


3

The RHS is direct to evaluate. The variance of $B_s$ is $s$ so: $$\int_0^t E[B_s^2] ds = \int_0^t s ds = \frac{t^2}{2}$$ You are right about use Ito's Lemma for the LHS. By Ito: $$B_t^2 = t + 2\int_0^t B_s dB_s$$ So \begin{eqnarray*} E[(\int_0^t B_s dB_s)^2] &=& E\left[\left(\frac{B_t^2 - t}{2}\right)^2 \right] \\ &=& ...


2

Recall the following well-known theorem: Let $B$ be a bilinear form (on some $\mathbb{R}$-vector space $X$) such that $B(x,y) = B(y,x)$ and $B(x,x) \geq 0$ for any $x,y \in X$. Then $$|B(x,y)| \leq \sqrt{B(x,x)} \sqrt{B(y,y)} \qquad \text{for all $x,y \in X$}.$$ Applying this to the bilinear form $B(X,Y) := [X,Y]_t$ gives $$|[X,Y]_t| \leq \sqrt{[X]_t} ...


1

By the definition of the Itô integral, we know that $$\sum_{j=1}^n V_j \cdot \Delta_j \to \int_0^t B_s \, dB_s. \tag{1}$$ Note that we can write $$I_1(n) = \sum_{j=1}^n V_{j+1} \Delta_j = \sum_{j=1}^n \underbrace{(V_{j+1}-V_j)}_{\Delta_j} \Delta_j + \sum_{j=1}^n V_j \Delta_j. \tag{2}$$ By $(1)$, the second term at the right-hand side converges to ...


1

Fix $\epsilon>0$. Define iteratively a sequence of stopping times by $$\tau_n := \inf\{t>\tau_{n-1}; |X(t)-X(\tau_{n-1})| \geq \epsilon\}, \qquad n \in \mathbb{N},$$ $\tau_0 := 0$. Then $$f(t,\omega) := \sum_{n \geq 1} X_{\tau_{n-1}}(\omega) 1_{[\tau_{n-1}(\omega),\tau_n(\omega))}(t)$$ satisfies $\|X-f\|_{\infty} \leq \epsilon$ (since $X$ has ...


1

To keep notation simple, we write $S_t$ instead of $S_t^i$, $\sigma_t^j$ instead of $\sigma_{t}^{ij}$ and so on. That's okay, because $i$ is a fixed number throughout this calculation. Suppose $(X_t)_{t \geq 0}$ is an Itô process of the form $$dX_t = b(t) \, dt + \eta(t) \, dW_t$$ where $\eta = (\eta_1,\ldots,\eta_m)$ and $(W_t)_{t \geq 0}$ is an ...


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Let $g$ be a predictable function such that $$\mathbb{E} \left( \int_0^t \!\! \int |g(s,y)| \, \nu(dy) \, ds \right)<\infty,$$ then one can show that the integral $$\int_0^t \!\! \int g(s,y) \, N(dy,ds)$$ is well-defined and $$\mathbb{E} \left( \int_0^t \!\! \int g(s,y) \, N(dy,ds) \right) = \mathbb{E} \left( \int_0^t \!\! \int g(s,y) \, \nu(dy) \, ...


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As Brenton stated in the comments, the $u$ could be pretty much anything, because it goes away after you apply the bounds on the definite integral. Perhaps a bit clearer would have been to designate it $t'$ instead of $u$: $$\mathbb E I^2(t)=\mathbb E\int_0^t \Delta^2(t') dt'.$$ This makes it more semantically clear that you're integrating over time.



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