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6

From the stochastic integral analogue of integration by part, which is a corollary of Ito's Lemma, $$d(W_t\ln t) = \frac{W_t}{t}dt+(\ln t)dW_t,$$ so $$I:=\int_0^t \frac{W_s}{s}ds=W_t\ln t-\int_0^t (\ln s)dW_s=\int_0^t \ln\frac{t}{s}dW_s$$ as $W_t\ln t\to 0$ in probability, as $t\to 0^+$. Since the random variable $I$ is a linear combination of independent ...


3

Let $$ X_1(t) = A(t) \sin (t) + B(t) \cos (t)\\ X_2(t) = A(t) \cos (t) - B(t) \sin (t) $$ then thanks to the Ito fornula: \begin{align} dX_1(t) &= dA(t) \sin (t) + A(t) \cos (t) dt + dB(t) \cos (t) - B(t)\sin(t) dt \\ &= X_2(t)dt + dA(t) \sin (t) + dB(t) \cos (t) \\ dX_2(t) &= dA(t) \cos (t) - A(t)\sin (t) dt - dB(t) \sin (t) - B(t) \cos (t) ...


2

Since $\cosh(x) \ge 1$ for all real $x$ you have $\frac{e^{t/2}}{\cosh X_t} \le e^{t/2}$. So on any finite time interval you have a bounded local martingale, which is a martingale. Alternatively, note that $\frac{\sinh(x)}{(\cosh(x))^2} \le \frac{1}{2}$ for all $x$. So for any $T$, $$\int_0^T E\left[\left|-\frac{e^{t/2} \sinh ...


2

This inequality is called the Grownwall inequality in the ode theory. Following the hint, you may compute the first order derivative of the both sides of the inhomogeneous integral equation, and convert the integral equation to an ordinary diffrential equation. Then you can solve the ODE and obtain the analytic solution. The following is another way to show ...


2

For an Itô process $(X_t)_{t \geq 0}$ of the form $$X_t-X_0 = \int_0^t \sigma(s) \, dB_s + \int_0^t b(s) \, ds \tag{1}$$ Itô's formula reads $$f(X_t)-f(X_0) = \int_0^t f'(X_s) \sigma(s) \, dB_s + \int_0^t \left( \frac{1}{2} f''(X_s) \sigma^2(s) + f'(X_s) b(s) \right) \, ds.$$ Now: $X_t := \int_0^t s \, dB_s - \frac{t^3}{6}$ is an Itô process. Choose ...


2

I'm also studying a course where things like this appear, so I'll present an attempt but corrections are welcome. I know stochastic calculus doesn't always play by the rules. I get that: $$\mathbb{E}[W_T \cdot \int_0^Tf(s)dW_s] = \int_0^Tf(s)ds$$ If $\mathbf{P} = \{t_0 = 0, t_1,t_2,\dots,t_n = T\} $ is a tagged partition, $||\mathbf{P}|| = ...


2

Hint: Using Itô's formula, show that $X_t := \sin(W_t) e^{t/2}$ satisfies $$dX_t = \cos(W_t) e^{t/2} \, dW_t.$$ Conclude that $$\sin(W_T) = \int_0^T e^{-(T-t)/2} \cos(W_t) \, dW_t.$$ Remark: The idea is to find a determinstic function $f$ such that $X_t := f(t) \sin(W_t)$ is a martingale. By Itô's formula, a sufficient condition is $$\frac{1}{2} ...


1

Ignoring for the moment the fact that $\sqrt{2t+B_t}$ will sometimes be negative, we have $F(t,x) = \frac23(2t+x)^\frac32,\qquad$ $\frac{\partial F}{\partial t}=2\sqrt{2t+x},\qquad$ $\frac{\partial f}{\partial x} = \frac{1}{2\sqrt{2t+x}}$, so $\int_0^T\sqrt{2t+B_t}\;dB_t = \frac23(2T+B_T)^\frac32 -\int_0^T2\sqrt{2t+B_t}+\frac{1}{2\sqrt{2t+B_t}}\;dt$. We ...


1

Let $(X_t)_{t \geq 0}$ be a weak solution of the given SDE, then there exists a Brownian motion $(B_t)_{t \geq 0}$ such that $$X_t = X_0 + \int_0^t b(s,X_s) \, ds + \int_0^t \sigma(s,X_s) \, dB_s.$$ Using the elementary estimate $$(a+b+c)^p \leq 3^p (a^p+b^p+c^p), \qquad a,b,c \geq 0,$$ we find $$|X_t|^p \leq 3^p |X_0|^p + 3^p \left| \int_0^t b(s,X_s) ...


1

Stochastic calculus is to do with mathematics that operates on stochastic processes. The best known stochastic process is the Wiener process used for modelling Brownian motion. Other key components are Ito calculus & Malliavin calculus. Stochastic calculus is used in finance where prices can be modelled to follow SDEs. In the Black-Scholes model, ...


1

Let $Y_t=ln(X_t)$, by Ito's lemma we derive that the process $Y_t$ follows the SDE $ dY_t=\sigma dW_t,\quad Y_0=0 $ which has solution $Y_t= \sigma W_t$. Because $W_t$ has distribution $N(0,t)$, $Y_t$ has distribution $N(0,\sigma^2t)$. Since $X_t=e^{Y_t}$ and $X_t$ has normal distribution, we conclude that $X_t$ has log-normal distribution, so the ...


1

I did that like that Let $f(t)$ and $g(t)$ be integrating factors By Ito's Formula $$d(f(t)X1(t))=f(t)dX1(t)+f'(t)X1(t)dt$$ $$d(f(t)X1(t))=(f(t)X2(t)+f'(t)X1(t))dt+af(t)dW_{1}(t)$$ Similarly $$d(g(t)X2(t))=(-g(t)X1(t)+g'(t)X2(t))dt+bg(t)dW_{2}(t)$$ Need to find $f(t)$and $g(t)$ such that $f(t)X2(t)+f'(t)X1(t)=0$ and $-g(t)X1(t)+g'(t)X2(t)$ So, ...


1

Fix $\epsilon>0$. Since $\|H\|^2 < \infty$, we can choose $R>0$ sufficiently large such that $$\mathbb{E} \left( \int_R^{\infty} H_t^2 \, dt \right) < \epsilon.$$ By assumption, there exists $H^n \in \mathcal{E}$ such that $$\mathbb{E} \left( \int_0^R |H_t^n-H_t|^2 \, dt \right) \leq \epsilon.$$ Without loss of generality, we may assume ...


1

Define a martingale by $$M_t := \begin{cases} \mathbb{E}(\phi(W_T) \mid \mathcal{F}_t), & t \leq T, \\ \phi(W_T), & t>T. \end{cases}$$ Then, by the martingale representation theorem, there exists a representation of the form $$M_t = c+ \int_0^t \beta_s \, dW_s.$$ For $t=T$, this proves the claim. Remark: One possibility to prove the mentioned ...


1

Integrating any locally bounded predictable process (in particular a continuous adapted process) with respect to a local martingale (resp locally square integrable martingale) results in a new process which is a local martingale (resp locally square integrable). A proof of this can be found in Jacod & Shirjaev (Chapter 1; Theorem 4.31). (It basically ...


1

the answer to this is quite simple if you look carefully to the definition of H, you should notice that $h_{i-1}$ is an $\mathcal{F}_{t_{i-1}}$-measurable random variable. From this and the following elementary properties on conditional expectation we get the result : If X is $\mathcal{F}$-measurable and for any bounded random variable $Y$, we have a.s. : ...


1

You can determine the sde that $X_t^2$ satisfies using Ito's lemma. Then you can find the pdf by solving the forward Kolmogorov equations (Fokker-Planck).



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