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6

Consider $$X_t=a(t-1)+bt+(1-t)Y_t,\qquad Y_t=\int_{0}^{t}\frac{\mathrm dB_s}{1-s},$$ then $$\mathrm dY_t=\frac{dB_t}{1-t},\qquad \mathrm dX_t=a\mathrm dt+b\mathrm dt+(1-t)\mathrm dY_t-Y_t\mathrm dt,$$ hence $$\mathrm dX_t=\mathrm dB_t+\left(a+b-\int_{0}^{t}\frac{\mathrm dB_s}{1-s}\right)\mathrm dt,$$ and $$\mathrm d\langle X\rangle_t=\mathrm dt.$$ Edit: In ...


3

Firstly, by the strong law of large number, the limit should exist. Consider $$\theta_k = \inf\{t\geq 0, W_t = k\pi\}$$ Then the integral can be written as the sum of $\int_{\theta_{k}}^{\theta_{k+1}}\sin^2(W_s)ds$, we have \begin{align} \frac{1}{t}\int_0^t \sin^2(W_s)ds = \dfrac{\sum_{k=1}^{+\infty}1_{\theta_k < t}}{t} ...


2

I just realized I can use the fact that for radon measure $\mu$ on $[0,1]$, $f(t) = \mu([0,t])$ is of finite variation. So in my question, if for $\omega$ in $\Omega_0$ with $P(\Omega_0) = 1$, $h(\cdot)(\omega)$ is a Radon measure on $[0,1]$, then $h([0,t]) = W_t$ is of finite variation. This is contradictory with the fact that Brownian motion has infinite ...


2

Let $X_t=\displaystyle\int_0^ts\mathrm dW_s$ and $Y_t=\displaystyle\int_0^tW_s\mathrm ds$, then $X_t+Y_t=tW_t$ hence $E(X_t\mid W_t)=tW_t-E(Y_t\mid W_t)$. Furthermore, $E(W_s\mid W_t)=sW_t/t$ for every $s$ in $(0,t)$ hence $$E(Y_t\mid W_t)=\displaystyle\int_0^tE(W_s\mid W_t)\mathrm ds=(W_t/t)\displaystyle\int_0^ts\mathrm ds=tW_t/2.$$ Thus, $E(X_t\mid ...


1

This follows from the following theorem: Theorem: Let $(M_t,\mathcal{F}_t)_{t \geq 0}$ be a martingale such that the sample paths are (almost surely) continuous and of bounded variation. Then $M_t = M_0$ almost surely. (For a proof see e.g. René Schilling/Lothar Partzsch: Brownian motion - An Introduction to Stochastic Processes.) Since $$M_t := ...


1

Hint: use the Ito formula in the form: $$ d\left[\left(\alpha(t) +\int_0^t a_s dB_s\right) \left(\beta(t) + \int_0^t b_s dB_s\right)\right] \\= \left(\alpha(t)+\int_0^t a_s dB_s\right)\left(d\beta(t) + b_t dB_t\right) + \left( d\alpha(t) + a_tdB_t \right)\left(\beta(t) + \int_0^t b_s dB_s\right) + a_sb_s ds $$ where $\alpha,\beta:\Bbb R^+\to ...


1

Lets assume that we are in the regime in which $V_t > 0$ for every $t>0$. Consider the sequence of stopping times $$\tau_n = \inf \left\{ t: V_t > n \right\}, \quad n \in \mathbf{N},$$ then, as you mentioned before, if we can show that $\tau_n \to \infty$ as $n \to \infty$ in probability the result follows by standard localization: the expectation ...


1

It simply depends whether one wants to consider integrals with infinite time-horizon or not. Usually, stochastic integrals of the form $$\int_0^T K_s \, dW_s \tag{1}$$ are considered where $T<\infty$. In this case, $$\mathbb{E} \left[ \left( \int_0^T K_s \, dW_s \right)^2 \right] = \mathbb{E}\left( \int_0^T K_s^2 \, ds \right) \tag{2}$$ is called Itô's ...


1

Hi as mentioned in my comment the proof given does not allows us to conclude by localization. I have finally found a rigorous and elementary proof of the fact that a CIR process possesses moments of all orders which is necessary to get the result (order 1 is enough for our need) . First two obersvations : by Yamada-Watanabe's theorem that the CIR's SDE ...


1

First of all, the statement $$dH_{n+1}(t,B_t) = H_n(t,B_t)$$ doesn't make sense. Or can you explain what you mean by it? I guess, it should read $$dH_{n+1}(t,B_t) = H_n(t,B_t) \, dB_t;$$ at least that's what I going to prove in this answer. Lemma 1 $$H_{n+1}(t,x) = \frac{x}{n+1} H_n(t,x) - \frac{t}{n+1} H_{n-1}(t,x). \tag{0}$$ Proof: Denote by ...


1

Hints Recall that the independence of two stochastic processes $(M_t)_{t \geq 0}$ and $(N_t)_{t \geq 0}$ is equivalent to the independence of the corresponding canonical $\sigma$-algebras $\mathcal{F}_{\infty}^M$ and $\mathcal{F}_{\infty}^N$, $$\mathcal{F}_{\infty}^M := \sigma(M_s; s \geq 0).$$ Let $f$ be an $\mathcal{F}_t^M$-adapted process such that ...


1

Via Ito integral and using Ito Lemma (third and penultimate steps): $$ \int_0^t \sin(B_u)\circ dB_u = \int_0^t \sin(B_u) dB_u +\frac{1}{2}\int_0^t d(\sin(B_u))dB_u$$ $$ = \int_0^t \sin(B_u) dB_u + \frac{1}{2}\int_0^t\left(\sin'(B_u) dB_u +\frac{1}{2}\sin''(B_u) du\right)dB_u $$ $$ = \int_0^t \sin(B_u) dB_u + \frac{1}{2}\int_0^t\sin'(B_u) du $$ $$ = ...


1

Hints: First, solve the homogeneous SDE $$dX_t = \delta \mu X_t \, dt + \delta X_t \, dB_t. \tag{1}$$ To this end, apply Itô's formula to $f(x) := \log x$ and $(X_t)_{t \geq 0}$. In order to solve the SDE $$dX_t = (1+\delta \mu X_t) \, dt + \delta X_t \, dB_t,$$ we use a variation of constants, i.e. we make the Ansatz $$X_t = Z_t X_t^0$$ where $X_t^0$ is a ...



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