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3

The last step is not correct. Note that $$\mathbb{E}(W_2 \mid F_1) = W_1$$ since $(W_t)_{t \geq 0}$ is a martingale. Therefore $$ \frac{1}{3} \mathbb{E}(8 W_2-W_1 \mid F_1) = \frac{7}{3} W_1.$$ The correct result is $$\frac{7}{3} W_1 + \frac{15}{4}.$$ Instead of applying Itô's formula (which is rather overkill, I would say), one can use the fact that ...


3

By linearity of the conditional expectation, $$\mathbb{E}\left(\left.\int_1^2 (t^2W_t+t^3 )\,\mathrm dt\,\right|F_1\right)=\int_1^2 (t^2\mathbb{E}(W_t\mid F_1)+t^3 )\,\mathrm dt,$$ hence $$\mathbb{E}\left(\left.\int_1^2 (t^2W_t+t^3 )\,\mathrm dt\,\right|F_1\right)=\int_1^2 (t^2W_1+t^3 )\,\mathrm dt=\int_1^2 t^2\,\mathrm dt\cdot W_1+\int_1^2t^3 \,\mathrm ...


2

A filtration generated by Brownian motion simply means the smallest filtration with respect to which Brownian motion is adapted i.e. $$\mathcal{F}^B_t = \sigma \{B_s, \: s\leq t\}$$ On the other hand, a filtration for the Brownian motion, $\mathcal{F}_t$ is one to which the Brownian motion is adapted AND we have that for all $s<t$, $B_t - B_s$ is ...


2

Let $f(x) := x^a$ for some fixed $a>0$. Then $$f'(x) = a x^{a-1} \qquad f''(x) = a (a-1) x^{a-2}.$$ Since by Itô's formula $$f(X_t)-f(X_0) = \int_0^t f'(X_s) \, dX_s+ \frac{1}{2} \int_0^t f''(X_s) \, d\langle X \rangle_s$$ we get $$\begin{align*} Y_t - Y_0 &= f(X_t)-f(X_0) \\ &= a \int_0^t X_s^{a-1} \,d X_s + \frac{1}{2} a (a-1) \int_0^t ...


2

Every continuous local martingale is a continuous time change of a Brownian Motion in a possibly extended probability space. So, modulo some technicalities, all the properties obtained from integration w.r.t. Brownian Motion translates, one way or another, to properties when integrating w.r.t. to continuous time local martingales.


2

It is not difficult to see that the process $$W_t := \begin{cases} t B_{1/t}, & t > 0, \\ 0, & t=0 \end{cases}$$ defines a Brownian motion and $$\mathcal{G}_t := \sigma(B_u; u \geq t) = \sigma\left(W_u; u \leq \frac{1}{t} \right) =: \mathcal{H}_{\tilde{t}}$$ for ${\tilde{t}} := 1/t$ is the canonical filtration of $(W_\tilde{t})_{\tilde{t} \geq ...


2

First of all, there are several typos in your calculations (e.g. it should read $\int_0^t W_s^2 \,ds$ instead of $\int_0^t W_t^2 \, ds$). Your calculation goes wrong when you write $$\mathbb{E} \left( \int_0^t W_s^3 \, dW_s \right) = \frac{\mathbb{E}(W_t^4)}{4} - \frac{3}{2} \int_0^t V(W_s) \, ds = \frac{\color{red}{3t^4}}{4} - \frac{3t^2}{4}.$$ (I don't ...


1

Lemma: On the space $\tilde{\Omega} := \{f: [0,\infty) \times \Omega \to \mathbb{R}; f$ càdlàg$\}$ we define a norm by $$\|f\|_{L^1} := \mathbb{E}(\|f\|_{\infty}) := \mathbb{E} \left( \sup_{t \geq 0} |f(t)| \right).$$ Then $$L^1(\Omega; D[0,\infty)) := \{f \in \tilde{\Omega}; \|f\|_{L^1}<\infty\}$$ is a complete normed space. Proof: As a composition of ...


1

Seeing as your process is made up of Riemann integrals and the function $G$ is smooth, I would try differentiating under the integral sign to get the relevant partial derivatives, which can then be plugged into Ito's lemma. So, for instance, if we label the process $X_t$, then $$ \begin{eqnarray*} \dfrac{\partial}{\partial ...


1

See here for the fact that $\int_0^t B_u du$ follows the normal distribution with zero mean and variance $\dfrac{t^3}{3}$. See here for moment generating function(MGF) of normal distribution


1

Answer is 1/9.for example you take a 5 digit number then there are 9*10*10*10*10 ways then for successive number to have same digit there are 1*10*10*10*10 ways.probability is 1/9 by dividing


1

Note that $$\begin{align*}B_T^3 &=\sum_{j=1}^n B_{s_j}^3 - B_{s_{j-1}}^3 = \sum_{j=1}^n (B_{s_{j-1}}+(B_{s_j}-B_{s_{j-1}}))^3-B_{s_{j-1}}^3 \notag \\ &= \underbrace{\sum_{j=1}^n (B_{s_j}-B_{s_{j-1}})^3}_{=: I_1}+ 3 \underbrace{\sum_{j=1}^n \cdot B_{s_{j-1}} \cdot (B_{s_j}-B_{s_{j-1}})^2}_{=:I_2} + 3 \underbrace{\sum_{j=1}^n B_{s_{j-1}}^2 \cdot ...


1

The principle is clear, one considers $Y_t=\exp((1-2t)W_t^2)$, then Itô's formula yields $$\mathrm dY_t=a(t,W_t)\mathrm dW_t+b(t,W_t)\mathrm dt,$$ for some functions $a$ and $b$, hence $$\mathrm dX_t=\mathrm dY_t+f(t,W_t)\mathrm dt=a(t,W_t)\mathrm dW_t+(b(t,W_t)+f(t,W_t))\mathrm dt,$$ thus, $X$ is a martingale if the drift term is zero, that is, for ...


1

$$ Y_t = \mathrm{e}^{2B_t-\alpha t} $$ using Ito $$ dY_t = -\alpha Y_t dt + 2Y_tdB_t + \frac{4}{2}Y_t dB_t^2 = -\alpha Y_t dt + 2Y_tdB_t + 2Y_t dt = (2-\alpha)Y_tdt + 2Y_tdB_t $$ thus martingale requires a driftless SDE thus $\alpha = 2$. Was this the way you performed it?


1

It is more intuitive to think $\mu(dx)$ in the form of $p(x) dx$. $\mu$ is a probability measure and generally it is assumed to be absolutely continuous with respect to Lebesgue measure $dx$. So let's put (informally) $\mu(dx) = p(x) dx$ where $p(x)$ is the probability density function of this probability measure and $dx$ is the Lebesgue measure. Then ...


1

This is a consequence from the Clarke-Ocone Theorem, and uses Malliavin derivative. See also the Clarke-Ocone formula paragraphe here. If you want a technical reference, see this introductory course, mainly theorem 1 p. 18.


1

Hint: The stochastic integral $$M_t := \int_0^t e^{-as} \, dB_s$$ is a martingale. Hence, $\mathbb{E}M_t = \mathbb{E}M_0=0$. In order to calculate the variance of $X_t$ use Itô's isometry. Alternative approach: Since stochastic integrals are martingales, we have $$\mathbb{E}X_t- \mathbb{E}X_0 = \int_0^t a \cdot \mathbb{E}X_s \, ds,$$ i.e. $m(t) := ...



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