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4

First of all, note that $\mathbb{E}(B_T^3)=0$, i.e. we have to find $F_t$ such that $$B_T^3 = \int_0^T F_s \, dB_s. \tag{1}$$ It follows from Itô's formula that $$B_t^3 = 3 \int_0^t B_s^2 \, dB_s + 3 \int_0^t B_s \, ds \tag{2}$$ (please compare this with what you wrote in your question; there are so many typos in there that I'm really not sure whether ...

3

For fixed $t_0 \geq 0$, we have $$\{\omega; t_0 \, \text{is not a local maximum for t \mapsto B_t(\omega)}\} \supseteq \{\omega; \forall \delta>0 \exists t \in [t_0,t_0+\delta]: B_{t}(\omega)>B_{t_0}(\omega)\}.$$ If we set $W_s := B_{s+t_0}-B_{t_0}$, $s \geq 0$, then we can rewrite this as \begin{align*} \{\omega; t_0 \, \text{is not a local ... 2 Just consider H:=\{1\} and suppose that a sample path t \mapsto B_t(\omega) of the Brownian motion (which starts at time 0 at 0) satisfiesB_T(\omega)=1$$for a certain T>0 and$$B_t(\omega)<1$$for all t \in [0,T), i.e. the sample path hits the set H=\{1\} at time T for the first time. Then \omega \in \{\tau \leq T\}, but ... 2 If \mu and \sigma are constant, dP = \mu\; dt + \sigma\; dW(t) is equivalent to P(t) = P(0) + \mu t + \sigma W(t) where W is Brownian motion with W(0)=0. Then \Delta P = P(T) - P(0) = \mu T + \sigma W(T) has a normal distribution with mean \mu T and standard deviation \sigma \sqrt{T}. The density at a given \Delta P is maximized with ... 1 Your stochastic integral formula is correct, as are your values for the variances of the two integrals on the right side. This implies that the covariance of those integrals is t^3/6. To see this directly, let M_t denote the first stochastic integral on the right; this is a martingale. Therefore,$$ \Bbb E\left[M_t\cdot\int_0^t ...

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Let's work it out (also let's finish the calculation): \begin{aligned} E[B_s(B_t^2-t)]&=E[E[B_s (B_t^2-t) \mid \mathcal{F}_s]] \\ &=E[B_s E[(B_t^2-t) \mid \mathcal{F}_s]] \\ &=E[B_s (B_s^2-s)] \\ &=E[B_s^3]-E[sB_s] \\ &=0. \end{aligned} The first equality is the tower property. The second equality is "factoring out what is known". ...

1

Essentials of Brownian Motion by Frank Knight has what you are looking for, and lots more.

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For a discrete version of Girsanov's theorem with adapted drift you need to consider a sequence $$X_n = X_{n-1} + \mu_n +\sigma_n \epsilon_n,$$ where $\{\epsilon_n\}$ are iid standard Gaussian variables, and $\{\mu_n,\sigma_n\}$ are predictable (for simplicity, let $\mu_n$ and $\mu_n/sigma_n$ also be bounded). In order to construct the martingale density, ...

1

Consider $\tilde B_s:=B_{s+1}$, $s\ge 0$, and notice that $R-1=\inf\{s>0:\tilde B_s=0\}=T_0(\tilde B)$. Now condition on where $B$ is at time $1$: the density of $B_1$ is $y\mapsto p_1(x,y)$ (as you surmise) and the conditional distribution of $\tilde B$, given that $B_1=y$, is $P_y$. Putting these thoughts together, \eqalign{ P_x(R>1+t) ... 1 For t>0 denote by n=n(t) \in \mathbb{N}_0 the unique number such that t \in (n,n+1]. Then, by the triangle inequality,\left| \frac{B_t}{t} \right| \leq \frac{|B_t-B_n|}{t} + \frac{|B_n|}{t} \leq \frac{1}{n} \sup_{s \in [n,n+1]} |B_s-B_n| + \frac{|B_n|}{n}. \tag{1}$$We already know from the law of large numbers that the second term converges to ... 1 First note that$$f(x_2,t+\Delta t \ | \ x_1,t) = \frac{1}{\sqrt{2\pi \Delta t}} \exp\left\{\frac{-1}{2}\frac{(x_2-x_1)^2}{\Delta t}\right\}, Hence we can write (let $u = x_2-x_1$) \begin{align} &\int_{|x_2-x_1|>\epsilon} \frac{1}{\sqrt{2\pi \Delta t}} \exp\left\{\frac{-1}{2}\frac{(x_2-x_1)^2}{\Delta t}\right\}dx_2 \\ &= \int_{|u|>\epsilon} ...

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The function $u(t,x)=e^{t/2}\,\cos(x)$ satisfies the heat equation ${d\over dt}u+{1\over 2}\Delta_x u=0,$ so that, by Ito's formula, $u(t,W_t)=e^{t/2}\,\cos(W_t)$ is a martingale.

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