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Since $B_t$ is Gaussian with mean $0$ and variance $t$, the moments of $B_t$ can be calculated explicitly. For any $k \in \mathbb{N}$ we have $$\mathbb{E}(B_t^{2k+1}) = 0 \tag{1}$$ and $$\mathbb{E}(B_t^{2k}) = t^k \frac{2^k \Gamma(k+1/2)}{\sqrt{\pi}},$$ in particular, $$\mathbb{E}(B_t^2) = t \qquad \mathbb{E}(B_t^4) = 3t^2 \qquad \mathbb{E}(B_t^6) = 15 ... 3 T=\min(t,n) is a bounded stopping time. By Optional Stopping, the expectation is$$E(B^2_T -T) = E(B^2_0 -0) = 0$$Conclude by linearity of expectation. 3 This is a partial answer, just too long for a comment. There is a nice lemma from Dellacherie and Meyer: Lemma If \{A_s,s\in [0,t]\} is a continuous non-decreasing adapted process such that E[A_t - A_s|\mathcal F_s]\le K for all s\in[0,t], then for any \lambda < 1/K,$$ E[e^{\lambda A_t}]< (1-\lambda K)^{-1}. $$Now take A_s = ... 3 Intuitive argument: by symmetry we should have P(X_1 \ge 0 \mid W_1 = 0) = \frac{1}{2}. However, P(f(1, W_1) \ge 0 \mid W_1 = 0) is either 0 or 1 depending on whether f(1,0) \ge 0. Of course this is not really a proof because I conditioned on an event of probability 0. So let's try to use the same idea in an actual proof. Consider the ... 2 Guess: \xi is independent of (W_t)_{t\ge 0}, and \tilde{\mathscr F}_t should be \sigma(\xi)\vee\mathscr F_t 2 There are lots of random variables that are positive with probability 1/2 and negative with probability 1/2, but their expected value is not 0. But the point about \sin being odd is a good one. What it means is that the distribution of your random variable is symmetric about 0, and that does imply that the expected value (if it exists) is 0. 2 There are (at least) two possibilites to prove this. The first one, the quick one, uses Itô's formula. A straightforward application of Itô's formula shows that$$M_t^u := u(t,W_t)-u(0,x) - \int_0^t \left( \frac{\partial}{\partial t} + \frac{1}{2} \Delta_x \right) u(s,W_s) \, ds$$is a stochastic integral (with respect to Brownian motion) and, moreover, ... 2 By independence, the middle term disappears. That's not correct - note that W_{t+r} and W_{s+r} are not independent. The increments of Brownian motion are independent, i.e. W_{t+r}-W_{s+r} and W_{s+r}-W_0 = W_{s+r} are independent for s<t. Therefore,$$\mathbb{E}(W_{t+r} W_{s+r}) = \mathbb{E}((W_{t+r}-W_{s+r}) W_{s+r}) + ...

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Note that $$Y_t = \alpha A e^{t^2/2} + e^{t^2/2} B_t$$ has to satisfy the initial condition $Y_0 = \alpha$. Thus, $$Y_0 = \alpha A \stackrel{!}{=} \alpha,$$ i.e. $A=1$. Consequently, $$Y_t = e^{t^2/2} (\alpha+B_t).$$ If you want to check whether this is indeed a solution to the given SDE, then just apply Itô's formula with $$f(t,x) := e^{t^2/2} ... 1 [Correction: Proof of Theorem 5.13] It looks like there is a typo in the text, and the claim \mathcal F^+(T_n)\subset\mathcal F^+(T) should read \mathcal F^+(T)\subset\mathcal F^+(T_n). This implies that \mathcal F^+(T) is independent of \{W_{s+T_n}-W_{T_n}: s\ge 0\} for each n, hence independent of the limit process \{W_{s+T}-W_{T}: s\ge 0\}. ... 1 The process X in this case is a squared Bessel process of dimension two, which means that it is a (weak) solution to the SDE$$ \text{d}X_t=2\,\text{d}t+2\sqrt{X_t}\,\text{d}B_t, $$for all t\geq 0, where B is some Brownian motion. An application of Ito's formula now yields$$ \text{d}\ln X_t=\frac{2}{\sqrt{X_t}}\,\text{d}B_t, $$for all t\geq 0. In ... 1 Since p>\frac{1}{2}, we can write p= \frac{1}{2}+q for some q>0. Now$$\frac{(n-1)^p}{\sqrt{n}} = \sqrt{\frac{n-1}{n}} (n-1)^q \geq \frac{1}{2} (n-1)^q, \qquad n \in \mathbb{N},$$implies$$\mathbb{P} \left(B_1 > C \frac{(n-1)^p}{\sqrt{n}} \right) \leq \mathbb{P} \left(B_1 > \frac{C}{2} (n-1)^q\right).$$Choose k \in \mathbb{N} ... 1 What you write doesn't make sense and I think you have a lot of misunderstandings. Your double integral makes no sense, I don't know where you got that. You don't define T or X. I have no idea where you get the idea of "temporal" or "spatial" random variable. Let me try to clear things up for you. X_t is a collection of random variables. Each t ... 1 Try to use the fact that B_t/\sqrt{t} has the same distribution as B_1 and the fact that \sup_{s\le t }B_s/\sqrt{s} is monotonic w.r.t. t [Warning: more details below] For any constant C>0, let$$ A = (\lim_{t\to 0} \sup_{0<s\le t}\frac{B_s}{\sqrt{s}}<C )$$We want to show that P(A)=0. Since A\in {\mathcal F}_{0+}, we only have to ... 1 The position of the maximum is uniformly distributed. Let (X_t)_{t \in [0, 1)} be a Brownian bridge. Fix k \in [0, 1) and define the process (Y_t)_{t\in[0,1)} by$$Y_t\equiv X_{(t+k) \mod 1} - X_k.$$Claim: Y is a Brownian bridge. Proof: Y is a continuous Gaussian process. Y is zero-mean, like X. After some computation, we see that ... 1 Yes, you are right - the key idea is to rewrite the set \{T_A \leq t\} in a clever way using the continuity of sample paths of Brownian motion. Hints (for T_A): Show that$$\{T_A \leq t\} = \left\{ \omega \in \Omega; \inf_{r \in \mathbb{Q} \cap [0,t]} d(A, B_r(\omega)) = 0 \right\}$$using the continuity of the sample paths. Here d(A,x) := ... 1 Assume B(0)=0. On \{\tau_a<\infty\},$$ \{\omega:\tau_{a+b}<\infty\}=\{\omega: \tau_{a+b}-\tau_a<\infty\}=\{\omega: \tau_{a+b}\circ \theta_{\tau_a}<\infty\}. $$Thus,$$ \mathbb{E}_0\left[1\{\tau_{a+b}<\infty \}\times 1\{\tau_a<\infty\}\right] =\mathbb{E_0}\left[\mathbb{E_0}[1\{\tau_{a+b}<\infty\}\circ \theta_{\tau_a}\mid ...

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There are two way to think about this problem. First, let $N_t$ be the number of jumps by the time $t$ (clearly you only need to consider $N_t\ge n$) and let $\bar n=n-1/2$. Then you are interested in $$EV(n)=E(\sum_{i=1}^{N_t}(\frac 1 2 P(S_{i-1}=n-1)+\frac 1 2P(S_{i-1}=n)))=E(\sum_{i=n}^{N_t}\frac 1 2\binom {i-1}{\lfloor\frac {i-n}2\rfloor})$$ Now you can ...

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