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1

The first "=" in (*) fails to be true. Note that $$\left\{ \sup_{0 \leq s \leq t} |W_s|> a \right\} \neq \left\{ \sup_{0 \leq s \leq t} W_s > a \right\} \cup \left\{ \sup_{0 \leq s \leq t} W_s < -a \right\}. \tag{1}$$ Say, for instance, we have a sample path with $\sup_{s \leq t} W_s(\omega)<a$ but $\inf_{s \leq t} W_s(\omega)<-a$, then $\... 1 Neither expression is correct, since interchanging$X$with$Y$corresponds to negating the expression$\log(X/Y)$, and hence negating the mean. In particular, the mean must be an antisymmetric function of$X$and$Y$, which it is not in either of your two expressions. With that being said, clearly the second expression is closer to being correct, where the ... 2 Since stopping times typically depend on a path of a process (and not only the value at some particular time), your formula for the conditional expectation is not too useful. However, there is a " functional" version of this formula which comes in handy; you can find it for instance in the book Brownian motion - an introduction to stochastic processes by ... 1 Assuming that$B$is a Brownian motion and$f \in L_2[0,T]$. First method: We know that$\forall t \in [0,T]$,$\int_0^t f(s)dB_s$is a Wiener integral. Therefore, we can use the fact that$T\geq s \geq t$,$\int_t^s f(u)dB_u$is independent of$\mathcal{F}_t$(1) and follows a centered Gaussian distribution with variance$\int_s^t |f(u)|^2du$(2). Thus, ... 0 For$t\geq 0, we have : \begin{align} E[H_t] &= E\left[\exp\left(\frac{W_t}{1+t}\right)\right] \\ &= E\left[\exp\left(\underbrace{\frac{\sqrt{t}}{1+t}}_\alpha X\right)\right] \quad X\sim\mathcal{N}(0,1)\\ &=\int_R \frac{1}{\sqrt{2\pi}}\exp\left(\alpha x -\frac12x^2\right)dx \\ &=\underbrace{\int_R \frac{1}{\sqrt{2\pi}}\exp\left(-\frac12(x^2-\... 2 You forgot thatM_t \geq B_t$. The joint density you have obtained is only for$m \geq w$so the integral for computing$f_{M_t}(m)$is from$-\infty$to$m$. 3 Note that$X_n$denotes the direction you move on the$n$th step, not at time$n$. So the fact that the sum goes up to$[t/\Delta t]$means that you take$[t/\Delta t]$steps. Why does this make sense? Well, you take a step every$\Delta t$units of time. So by time$t$,$[t/\Delta t]$is exactly the number of steps you will have taken. 1 We use the strong Markov property to remove the condition on$\mathcal{F}^W_{\tau_a}$and say that$X(t-\tau_a)$is just another Brownian motion, hence $$\mathbb{P}(X(t-\tau_a)<0\mid\mathcal{F}^W_{\tau_a})=\frac12$$ and we conclude $$\mathbb{E}\left[\chi_{\sup_{s\in[0,t]} W(s)\geq a}\mathbb{P}(X(t-\tau_a)<0\mid\mathcal{F}^W_{\tau_a})\right]= \mathbb{... 1 If you take f(x) = x, you see that W_t is a continuous local martingale. Then, taking f(x) = x^2 gives us that W_t^2 - W_0^2 - t is also a continuous local martingale so that W_t has quadratic variation at time t equal to t. Now apply Levy's characterisation of Brownian motion to conclude. 3 W(1) is a random variable and you cannot just suppose that it is equal to a. In fact, you know its distribution. If W is a standard Brownian motion then W(t) \sim \mathcal{N}(0,t) for every t. In particular, W(1) \sim \mathcal{N}(0,1) which is why \mathbb{E}[W(1)] = 0. 0 If I and J are two disjoint intervals, then 1_I 1_J =0 and therefore$$\big(x 1_{I}(t) + y 1_{J}(t) \big)^2 = x^2 1_I(t) + y^2 1_J(t)$$for any real numbers x,y \in \mathbb{R}. This implies$$f(t)^2 = (W_2-W_1)^2 1_{[2,3)}(t) + (W_3-W_1)^2 1_{[3,5)}.$$Hence,$$\int_0^{\infty} \mathbb{E}(f(t)^2) \, dt = \int_2^3 \mathbb{E}((W_2-W_1)^2) \, dt + \... 2 You didn't specified any filtration... I guess you talk about$\mathcal F_t=\sigma (X_t,Y_t)$. Hint Let$s<t$. You have that $$\mathbb E[X_tY_t\mid \mathcal F_s]=\mathbb E[(X_t-X_s)(Y_t-Y_s)\mid \mathcal F_s]+\mathbb E[(X_t-X_s)Y_s\mid \mathcal F_s]+\mathbb E[(X_t-X_s)Y_s\mid \mathcal F_s]+\mathbb E[X_sY_s\mid \mathcal F_s].$$ Now, using$X_s$and$...

2

Here are two methods that don't use any particularly advanced facts about B.M. (which you prefer will depend on what things you know about B.M.) (Via the reflection principle) Let $M(t) = \sup_{0 \leq s \leq t} B(t)$. The Reflection principle states that $$\mathbb{P}(M(t) \geq a) = 2 \mathbb{P}(B(t) > a) = 2 - 2\Phi(\frac{a}{\sqrt{t}})$$ where $\Phi$ is ...

1

This may be criticized as an overkill but a one line answer to your question comes from the Law of Iterated Logarithm for BM: $\lim \inf_{ t\to 0} \frac {B_t} {\sqrt {2t\log\, \log\, (\frac 1 t)}} =-1$ a.s.. This implies that every interval $(0,\delta)$ contains points $t$ with $B_t <0$.

0

Recall that the quadratic variation of Brownian motion up to time $t$ is simply given by $t$. It follows that the first variation of Brownian motion is infinite since processes of finite first variation have $0$ quadratic variation. Indeed, if $X$ is a continuous process with finite first variation then for any partition $\mathcal{P}$, \begin{align*}\sum_\...

2

The (simple) Markov property $$\mathbb{P}(X_t \in A \mid \mathcal{F}_s) = \mathbb{P}(X_t \in A \mid X_s) \tag{1}$$ makes perfect sense in any dimension $n \geq 1$. If, say, $(X_t)_{t \geq 0}$ is a continuous stochastic process which takes values in $\mathbb{R}^n$, then $(1)$ is well-defined for any Borel set $A \in \mathcal{B}(\mathbb{R}^n)$. The ...

4

It's not a martingale. Basically, a process of bounded variation can never be a continuous-time martingale (unless it's constant), so intuitively you can see immediately that $W_t^3$ is not a martingale because the bounded variation part of the Ito decomposition doesn't vanish. The error in your reasoning is here: $$\mathbf{E}\Big( \int_0^\tau W_s~ds \... 2 No, the statement is C[0,\infty) \notin \mathcal{B}(\mathbb{R}^{[0,\infty)}). Note that's a \notin, not \not\subset. Let's unpack this. \mathbb{R}^{[0,\infty)} is the set of all functions from [0,\infty) to \mathbb{R}. \mathcal{B}(\mathbb{R}^{[0,\infty)}) is a \sigma-field on this set; a set of functions A which is in \mathcal{B}(\... 2 Suppose s\geq t.$$E(W_s^2W_t^2) = E((W_s-W_t)^2W_t^2+2W_t^3W_s-W_t^4) \\ = E(W_{s-t}^2)E(W_t^2)+2E(W_t^3W_s)-E(W_t^4) =\\ (s-t)t+2E(W_t^3(W_s-W_t))+E(W_t^4) = \\ (s-t)t+2E(W_t^3)E(W_{s-t})+3t^2 = (s-t)t+3t^2 = 2t^2+st 1 For 0\le s<t, \begin{align} E[X_t|B_r]&=&E[e^{-t\sigma^2/2+\sigma B_t}|B_r, 0\le r\le s]\\ &= & e^{-t\sigma^2/2} E[e^{\sigma (B_t-B_s+B_s}|B_r, 0\le r\le s]\\ &= & e^{-t\sigma^2/2} E[e^{\sigma B_{t-s}}] e^{\sigma B_{s}}\\ &= & e^{-t\sigma^2/2} e^{(t-s)\sigma^2/2} e^{\sigma B_{s}}\\ &=& e^{-s\sigma^2/2+\sigma B_{... 3 I'm very new in this field, but I'm giving a try anyways. So please be critical and let me know if something is wrong (We are here to learn so...). The assertion is not true in general, indeed we have a couple of counterexamples. A simple counterexample. Let \mu\equiv 1 and \sigma\equiv 0. DefineA:=(0,1)$$Then we have for X_0=1, the explicit ... 3 Hints: Since (B_t)_{t \geq 0} is a martingale it follows from the optional stopping theorem that$$\mathbb{E}(B_{T_R \wedge T_0}) = x.$$Deduce from B_{T_R \wedge T_0} \in \{0,R\} that$$\mathbb{P}(B_{T_R \wedge T_0}=R) = \frac{x}{R} \qquad \mathbb{P}(B_{T_0 \wedge T_R} = 0) = \frac{R-x}{R}. \tag{1}$$Conclude that$$\mathbb{P}(T_R<T_0) = \frac{x}{R}....

0

No! You need much more than just $W_t-W_{k-1}\sim N(0,t-k+1)$. You want to show that $W_{t+k}-W_k$ is another Wiener process, i.e., you want to show it is almost surely $=0$ at $t=0$, independent Gaussian increments, and continuous paths with probability 1 These follows from $W_t$ being a Wiener process (why?).

1

Feller obtains in his book An Introduction to Probability Theory. Vol II (p.342) the following result (... unfortunately he does not give a detailed proof): Let $(B_t^x)_{t \geq 0}$ be a Brownian motion started at $x \in \mathbb{R}^d$ (i.e. $B_t^x = x+B_t$ where $(B_t)_{t \geq 0}$ is a standard Brownian motion). For fixed $a>0$ define a stopping time \... 2 First of all, note that Erfc does not denote the Error function but its "complement", i.e. $$\DeclareMathOperator{\erfc}{Erfc}\DeclareMathOperator{\erf}{Erf} \erfc(x) := \frac{2}{\sqrt{\pi}} \int_x^{\infty} \exp (-y^2) \, dy = 1- \erf(x)$$ where $$\erf(x) = \frac{2}{\sqrt{\pi}} \int_0^x \exp (-y^2) \, dy$$ is the Error function. Since $$|2(-k)+1| = |... 1 They are not that complicatedly nested. You just need the following result: Let T be a stopping time, r be a \mathcal F_T-measurable positive random variable and$$ S=\inf\{t>T\mid d(B_t,B_T)=r\}. Then S is a stopping time. Proof: \begin{align*} \{S\le t\}&=\{T<t\}\cap\left(\bigcup_{u\in]T,t]}\{d(B_u,B_T)=r\}\right)\\ &=\{T<t\... 4 Approach I (via characteristic functions): The identityE(1_A g(B_{t_1},\ldots,B_{t_n})) = P(A) E(g(B_{t_1},\ldots,B_{t_n})) \tag{1}$$can be easily extended to complex-valued continuous functions (just write g= \text{Re} g + i \, \text{Im g} and apply (1) separately to the real and imaginary part of g). Choosing$$g(x_1,\ldots,x_n) := \exp \left( i ... 1 Please feel free to amend the solution to improve it or correct some mistakes or you can even propose a new one of your own. Solution to Exercise 1.19 : 1 . Almost sure continuity results from the composition of the continuous mapping<x,.>$with the a.s. continuous trajectories of$X_t(\omega)$. So we only have to prove that increments over the ... 0 Take$m=2$. Let$\mathcal F_t$be the natural filtration and consider a variable$X$such that $$F_{X,B_{1,1},B_{2,1}}(x,y,z) = 2F(x)F(y)F(z) \text{ if } xyz>0, 0 \text{ otherwise}$$ where$F$stands for the standard normal cdf, and this definition is extended by "independence" (i.e. you build the rest of your Brownian motions by drawing mutually ... 0 To show the Markov property there is a direct approach. Let$u \in \mathcal{B}_b(\mathbb{R})$, and$s,t \ge 0. Then by the independent and stationary increment properties of BM: \begin{align*} \mathbb{E}[u(|B_{t+s}|) | \mathcal{F_s}] &= \mathbb{E}[u(|B_{t+s}-B_s +B_s|)| \mathcal{F_s}]\\ &= \mathbb{E}u(|B_{t+s}-B_s +y|)\bigg|_{y=B_s}\\ &= \... 3 We can use this proposition. Proposition. LetX=(X_t)_{t\geq0}$be an adapted process with values in a metric space$(E, d)$. Assume that the sample paths of$X$are continuous, and$F$be a closed subset of$E$. Then $$T_F=\inf\{t\geq0: X_t \in F\}$$ is a stopping time. Clearly$X_t$has continuous sample paths. Since$(-\infty, e^{-b}]$is ... 0 I believe that there is a typo in the definition of$X_t$; it should read $$X_t := \frac{t}{\sqrt{\pi}} \xi_0 + \frac{\color{red}{\sqrt{2}}}{\sqrt{\pi}} \sum_{k \geq 1} \frac{\sin(kt)}{k} \xi_k. \tag{1}$$ Moreover, it should be mentioned that this process is only a Brownian motion on the time interval$[0,1]$. It is known that $$\varphi_0 := 1 \qquad \... 0 For the second one use integration by parts formula. The integration by parts formula states that if X and Y are two processes then$$ d(X_t Y_t) = X_t dY_t+ Y_t dX_t + dX_t\cdot dY_t$$In your case, we set X=W_t and Y_t=t. Then$$ d(W_t t) = W_t dt + t dW_t + dt\cdot dW_t$$From "multiplication table," we know that dt\cdot dW_t=0. So that$$ ... 1 a)$P(S_5\leq x)=P(Z\leq \frac{lnx-25}{3\sqrt{5}})=\int_{-\infty}^{\frac{lnx-25}{3\sqrt{5}}}\frac{1}{\sqrt{2\pi}}e^{-\frac{y^2}{2}}dy=\int_{-\infty}^{\frac{lnx-25}{3\sqrt{5}}}I(y)dy$. So, thanks to fundamental theorem of calculus$f_{S_5}(x)=I(\frac{lnx-25}{3\sqrt{5}})\frac{\partial}{\partial x}(\frac{lnx-25}{3\sqrt{5}})$is the density of$S_5$. c)$P(S_3-...

2

No generality is lost by assuming that all of the random variables $\xi_n$, $\eta_n$ are bounded by one fixed constant. If not, replace them by $\arctan\xi_n$ and $\arctan\eta_n$ and proceed. The continuity of $\eta$ implies that $(\omega,n)\to\eta_n(\omega)$ is jointly measurable, permitting the use of Fubini's theorem. Let $c=\Bbb E[\eta_n]=\Bbb E[\xi_n]$...

1

If your variables are bounded, you may easily get the implication by showing that covariance converges to variance by dominated convergence, so variance must be zero. But in the general case where $\xi_0$ may not even have a variance. Take $I$ and $J$ two disjoint closed intervals, and $t_n, s_n$ two sequences both converging to the same $t$, with disjoint ...

0

I believe the $\mathcal N(0, s)$ notation is more common ($s$ is the variance). A Wiener process is Gaussian, therefore $(W_s, W_u)$ is jointly normal. The condition $$\operatorname{Var}(2 W_s + W_u) = \begin{pmatrix} 2 & 1 \end{pmatrix} \begin{pmatrix} s & \sigma \\ \sigma & u \end{pmatrix} \begin{pmatrix} 2 & 1 \end{pmatrix}^t = 8 s + u$$ ...

2

Do not get confused between the $a$ in the title and the $a$ in the reflection principle. Take $a=0$ in the reflection principle. Note that $2P(B_t \geq b)=2P(X \geq b/\sqrt t) \to 1$ as $t \to \infty$ (where $X$ is a standard normal variable). Hence $P(B_s \geq b \, \text{for some} \, s\leq t) \to 1$ as $t \to \infty$ which is what we want to prove.

1

I think the wording in the problem is wrong. We are looking for the "quadratic variation"(co-variation) of the following processes : $\forall t\in \mathbb{R^+}$, $X_t^1=e^{B_t}, X_t^2 = \ln(1+B_t^2), X_t^3=\cos^2(B_t)+\sin^2(B_t)$. Let's do it for $X_t^1$ : First, we express $X_t^1$ as an Itô process by applying the Itô's lemma to the function $F(x) = e^x$....

0

The formula is $$\operatorname{Var}(X_{s+t} - X_s) = \operatorname{Var}(X_{s+t}) + \operatorname{Var}(-X_s) + 2 \operatorname{Cov}(X_{s+t}, -X_s) = s + t + s - 2 \operatorname{Cov}(X_{s+t}, X_s).$$ Presumably, $\operatorname{Cov}(X_{s+t}, X_s) = \sqrt{s(s+t)}$, but we cannot deduce this from what you have written in your question. With the context that you ...

1

I think, I have an easy way to proof Formular I. Using the the law of total expectation: $E <X>_n = \sum_{i=1}^n [E(E(X_i^2|F_{i-1}) - E(X_{i-1}^2)] = \sum_{i=1}^n [E(X_i^2) - E(X_{i-1}^2)] = E(X_n^2) - E(X_0^2)$ Finally, notice that: \$Var(X_n - X_0) = E(X_n - X_0)^2 = E(X_n^2) - 2E(X_n X_0) + E(X_0^2) = E(X_n^2) - 2E(E(X_n X_0 | F_0)) + E(X_0^2) = ...

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