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3

For $c=0$ this result is knows as reflection principle (see e.g. René Schilling/Lothar Partzsch: Brownian Motion - An Introdction to Stochastic Processes, Chapter 6) and follows from the Markov property and symmetry of Brownian motion. However, for $c>0$ the proof is more involved since we have to get rid of the drift term. Since by definition $$[H_a ...


3

We have that $$E(B(t)W(1))=E \left( (W(t)-tW(1))W(1) \right) =E(W(t)W(1)-tW(1)W(1)) $$ Using the linearity of the expectation we also have that this expression is equal to $$ E(W(t)W(1))-E(tW(1)W(1))=E(W(t)W(1))-tE(W(1)W(1))$$ As @PetiteEtincelle said, $E(W(t)W(s))=\min(t,s)$, so finally we conclude that $$ E(B(t)W(1))= \min (t,1)-t\min(1,1)=t-t=0$$


3

If $D$ is a domain in $\mathbb{R}^n$, and $x_0 \in D$ is a point, then the distribution of Brownian motion started at $x_0$ stopped at the time when it leaves $D$ is the harmonic measure of $\partial D$ with base point $x_0$. In the plane you can often use conformal invariance to calculate harmonic measure. Since you know that the harmonic measure of the ...


2

Hint: Set $Y_k = B_{t_k + s} - B_{t_{k-1}+s}$ where $t_0 = 0$. Note that $X_k = Y_1 + \dots + Y_k$, so you can express $f(X_1, \dots, X_n)$ as a Borel function of $Y_1, \dots, Y_n$. Now observe that $\{\mathcal{F}_s, \sigma(Y_1), \dots, \sigma(Y_n)\}$ are mutually independent $\sigma$-fields. Use a Dynkin lemma argument to conclude that $\mathcal{F}_s$ ...


2

Define for fixed $s,t\in\mathbb R_+$ the random variable $$Y_N:=\frac{t}{\sqrt{\pi}}X_0+\sqrt{\frac 2{\pi}}\sum_{n=1}^NX_n\frac{\sin(nt)-\sin(ns)}n.$$ For a fixed $N$, the random variable $Y_N$ has Gaussian distribution has a linear combination of the Gaussian vector $(X_0,\dots,X_N)$. By the results mentioned in the OP, the sequence $(Y_N)_{N\geqslant ...


2

Consider the functions $g=f\mathbf 1_{f\geqslant1}$ and $h=f\mathbf 1_{f\lt1}$, hence $f=g+h$. The function $g$ is nonzero only on the ball centered at $-z$ with radius $1$ and the density of $B_t$ is uniformly bounded by $c(t)=(2\pi t)^{-3/2}$ hence $$E(g(B_t))\leqslant c(t)\int_{\|x\|\leqslant1}\frac{\mathrm dx}{\|x\|}=c(t)4\pi\int_0^1\frac{r^2\mathrm ...


2

Brownian motion is often defined as a probability measure on the space $C([0,\infty),\mathbb R)$ of real valued continuous functions defined on $[0,\infty)$. As such, a change of probability measure on this space can yield processes pretty different from the ones you have in mind: the means can change, the variances can change, the gaussianity can fail, etc. ...


2

Basically, you can simply apply Itô's formula: $$\begin{align*} f(W_t^1, W_t^2)-f(0,0) &= \int_0^t f_x(W_s^1,W_s^2) \, dW_s^1 + \int_0^t f_y(W_s^1,W_s^2) \, dW_s^1+ \\ &\quad \frac{1}{2} \bigg( \int_0^t f_{xx} (W_s^1,W_s^2) \, \underbrace{dW_s^1 dW_s^1}_{ds} + 2 \int_0^t f_{xy}(W_s^1,W_s^2) \, \underbrace{dW_s^1 dW_s^2}_{\varrho \, ds} \\ &\quad ...


2

You are confusing the measure on path space with Lebesgue measure. The "almost everywhere" refers to the former: almost every individual path can be taken to be continuous everywhere. Indeed, the Wikipedia page you link to says that Brownian motion is "almost surely everywhere continuous". In other words, if $\mathbb{P}$ is Wiener measure on a suitable ...


2

Suppose that $S_t$ and $S_t'$ are perfectly correlated for each $t>0$ with correlation $1$. Then it follows from the very definition that $S_t = S_t'$ for all $t>0$. Applying Itô's formula (with $f(x)=x^2$ yields $$\begin{align*} \underbrace{(S_T-S_T')^2}_{0} &= 2 \int_0^T \underbrace{(S_t-S_t')}_0 d(S_t-S_t') + \int_0^T \underbrace{\langle S-S' ...


1

Sorry I cant leave this as a comment. As a first idea, you can approximate the continuous function $f$ with a piece-wise linear continuous function (discretise the interval $[0,1]$. And on each subinterval $[t_k, t_{k-1}]$, you can use a brownian bridge to calculate the probability and show it's positive.


1

Let $f: [0,1] \to \mathbb{R}$ be a continuous function. Since $[0,1]$ is compact, $f$ is uniformly continuous, i.e. we can choose $n \in \mathbb{N}$ such that $$|f(s)-f(t)| < \frac{\varepsilon}{2} \quad \text{for all $|s-t| \leq \frac{1}{n}$.}$$ If we se set $t_j := j/n$, $j=0,\ldots,n$, then $$\begin{align*} \mathbb{P} \left( \sup_{0 \leq t \leq 1} ...


1

Hint: Use $$\left\{ \limsup_{t \to 0+} \frac{B_t}{\sqrt{t}} > k \right\} = \bigcup_{\ell \in \mathbb{N}} \left\{ \limsup_{t \to 0+} \frac{B_t}{\sqrt{t}} > k+ \frac{1}{\ell} \right\}.$$


1

Not sure I understand the bafflement here... Let $X=\limsup\limits_{t \to 0^{+} } B_t/\sqrt{t}$. For every positive $s$, $X$ is equal to the same limsup restricted to the values of $t$ in the interval $(0,s)$ hence $X$ is $\mathcal F_s$-measurable, that is, for every Borel subset $A$, $[X\in A]$ belongs to $\mathcal F_s$. Thus, $[X\in A]$ belongs to ...


1

For #1: Let's call your isomorphism $T$, the isometry from $\mathcal{H}_K$ to $L^2(P)$ which maps $K_t$ to $B_t$. Yes, $T$ really is a stochastic integral: $Tf$ is the Itô integral, not of $f$, but of $f'$. So $$Tf = \int_0^1 f'(t)\,dB_t$$ or in other words, for deterministic $g \in L^2([0,1])$, $$\int_0^1 g(t)\,dB_t = T\left(\int_0^\cdot ...


1

The joint distribution of the running maximum $M_1 = \max_{0 \leq t \leq 1}B_t$ and $B_1$ is $$f(m,b) = \frac{2(2m-b)}{\sqrt{2\pi}}\exp(-\frac{(2m-b)^2}{2}), \qquad m\geq 0, b \leq m $$ Hence, $P(M_1 = B_1) = 0$ Note : In your approach, you replaced $M_1$ with $\vert B_1 \vert $ which is not true! They have same distribution but the random variables are ...


1

$P(B_1 = \max_{t\in[0,1]}B_t) = P(\max_{t\in[0,1]}(B_t-B_1) = 0)= P(\max_{t\in[0,1]}(B_{1-t}-B_1) = 0)$ You can easily check that $(B_{1-t} - B_1)_{t\in [0,1]}$ is itself a Brownian motion, so $P(\max_{t\in[0,1]}(B_{1-t}-B_1) = 0) = P(\max_{t\in[0,1]}B_t = 0) = P(|B_1| = 0) = 0$


1

As @Did wrote, let's talk about the space $\Omega = C([0,\infty),\Bbb R)$. This is the space of continuous trajectories, so if we define a measure on $\Omega$ we are saying which of the trajectories are more likely than others. For example, we can define a Wiener measure $P$ on $\Omega$ - that is what you call a Brownian motion. If we define another measure ...


1

Let $f: \mathbb{R}^2 \to \mathbb{R}$ be a bounded continuous function. In order to show that $$M_t := (I_t,W_t) := \left( \int_0^t W_s \, ds, W_t \right)$$ is Markov, we have to show that there exists $g: \mathbb{R}^2 \to \mathbb{R}$ such that $$\mathbb{E}^x(f(M_t) \mid \mathcal{F}_s) = g(M_s)$$ for any $s \leq t$. To this end, write $$\begin{align*} ...



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