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7

I will complement saz's answerby proving that $L$ (to use saz's terminology) is a random variable. EDIT: A simpler and more powerful proof was given later by saz here (see his remark at the end). Introduction & proof outline 0 Following is a proof that the measure of the path of an $n$-dimensional ($n \geq 1$) Brownian motion is itself a measurable ...

6

Unfortunately, I'm not aware of an alternative proof (or a more detailed version of the proof by Mörters/Peres). So, instead of providing you with a source, I'll follow the lines of Paul Lévy; hopefully filling the gaps in his proof. Let $(B(t))_{t \geq 0}=((B^1(t),B^2(t)))_{t \geq 0}$ be a two-dimensional Brownian motion and $L := \lambda(B([0,1]))$. ...

3

Let $$W_t := \sigma B_t - \mu \cdot t$$ be a Brownian motion with drift $\mu>0$ and $$T_x := \inf\{t \geq 0; W_t \geq x\}.$$ Using the stationarity and independence of the increments of the Brownian motion $(B_t)_{t \geq 0}$, it is not difficult to see that $\left(\exp\left[\frac{2\mu W_t}{\sigma^2}\right]\right)_{t \geq 0}$ is a martingale. In ...

2

Since the function $f$ depends on the time $t$, we have to apply the time-dependent Itô formula which states that $$f(t,B_t)-f(0,B_0) = \int_0^t \frac{\partial}{\partial x} f(s,B_s) \, dB_s + \int_0^t \left( \frac{\partial}{\partial t}f(s,B_s) + \frac{1}{2} \frac{\partial^2}{\partial x^2} f(s,B_s) \right) \, ds.$$ For $f(t,x) := t \cdot x$ we see that $$t ... 2 Yes, your approach is correct. Your calculation simplifies (a bit) if you use the symmetry, i.e. the fact that$$\text{var} \left( \sum_{i=1}^n W_{t_i} \right) = \sum_{i=1}^n \text{var}(W_{t_i}) + 2 \sum_{i=1}^n \sum_{j=1}^{i-1} \text{cov}(W_{t_i},W_{t_j}).$$An alternative approach is the following: The vector (W_{t_1},\ldots,W_{t_n}) is a Gaussian ... 2 For simplicity (of notation), we assume n=2, i.e. that (B_s)_{s \geq 0} is a 2-dimensional Brownian motion, and t=1. For k,j \in \mathbb{Z} and m \in \mathbb{N} set$$A_{k,j}^m := \left[ \frac{k}{2^m}, \frac{k+1}{2^m} \right) \times \left[ \frac{j}{2^m}, \frac{j+1}{2^m} \right)$$and$$B_{k,j}^m := \left[ \frac{k-1}{2^m}, \frac{k+2}{2^m} ...

2

$$E[ (W_{n\Delta} - W_{(n-1)\Delta})^2] = E[W_{n\Delta}^2] - 2 E[W_{n\Delta}W_{(n-1)\Delta}] + E[W_{(n-1)\Delta}^2]$$ Use what we know about the second moment of $W_t$, and rewrite the second term. $$=n \Delta - 2 E\left[\left(W_{(n-1)\Delta} + (W_{n\Delta}-W_{(n-1)\Delta})\right)W_{(n-1)\Delta}\right] + (n-1)\Delta$$ Expand second term and use ...

2

Since $X_t$ is $N(0,t)$ for every $t$, one knows that $X_0=0$. Since the distribution of $X_t-X_s$ depends only on $t-s$, one knows that, for every $n$, $X_{n\Delta}-X_{(n-1)\Delta}$ is distributed like $X_{\Delta}-X_0=X_\Delta$, which is $N(0,\Delta)$. In particular, $X_{n\Delta}-X_{(n-1)\Delta}$ is centered with variance $\Delta$, QED.

2

I think you are on the right way. Consider $\ \varphi _{\lambda} (s) = \lambda e ^{-\lambda(t-s)}$ , $s \in [0;t]$. Then $||\varphi _{\lambda}||_{L_1} = 1 - e^{-\lambda t}$, so $||\varphi _{\lambda}||_{L_1} \to 1$ as $\lambda \to \infty$, and at the same time for any $\varepsilon \in (0;t)$ $$\int _0 ^{t-\varepsilon} \varphi _{\lambda} (s) ds \to 0,\ ... 1 Write that B_t - B_s \sim N(0,t-s) and the bilinear form as a sum of squares:$$ E\exp (a(B_t - B_s))= \int \frac{dx}{\sqrt{2\pi}} \exp \left(a\sqrt{t-s} x - \frac{x^2}2\right) \\= \int \frac{dx}{\sqrt{2\pi}} \exp \left(-\frac 12\left(x-a\sqrt{t-s}\right)^2\right) \exp\left(\frac 12 a^2\left(t-s\right)\right) = \exp\left(\frac 12 ...

1

Taken for granted that $\rho \leq \tau$, let us see why the derivation you gave makes sense. Recall the equation chain: \begin{aligned} u(x) & \stackrel{A}{=} E_x\left[E_x\left[\varphi(B(\tau))\mathbb{1}_{\{\tau < \infty\}} \mid \mathcal{F}^+(\rho)\right]\right] \\ & \stackrel{B}{=} E_x\left[u(B(\rho))\right] \\ & \stackrel{C}{=} ... 1 First we consider the case that f is bounded, i.e.\sup_{x \in \mathbb{R}} |f(x)| = \|f\|_{\infty}<\infty.$$By definition, we have$$S^{(n)}-S = \sum_{k=1}^{k_n} f(B_{t_{k-1}^{(n)}}) \cdot \bigg[ \left( B_{t_k^{(n)}}-B_{t_{k-1}^{(n)}} \right)^2 - (t_k^{(n)}-t_{k-1}^{(n)}) \bigg].$$Squaring this expression yields$$(S^{(n)}-S)^2 = ...

1

Suppose that $$dX_t = - \mu X_t \, dt + \sigma \, dW_t \tag{1}$$ and set $Y_t := e^{\mu t} X_t$. Itô's formula states that $$df(t,X_t)= \frac{\partial}{\partial x} f(t,X_t) \, dX_t + \frac{\partial}{\partial t} f(t,X_t) \, dt + \frac{1}{2} \frac{\partial^2}{\partial x^2} f(t,X_t) \, d\langle X \rangle_t. \tag{2}$$ Here, we have $f(t,x) = e^{\mu t} \cdot ... 1 I think the trick is usually to find an Ito diffusion that is relevant and augment it by$t$; then you can use the results for Ito diffusions. So for example$\mathrm{d}Y_t = \mathrm{d}B_t + \mathrm{d}t$is an Ito diffusion and we have$Y_t = B_t + t\$. So then lets look at $$Z_t = \left( \begin{array}{c} Y_t \\ t \\ \end{array} \right)$$ We know the ...

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