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4

Using Ito's Lemma we have $$d\log X_t=\left(v-\frac12\sigma^2\right)dt+\sigma dW_t \tag 1$$ Integrating $(1)$ between $t_1$ and $t_2$ yields \begin{align} \log(X_{t_2}/X_{t_1})&=\left(v-\frac12\sigma^2\right)\left(t_2-t_1\right)+\sigma\int_{t_1}^{t_2}dW_t\\\\ &=\left(v-\frac12\sigma^2\right)\left(t_2-t_1\right)+\sigma\left(W_{t_2}-W_{t_1}\right) ... 4 I will treat the case where M is a continuous semimartingale. Unfortunately it is generally not the Riemann Stieltjes integral. You know that the Stieltjes measure of g, is only defined if g has finite variation. However, as you know, many stochastic processes does not have sample paths with finite variation, and therefore such an integral does not exist. ... 4 Brownian motion, Solution I Since W_t \sim N(0,t), we have\mathbb{E}(|W_t| 1_{\{|W_t|>K\}}) = \frac{2}{\sqrt{2\pi t}} \int_K^{\infty} x \exp \left( -\frac{x^2}{2t} \right) \, dx = \sqrt{\frac{2}{\pi}} \sqrt{t} \exp \left(-\frac{K^2}{2t} \right)$$and therefore$$\sup_{t \geq 0} \mathbb{E}(|W_t| 1_{\{|W_t|>K\}})=\infty.$$Brownian motion, ... 3 The RHS is direct to evaluate. The variance of B_s is s so:$$\int_0^t E[B_s^2] ds = \int_0^t s ds = \frac{t^2}{2}$$You are right about use Ito's Lemma for the LHS. By Ito:$$B_t^2 = t + 2\int_0^t B_s dB_s$$So \begin{eqnarray*} E[(\int_0^t B_s dB_s)^2] &=& E\left[\left(\frac{B_t^2 - t}{2}\right)^2 \right] \\ &=& ... 3 Let's talk a little bit about the general situation, because in a sense this result should've been expected even if we knew very little about Brownian motion. In general, Chebyshev's inequality tells you that if you have a random variable X with finite mean m and finite variance \sigma^2, then for each a>0,$$P(|X-m|>a) \leq ...

3

Brownian Motion is defined to have Gaussian increments $B_u - B_v$ with variance $u - v$ and mean $0$. Since $B_0$ is $0$ that means $B_t$ is a Gaussian with variance $t$ and mean $0$. $$E[Var(B_t)] = E[B_t^2] - E[B_t]^2 = t - 0$$ I'll interpret your second statement as saying that the outcome of $B_t$ tends to be on the order of $\sqrt{t}$. I can state ...

3

It is true that the second property can be deduced from the first one. Indeed, the first property implies that $(B_t)$ has stationary, independent increments. Hence, the following statement would give you the implication in the forward direction: Proposition: Any real-valued stochastic process with stationary, independent increments has the elementary ...

3

You have correctly found $x_t$ as $$x_t=x_0e^{-\theta t}+\mu (1-e^{-\theta t})+\sigma e^{-\theta t}\int_0^te^{\theta s}dW_s$$ We can rewrite this as $$x_t=a_t-b_tc_t$$ where \begin{align} &a_t=x_0e^{-\theta t}+\mu (1-e^{-\theta t})\\\\ &b_t=-\sigma e^{-\theta t}\\\\ &c_t=\int_0^te^{\theta s}dW_s \end{align} Note, ...

3

EDITED to meet edit of question The first equation is (after the edit) true. Consider the twodimensional continuous semimartingale $\left( t,B_t\right)$, and function $f(x,y)=xy$ we get $$D_xf(x,y)=y\quad D_yf(x,y)=x\quad D_1D_1f=D_2D_2f=0\quad D_1D_2f=D_2D_1f=1$$ And therefore ITO's formula gives $$tB_t=0+\int_0^t s\; dB_s+\int_0^t B_s\; ds.$$ The 2nd ...

3

Hints: Since $(W_t)_{t \geq 0}$ is a Brownian motion, it has independent normally distributed increments; in particular, $W_3-W_2, W_2-W_1,W_1$ are independent random variables which are Gaussian with mean $0$ and variance $1$. This implies that $(W_1, W_2-W_1,W_3-W_2)$ is (jointly) Gaussian with mean $\mu=0$ and covariance matrix $$C = \begin{pmatrix} 1 ... 2 Your first solution is correct; you didn't apply Itô's formula correctly. Recall that for$$X_t = \sigma(t) \, dB_t + b(t) \, dt$$we have$$f(X_t)-f(X_0) = \int_0^t f'(X_s) \, dX_s + \frac{1}{2} \int_0^t f''(X_s) \sigma^2(s) \, ds.$$Applying this with \sigma(t) = \sigma X_t, b(t) = - X_t and f(x) = x^3 gives$$\begin{align*} Y_t - Y_0 &= 3 ...

1

I would suggest the following method: 1) First show that $T_{\alpha} \stackrel{d}{=} \frac{\alpha^2}{Z^2}$ where $Z = N(0,1)$ r.v. This will require using the reflection principle 2) Use part 1 to compute the density function from the hitting time distribution by differentiating with respect to time. 3) You can now easily compute the expectation once you ...

1

The increments are stationary. Since $B_t - B_s$ is the increment over the interval $[s, t]$, it is the same in distribution as the incremeent over the interval $[s-s, t-s] = [0,t-s]$. Hence, $$B_t-B_s \sim B_{t-s}-B_0.$$ But $B_0 = 0$ almost surely, so that: $$B_t-B_s \sim B_{t-s}.$$ Finally, $B_{t-s} \sim \mathcal{N} (0,t-s)$. We didn't use the ...

1

"Invariance" here means that the result does not depend on the law of $X_1$ provided that $X_1$ is centered and has a finite moment. That is, the convergence $$\left(n^{-1/2} \sum_{j=1}^{[nt]}X_j\right)_{t\in [0,1]} \to (W_t)_{t\in[0,1]}$$ holds if $X_1$ is centered and has a finite moment. We get a Wiener process as a limit even if the law of $X_1$ ...

1

As @muaddib pointed out, you have simply rewritten the definition of $\hat{W}_t$ - but this doesn't show that $(\hat{W}_t)_{t \geq 0}$ is a martingale. Hints: Show that $(X_t)_{t \geq 0}$ is a martingale with respect to its canonical filtration $$\mathcal{F}_t := \sigma(X_s; s \leq t) = \sigma(W_s; s \leq e^{\beta t}-1).$$ Conclude that $(\hat{W}_t)_{t ... 1 Yes indeed you have to find a,b such that $$cov(aW_1-W_2,W_3+bW_5)=0$$ Because you know that$(W_1,W_2,W_3,W_5)$follows a normal distribution. Hence$(aW_1-W_2,W_3+bW_5)$also follows a normal distribution. Now, we know that for normal distributed variables , independence holds if and only if the covariance between them is 0. 1 Set $$A := \{W_t < \sqrt{t} \, \text{for infinitely many t>0}\}.$$ Since, by the law of the iterated logarithm, $$\limsup_{t \to \infty} \frac{W_t}{\sqrt{t \log \log t}} = 1 \qquad \text{almost surely}$$ we have$A = \Omega \backslash N$for some null set$N$. So the trouble is basically: Is$N \in \mathcal{F}_{0+}$? Since$\mathcal{F}_{0+}$is (in ... 1 You have to start proving the claim on the rationals$s=j/n$and$t=k/n$. We we can so rewrite you expression as : $$E[B_t-B_s|B_1-B_s]=E\left[\sum_{i=j}^{k-1}B_{(i+1)/n}-B_{i/n}{\large\mid}\sum_{i=j}^{n-1}B_{(i+1)/n}-B_{i/n}\right].$$ Noting$Y_i=B_{(i+1)/n}-B_{i/n}$, and$Y=\sum\limits_{i=j}^{n-1}B_{(i+1)/n}-B_{i/n}=B_1-B_s$, the above expression ... 1 As$\mathcal{F}_{t_n} \supseteq \mathcal{F}_{t+}$, we know that$X_n$is independent of$\mathcal{F}_{t+}$. Therefore, $$(B_{t+s_1}-B_t, \ldots,B_{t+s_m}-B_{t}) = \lim_{n \to \infty} (B_{t_n+s}-B_{t_n},\ldots,B_{t_n+s_m}-B_{t_n}) = \lim_{n \to \infty} X_n$$ is also independent of$\mathcal{F}_{t+}$. Lemma: Let$\mathcal{F}$be a$\sigma$-algebra and ... 1 If$X_n\to X$almost surely and$X_n$is independent of$\mathcal G$, then$X$is independent of$\mathcal G$. We use actually this. Call$X$the left hand side of (2). We can show that for each continuous and bounded function$f\colon\mathbf R^n\to\mathbf R$and$E\in\mathcal F_t$, we have$\mathbb E\left[f(X)\mathbf 1(E)\right]=\mathbb E[f(X)]\mu(E)$... 1 For$m=2$, write $$\left(X_{t_1}^n,X_{t_2}^n -X_{t_2}^n\right)=\left(\frac 1{\sqrt n}\sum_{i=1}^{[nt_1]}Y_i, \frac 1{\sqrt n}\sum_{i=[nt_1]+1}^{[nt_2]}Y_i \right) +\left(\frac{nt_1-[nt_1]}{\sqrt n}Y_{[nt_1]+1} , \frac{nt_2-[nt_2]}{\sqrt n}Y_{[nt_2]+1}-\frac{nt_1-[nt_1]}{\sqrt n}Y_{[nt_1]+1} \right);$$ the second vector converges to$0$in probability. For ... 1 By the definition of the Itô integral, we know that $$\sum_{j=1}^n V_j \cdot \Delta_j \to \int_0^t B_s \, dB_s. \tag{1}$$ Note that we can write $$I_1(n) = \sum_{j=1}^n V_{j+1} \Delta_j = \sum_{j=1}^n \underbrace{(V_{j+1}-V_j)}_{\Delta_j} \Delta_j + \sum_{j=1}^n V_j \Delta_j. \tag{2}$$ By$(1), the second term at the right-hand side converges to ... 1 Yes, it is a consequence of Fubini's theorem. By Fubini, we have $$\text{var}(I(T)) = \mathbb{E} \left( \int_0^T W(u) \, du \int_0^T W_v \, dv \right) = \int_0^T \int_0^T \underbrace{\mathbb{E}(W_u W_v)}_{\text{cov}(W_u,W_v)} \, du \, dv.$$ Using again Fubini's theorem, we find \begin{align*} \int_0^T \int_0^T \text{cov}(W_u,W_v) \, du \, dv &= ... 1 Hint: Apply Tonelli's (or Fubini's) theorem and use that\mathbb{E}\exp(W_u) = \exp \left( \frac{1}{2} u \right)$$since W_u \sim N(0,u). 1 To show independence you just need to show that \Bbb{P} (B_t- B_s \in A_0, B_{r_1} \in A_1,B_{r_2} \in A_2\ldots B_{r_k} \in A_k) = \Bbb{P} (B_t- B_s \in A_0)\Bbb{P} ( B_{r_1} \in A_1,B_{r_2} \in A_2\ldots B_{r_k} \in A_k) for every r_1, \ldots r_k \leq s and A_1, \ldots, A_k \in \mathcal{B}. This is because \sigma(\{B_r, r \leq s\}) is generated ... 1 First of all, note that we have to ensure that$$\exp \left( \int_0^t X_s \, dB_s \right) \in L^1. \tag{1}$$If your claim is true, then$$\mathbb{E} \exp \left( \int_0^t X_s \, dB_s \right) = \mathbb{E}\exp \left( \frac{1}{2} \int_0^t X_s^2 \, ds \right),$$i.e. (1) holds if$$\mathbb{E}\exp \left( \frac{1}{2} \int_0^t X_s^2 \, ds \right)< \infty. ... 1 Doob says:B_{\tau \wedge t} = E[B_{t \wedge t}\mid \mathcal{F}_\tau]$. Jensen says: $$E[\lvert B_{\tau \wedge t} \rvert] = E[\lvert E[B_{t \wedge t}\mid \mathcal{F}_\tau]\rvert] \leq E[ E[\lvert B_{t \wedge t}\rvert\mid \mathcal{F}_\tau]] = E[\lvert B_t \rvert] < \infty$$ 1 Assuming$t > t_n$you can reduced your relationship to $$p\{-x\leq B_t\leq x \mid B_{t_n}=\pm x_n\}=p\{-x\leq B_t\leq x\mid B_{t_n}=x_n\}$$ because$B_t$is Markov. I'll proceed formally: \begin{eqnarray*} p\{-x\leq B_t\leq x \mid B_{t_n}=\pm x_n\} &=& \frac{p\{-x\leq B_t\leq x \text{ and } B_{t_n}=\pm x_n\}}{p\{B_{t_n}=\pm x_n\}} \\ ... 1 To keep notation simple, we write$S_t$instead of$S_t^i$,$\sigma_t^j$instead of$\sigma_{t}^{ij}$and so on. That's okay, because$i$is a fixed number throughout this calculation. Suppose$(X_t)_{t \geq 0}$is an Itô process of the form $$dX_t = b(t) \, dt + \eta(t) \, dW_t$$ where$\eta = (\eta_1,\ldots,\eta_m)$and$(W_t)_{t \geq 0}\$ is an ...

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