Tag Info

Hot answers tagged

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 ...



Only top voted, non community-wiki answers of a minimum length are eligible