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6

Let $(B_t)_{t \geq 0}$ be a Brownian motion on a probability space $(\Omega_1,F_1,P_1)$ and let $(\Omega_2,F_2,P_2)$ be an arbitrary probability space. Define a new probability space $(\Omega,F,P)$ by $$\Omega := \Omega_1 \times \Omega_2 \qquad F := F_1 \otimes F_2 \qquad P := P_1 \otimes P_2.$$ If we set $$\tilde{B}_t(\omega_1,\omega_2) := B_t(\omega_1) ...


4

The rules $$(dt)^2 = 0 \qquad dW_t \, dt = 0 \qquad (dW_t)^2 = dt \tag{1}$$ are heuristic rules to simplify calculations when applying Itô's formula. Mind that this is the only application; do not use them anywhere else. In Itô's formula expressions of the form $$\int_0^T f(X_t,Y_t) dX_t \, dY_t$$ pop up. Using $(1)$, we get $$dX_t \, dY_t = (\mu_t dt ...


3

If $M = \int h dX$ where $X$ is a continuous square integrable martingale, you don't need $E\int_0^\infty h_s^2d\langle X\rangle_s < \infty$ for $M$ to be a martingale, you just need $E\int_0^t h_s^2d\langle X\rangle_s < \infty$ for all $t$. You did show this, and hence your $M$ is a martingale and you can say $E M_t = E M_0$ to solve your problem.


3

Hints: Fix $x \in [a,b]$. Instead of $\mathbb{E}_{X_0=x}$ I will use the notation $\mathbb{E}^x$. Suppose that $f \in C_b^2$ solves the differential equation $$x f'(x) + \frac{\sigma^2}{2} f''(x)=0. \tag{1}$$ Show, using Itô's formula, that $(f(X_t))_{t \geq 0}$ is a martingale. Define $$f(x) := \Phi \left( \frac{2}{\sigma} x \right)$$ where $$\Phi(x) := ...


3

The sign in the first part of your solution is wrong. Note that you consider the case $\frac{k}{n} < t$, so the expectation is equal to $t - \frac{k}{n}$.


3

No, it is not correct; the identity $$\mathbb{E}(W(k/n) \cdot W(t))=0$$ (which you used in your calculation) does not hold true. Hint: A Brownian motion has stationary increments, i.e. $W_t-W_{k/n}\stackrel{d}{=} W_{t-k/n}$. So, $$\mathbb{E} \left( \left[ W \left( \frac{k}{n} \right)-W(t) \right]^2 \right) = \mathbb{E}\left( \left[ W \left( t-\frac{k}{n} ...


2

Consider the process $$X_s = W^{(1)}_s-W^{(2)}_s$$ At any given time, this is the difference of two mean zero, independent, normals with variance $s$. That means $X_s$ is a mean zero normal with variance $2s$. So the question is, what is the expected value of the absolute value of a normal random variable. This is called a Folded Normal Distribution. One ...


2

Actually, Durrett takes the expectation in the previous identity: $$Y_S(\theta_S \omega) = \begin{cases} 1, & \text{if} \, S(\omega)<t, B_t(\omega)>a \\ 0, & \text{otherwise} \end{cases}. \tag{1}$$ Indeed: By $(1)$, we have $$Y_S(\theta_S \omega)= 1_{\{S<t,B_t>a\}}(\omega). $$ Since $T_a(\omega) = S(\omega)$ for all $\omega \in ...


2

The distribution for the $$Y = \int_0^1 |W_s|ds$$ is worked on in On the Distribution of the Integral of the Absolute Value of the Brownian Motion Takacs (1993). It isn't very pretty: $$P(Y \leq x) = H(x)$$ where $$\frac{dH(x)}{dx} = \frac{\sqrt{3}}{x \sqrt{\pi}} \sum_{j=1}^\infty C_j e^{-v_j}v_j^{2/3} U(1/6, 4/3, v_j)$$ where $$C_j = \frac{1 + ...


2

Define $\tilde\tau_x=\inf\{t\ge 0:\tilde B(t)=x\}$ where $\tilde B(0)=0$. Using the translation invariance, symmetry, and the reflection principle for BM $$P\{\tau_0>T\}=P\{\tilde\tau_1>T\}=\frac{2}{\sqrt{\pi}}\int_{0}^{\frac{1}{\sqrt{2T}}} e^{-u^2}du$$ ...


2

I will follow the outline that I described in my first comment. Let's start with random walk. Fix $\alpha \in \mathbf{R}$ and consider the process $(M_j)$ defined as $$M_j = (\cosh{\alpha})^{-j}\cosh{(\alpha S_j)}$$ We want to show that $(M_j)$ is a martingale with respect to the natural filtration $(\mathcal{F}_j)$ generated by the constituent random ...


2

Your attempt: How do you conclude that $$\mathbb{E} \left( \int_0^t \frac{1}{1-s} \, dW_s \right) = \mathbb{E} \left( \frac{W_t}{1-t} \right)$$ ...? To me it looks as you used something of the form $$\int_0^t f(s) \, dW_s = f(t) W_t,$$ but this identity is, in general, not correct. Your professor's answer: Your professor uses the following well-known ...


1

To see what's happening, set $X = \frac{W(b) + W(a)}{2}$ and $Y = W(k) -X$. I agree that your computation shows $Y \sim \frac{1}{2} N(0,b-a)$, regardless of the value of $k$. And the distribution of $X$ certainly doesn't depend on $k$. But this does not imply that the distribution of $W(k) = X+Y$ is the same for all $k$, because the covariance of $X$ and ...


1

The question doesn't ask you to compute $E\int_0^\infty X_n^2(t)dt$, it asks you to show that it is finite. Do you need to know $\sum_{k=0}^{n-1}\frac{k}{n^2} = \frac{n-1}{2n}$ in order to realize this is a finite number? Nope!, finite sum of finite numbers is finite. Also you should mention something about why $X_n$ is adapted (trivial but since the ...


1

You have $$ (1)=\mathbb E[1_{\{\tau_a\leq t\}}1_{\{X_{t-\tau_a}+W_{\tau_a}\leq a\}}] = \mathbb E[\mathbb E[1_{\{\tau_a\leq t\}}1_{\{X_{t-\tau_a}+W_{\tau_a}\leq a\}}|F_{\tau_a}]] = \mathbb E[1_{\{\tau_a\leq t\}}\Pr(X_{t-\tau_a}\leq 0]|F_{\tau_a})],$$ where you use that $W_{\tau_a}=a$ and measurability of $\tau_a$ wrt. $F_{\tau_a}$. Moreover, ...


1

You write $E^Q[e^{-\lambda \tau}] = e^{\sqrt{2\lambda} m}$. What measure $Q$ is this? Under $Q_t$ we know $X$ is a brownian motion on $[0,t]$, but your stopping time's moment generating function was for a brownian motion on $[0,\infty)$. It is actually possible to get a measure $Q$ where $X$ is a brownian motion on $[0,\infty)$ assuming the underlying space ...


1

$$ dS_t = \Big(\mu + \frac{1}{t}\Big) S_t\,dt + \sigma dW_t $$ for example. Of course this comes with a restriction to $t > 0$.


1

To add onto nullUser's response. Of course $\int_0^t 1 dW_s=W_t$ is a martingale right? Well: $$ \int_0^{\infty} 1 ds=\infty$$ This is just to give a solid counterexample, the rest was explained in his post.


1

The reason continuity is relevant in showing $\{ \tau \leq t\} = \{M_t \geq b\}$ is because if $B_t$ were not continuous, then it could be that $B_t$ jumps over $b$ without hitting it, in which case $\tau$ might not occur even if $M_t > b$. As for $\{\tau \leq t\}$, this set is actually measurable with respect to $\mathcal{F}_{\tau \wedge t} = ...


1

Suppose $a< B(s) <b$ for all $s \in [0,1]$. Take $s=1$, so $a<B(1) < b$. The first requirement is more restrictive, and hence gives a smaller set.



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