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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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