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3

If you have infinite money and decide to stop doubling your bet and reset after each time you win, then your expected total profits will increase without bound with probability 1, in the limit as you keep playing more and more rounds. However, if you start with a FINITE amount of money, then the chances of losing everything eventually and going broke make it ...


2

For fixed $k \in \mathbb{N}$ define a bounded stopping time $\tau_k$ by $$\tau_k := \inf\{t \geq 0; M_t > x \} \wedge k = \tau \wedge k.$$ Then, by the optional stopping theorem, we have $$\mathbb{E}(M_{\tau_k} \mid \mathcal{F}_0) = M_0. \tag{1}$$ On the other hand, since $(M_t)_{t \geq 0}$ is a continuous martingale, we know that $$M_{\tau_k} = ...


2

Obviously, $$\ln X = \int_0^T \sigma(s) \, dW_s \qquad \text{for} \quad \sigma(s) := 1_{[0,T/2]}(s) + 1_{[0,T]}(s). \tag{1}$$ Define an Itô process $(Y_t)_{t \geq 0}$ by $$Y_t := \int_0^t \sigma(s) \, dW_s - \frac{1}{2} \int_0^t \sigma(s)^2 \, ds. \tag{2}$$ Then, by Itô's formula, $$\begin{align*} \exp(Y_t)-1 &= \int_0^t \exp(Y_s) \, dY_s + ...


2

Hint: Let $X$ be a (non-trivial) random variable such that $X$ is independent from $\mathcal{F}^W_{\infty} := \sigma(\mathcal{F}_t^W; t \geq 0)$. Consider the filtration defined by $$\mathcal{F}_t := \sigma(\mathcal{F}_t^W, X).$$


2

Hints: Since $(X_t)_{t \geq 0}$ is a martingale and $X_0=0$, we have $\mathbb{E}(X_t)=0$ for all $t \geq 0$. If a random variable $Y$ has finite second moment, then $$\mathbb{E}(Y) = \frac{1}{\imath} \frac{d}{d\xi} \chi(\xi) \bigg|_{\xi=0} \qquad \quad \mathbb{E}(Y^2) = - \frac{d^2}{d\xi^2} \chi(\xi) \bigg|_{\xi=0}$$ where $\chi$ denotes the ...


2

You example is correct. For brevity, you could have used Doob's Martingale convergence theorem. $$ E[|M_N|] \leq \sum_{i=1}^N E[|Y_i|] = \sum_{i=1}^N \frac{1}{i^2} < \frac{\pi^2}{6} $$ therefore $M_n \rightarrow M_\infty\, a.s.$ and $M_\infty \in L^1$. It does not imply convergence in $L^1$. For that you need stronger conditions such as uniform ...


2

Denote by $$p(t,x) := \frac{1}{\sqrt{2\pi t}} \exp \left(- \frac{x^2}{2t} \right), \qquad x \in \mathbb{R},$$ the density of the normal distribution with mean $0$ and variance $t$. As you already noted, this function solves the heat kernel equation, i.e. $$\frac{\partial}{\partial t} p(t,x) = \frac{1}{2} \frac{\partial^2}{\partial x^2} p(t,x).$$ For $f ...


2

Should the relationship in question be $$ \mathbb{E}[ M_T \mathbf{1}_{T>S} | \color{red}{\mathcal{F}_S} ] \overset{?}{=} \mathbb{E}[ M_T \mathbf{1}_{T>S} | \mathcal{F}_{T\color{red}{\wedge S}} ]\,, $$ in line with the primary problem of determining the general form of $\mathbb E[M_{T}|\mathcal F_{S}]$? If this is the case, notice that, on the event ...


1

first you can get $b$ simply be computing the time zero expectation: $$\mathbb{E}(e^{5B_t})= e^{0.5 \times 25 \times t}.$$ So $b = 12.5.$ With this value, your final equation $$\exp\{\frac{25(t-s)}{2}+5B_s-bt\} = \exp\{5B_s-bs\}$$ holds and we are done.


1

For any ($L^2$-)martingale $(M_t)_{t \geq 0}$ the quadratic variation $(\langle M \rangle_t)_{t \geq 0}$ is defined such that $$M_t^2 - \langle M \rangle_t$$ is a martingale. Apply this to $M_t = W_t^2-t$. (There is no need for Itô's formula!)


1

Let $\Pi$ be a partition of the interval $[0,t]$, i.e. $\Pi : 0 = t_0 <t_1 <\ldots<t_m = t$. Note that as $\text{mesh}(\Pi) \rightarrow 0$ $$S(\Pi) := \sum_{i=0}^{m-1}\lambda(t_i)(W_{t_{i+1}}-W_{t_i}) \rightarrow \int_0^t\lambda(s)dW_s$$ This convergence is in $L^2$. Next, you can see that $S(\Pi)$ is a linear combination of the increments of ...


1

Step 1: The assumption holds for all $f \in C_b^2$, i.e. functions which are twice differentiable and bounded derivatives. To this end, pick a function $\chi \in C^2_b$ such that $0 \leq \chi \leq 1$, $\chi(x) = 1$ for $x \in (-1,1)$ and $\chi(x)=0$ for all $|x|>2$. Then the function $$f_n(x) := f(x) \cdot \chi \left( \frac{x}{n} \right), \qquad x ...


1

\begin{align} E[B_t^3|\mathcal{F}_s]&=E[(B_t-B_s+B_s)^3|\mathcal{F}_s]\\ &=E[(B_t-B_s)^3+3(B_t-B_s)^2B_s+3(B_t-B_s)B_s^2+B_s^3|\mathcal{F}_s]\\ &=E[3(B_t-B_s)^2B_s|\mathcal{F}_s]+B_s^3\neq B_s^3\\ E[3tB_t|\mathcal{F}_s]&=E[3t(B_t-B_s+B_s)|\mathcal{F}_s]\\ &=E[3t(B_t-B_s)|\mathcal{F}_s]+E[3tB_s|\mathcal{F}_s]\\ &=3tB_s \neq 3sB_s ...


1

@Did's solution is the slick approach to proving $X_\infty$ equals $0$ or $1$ when $p<1$. But your approach is also valid, after some modification. There are some errors in your solution: The limit point $c$ depends on $\omega$. Given that $P(a < X_\infty < b) > 0$, it does not follow that there exists $c\in(a,b)$ such that $P(X_n\to c) > 0$ ...


1

Hi the Optimal Sampling Theorem works fine here because you don't need bounded stopping times if you know that the process $X$ is uniformly integrable. Indeed in that case Optimal Sampling Theorem works for every stopping times (even unbounded ones) and your argument is simplified. As you may know a (right-)continuous martingale $(X_t)$ is uniformly ...


1

The random variable $H_1$ has to be time $0$ measureable which makes it a constant.


1

The argument is correct. Maybe in the third line of the computation of $\mathbf{Var}\bigl[X_n - X_0\bigr]$, I would refer (and give a name to) the displayed equation where $\mathbf{E}\bigl[X_n X_0\bigr]$ is computed.


1

For $i\leqslant t-1$, we have $\mathbb E\left[X_i\mid \mathcal F_{t-1}\right]=X_i$, hence the sequence $(Z'_t)_{t\geqslant 1}$, where $Z'_t=\sum_{i=1}^t\alpha^{-i}X_i$ is a martingale with respect to the filtration $(\mathcal F_t)_{t\geqslant 1}$ as long as $\mathbb E\left[X_t\mid\mathcal F_{t-1}\right]=0$ (which is seemed to be assumed in the first ...


1

Using your argumentation, we get an exceptional null set $N = N((s_n),(t_n))$ such that $$X_{\infty}^s(\omega) = X_{\infty}^t(\omega) \qquad \text{for all} \, \omega \in \Omega \backslash N$$ for any two sequences $(s_n)_n$ and $(t_n)_n$. The problem is the following: There exist uncountably many sequences $(s_n)$ and $(t_n)$ such that $s_n \to \infty$ and ...


1

Question 1: Note that $[M]^{\tau_k}_{t \geq 0}$ is an increasing process. Therefore, $$|[M]^{\tau_k}_b - [M]^{\tau_k}_{b-\frac{1}{n}}| \leq 2 [M]_b.$$ Since $$\mathbb{E}([M]_b^{\tau_k}) = \mathbb{E}((M_b^{\tau_k})^2)<\infty,$$ we can apply the dominated convergence theorem to conclude $$\mathbb{E}([M]_b^{\tau_k}-[M]_{b-1/n}^{\tau_k}) \to 0 \qquad ...


1

What follows assumes that processes involved are well-defined for our immediate purposes so that, for instance, $(W_s)_{s\geqslant 0}$ is a regular Brownian-motion adapted to some underlying filtration, $(\lambda_s)_{s\geqslant 0}$ is previsible and adapted to the same filtration, and $\int_{0}^{t}\lambda_s^2\text ds<\infty$. We move by a sequence of ...



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