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Take $(X_n)_{n\geqslant 1}$ an i.i.d. sequence where $X_1$ is a non-degenerated zero-mean random variable and $\mathcal F_n :=\sigma(X_1,\dots,X_n)$ for $n\geqslant 1$. Then $\mathbb E[X_n\mid\mathcal F_{n-1}]=0$ for each $n$.
$X_T$ is the random variable with value $X_{T(\omega)}(\omega)$ at each point $\omega\in\Omega$ where $T(\omega)<\infty$. For $\omega\in\{T=\infty\}$ one can take $X_T(\omega)$ to equal $\lim\limits_{n\to\infty}X_n(\omega)$ for those $\omega$ for which the limit exists. $|X_{T\wedge n}|=\sum\limits_{k=0}^{n-1}1_{\{T=k\}}|X_k|+1_{T\ge n}|X_n|\le ... 2 Let's assume for simplicity that the probability$p$of winning a given spin is $$p=0.5$$ and then the probability of losing$q$on a given hand is $$q=1-p=0.5$$ With the martingale betting system, one win will yield a net profit because you always bet more than you've lost. The probability you will not win at least once after$n$spins is $$q^{n}=0.5^n$$ ... 2 There are (at least) two possibilites to prove this. The first one, the quick one, uses Itô's formula. A straightforward application of Itô's formula shows that $$M_t^u := u(t,W_t)-u(0,x) - \int_0^t \left( \frac{\partial}{\partial t} + \frac{1}{2} \Delta_x \right) u(s,W_s) \, ds$$ is a stochastic integral (with respect to Brownian motion) and, moreover, ... 2 Guess:$\xi$is independent of$(W_t)_{t\ge 0}$, and$\tilde{\mathscr F}_t$should be$\sigma(\xi)\vee\mathscr F_t$2 Hint: The expected value of the contract at time$N$will be$47$because of the value sequence being a martingale. But also we know that the actual value at time$N$will either be$0$or$100$. There is only one way to assign the probabilities for$S(N)$being$0$or$100$so as to satisfy the expected value condition. 1 There are two way to think about this problem. First, let$N_t$be the number of jumps by the time$t$(clearly you only need to consider$N_t\ge n$) and let$\bar n=n-1/2$. Then you are interested in $$EV(n)=E(\sum_{i=1}^{N_t}(\frac 1 2 P(S_{i-1}=n-1)+\frac 1 2P(S_{i-1}=n)))=E(\sum_{i=n}^{N_t}\frac 1 2\binom {i-1}{\lfloor\frac {i-n}2\rfloor})$$ Now you can ... 1 Hint: Doob's Optional Stopping Theorem. Consider the stopping time$T:=\min\{n: X_n<0\}$. By Optional Stopping,$0\le\Bbb E[X_0]\le\Bbb E[X_{\min(T,n)}]$for$n=1,2,\ldots$. The latter expectation is equal to $$\Bbb E[X_T; T\le n]+\Bbb E[X_n; T>n].$$ The second of these expectations is at most$\Bbb E[M; T>n]$, where$M:=\sup_nX_n$. What would ... 1 Got a hint from http://wt.iam.uni-bonn.de/fileadmin/WT/Inhalt/people/Karl-Theodor_Sturm/Lectures/vorlesungWS0809/sheet1.pdf Consider the problem in the canonical space. Define$\alpha_t(\omega(\cdot)) = \omega(\cdot \wedge t)$Using the monotone class theorem, we can show that the mapping:$\alpha_t: (\Omega,{\mathcal F}_t^X) \to (\Omega,\mathcal{F})$... 1 Looks good, as indeed $$X_{n+1} = X_n + R_{n+1},$$ where$R_{i}$denotes the indicator variable that takes value$1$if color of the$i$-th ball extracted is red, and$0$if green. By definition we have that the urn contains$X_n$red and$n+2-X_n$green balls after$n$extractions. Then the conditional probability given$X_n$of a red ball on the$n+1$-th ... 1 Yes, it is correct. So the sequence$(S_n)$is never a martingale unless in degenerated cases. We could note that the conclusion would not be the same if we replaced$P(X_i = 0) = q$by$P(X_i = -1) = q$. 1 Hints: Since$S$is a bounded stopping time, there exists$N \in \mathbb{N}$such that$S(\omega) \leq N$for all$\omega \in \Omega$. Show that $$\mathbb{E}(|Y_S|) = \sum_{k=0}^n \mathbb{E}(|Y_k| 1_{\{S=k\}}).$$ Conclude $$\mathbb{E}(|Y_S|) \leq \sum_{k=0}^N \mathbb{E}(|Y_k|) \leq (N+1) \max_{k=0,\ldots,N} \mathbb{E}(|Y_k|)<\infty.$$ Note that this ... 1 Fix a positive integer$n$. By the martingale property and$L^1$convergence of$\{X_n\}$we have for$m\geqslant n$, $$\mathbb E[|X_n - \mathbb E\left[X\mid \mathcal F_n]|\right] = \mathbb E\left[| \mathbb E[X_m-X\mid \mathcal F_n]|\right]\leqslant \mathbb E\left[|X_m-X|\right]\stackrel{m\to\infty}\longrightarrow 0,$$ so that $$X_n=\mathbb E[X\mid \mathcal ... 1 Example: Toss two fair dice. Let Y_i be the number showing on die i, i=1,2. Let X be 1 or 0 depending on whether the sum of the two dice is even or odd. Then Y_1 is independent of X, Y_2 is independent of X, but X is not independent of the pair (Y_1,Y_2). For joint Gaussian: What has to be true about the covariance matrix of ... 1 I think you have nearly answered your own question. The finite nature of the inequality in (ii) follows from the finite property of the expectation of the supremum, i.e.$$ \int\limits_{ \{ 0\leq T \leq t \} } |X_T| \ dP \leq E\left[ \sup_{ 0\leq u \leq t } |X_u|\right] < \infty, $$and hence,$$ E\left[ |X_{T\wedge t}| \right] \leq E\left[ \sup_{ 0\leq ... 1 The process$X$in this case is a squared Bessel process of dimension two, which means that it is a (weak) solution to the SDE $$\text{d}X_t=2\,\text{d}t+2\sqrt{X_t}\,\text{d}B_t,$$ for all$t\geq 0$, where$B$is some Brownian motion. An application of Ito's formula now yields $$\text{d}\ln X_t=\frac{2}{\sqrt{X_t}}\,\text{d}B_t,$$ for all$t\geq 0$. In ... 1 Hint: Choose the process$C$so that$(C\cdot X)_n=X_{T\wedge n}-X_0$. Intuition: To obtain the stopped-at-$T$process, you want to keep betting so long as time$T$hasn't yet passed. 1 There is the following general result: Let$f: [0,\infty) \to \mathbb{R}$be a continuous function and$g: [0,\infty) \to \mathbb{R}$be a function of bounded variation (on compact intervals). Then $$\langle f, g \rangle_t = 0$$ for all$t \geq 0$, i.e.$$\langle f,g \rangle_t = \lim_{|\Pi| \to 0} \sum_{t_j \in \Pi} (f(t_{j+1})-f(t_j)) ... 1 Is$F_n = \sigma(X_1, X_2, ..., X_n)$? I think you should be more explicit: Let$X_1, X_2, ...$be a iid random variables in a filtered probability space$(\Omega, \mathscr{F}, \{\mathscr{F}_n\}_{n \in \mathbb{N}}, \mathbb{P})$, where$\mathscr{F}_n = \mathscr{F}_n^X = \sigma(X_1, X_2, ..., X_n)$and$X_i$~$N(\mu, \sigma^2)$. We want to find$\lambda\$ ...