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Solution 1: Recall the following two statements. Lemma 1: Let $(Y_t)_{t \geq 0}$ be a supermartingale and $f$ an increasing concave function, then $(f(Y_t))_{t \geq 0}$ is a supermartingale. Lemma 2: Let $(Y_t)_{t \geq 0}$ be a locale martingale such that $Y_t \geq 0$ for all $t \geq 0$. Then $(Y_t)_{t \geq 0}$ is a supermartingale. Lemma 1 is a ...

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This is not a complete answer, just some ideas too long to fit into a comment. Since only cosines are involved, you can understand $\sum_{i=1}^n X_i$ as a random walk on a circle. Then it is not hard to see that your random walk visits any neighborhood of $0$ infinitely often. So the $\limsup$ of numerator is $1$, while the denominator converges to $0$, ...

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Assume without loss of generality that $r \leq s \leq t$. By conditioning with respect to $\mathcal{F}_s$, we find using the martingale property $$\mathbb{E}(M_r M_s M_t) = \mathbb{E}(M_r M_s \mathbb{E}(M_t \mid \mathcal{F}_s)) = \mathbb{E}(M_r M_s^2).$$ If we denote by $\langle M \rangle_t$ the quadratic variation (i.e. the unique increasing process ...

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I think everything you're doing is correct, but perhaps it would benefit from exploiting some cancellations so things work out more smoothly. In your second line, with six terms, I claim if you take the first, fourth, fifth, and sixth terms and sum them, you get $$\boxed{ \lambda^2 s^2 -\lambda s -2\lambda t X_s}.$$ I used this grouping because lots of ...

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You have to rewrite $X_T$ in a clever way before doing the computations. Hint: Show that $$X_T = X_0 + \sum_{j=1}^N (X_j-X_{j-1}) 1_{\{T \geq j\}}.$$ Hence, $$\mathbb{E}(X_T)= \mathbb{E}(X_0) + \sum_{j=1}^N \mathbb{E}( (X_j-X_{j-1}) 1_{\{T \geq j\}}).$$ Use the tower property to prove that $$\mathbb{E}(X_T) = \mathbb{E}(X_0) + \sum_{j=1}^N \mathbb{E} ... 2 First, we can show the function f:\mathbb{R}\to \mathbb{R} given by f(x)=\sqrt{1+x^2} is such that: \forall x,f(x)\geq 1. Also f(x)=1\iff x=0. x< 0 \implies 0<x+f(x)<1. Similarly x>0\implies -1<x-f(x)<0 Second, to prove$$ 1_{\{|S_n|<1\}}=\frac{-X_nsign(S_{n-1})+1}{2} $$note this: \begin{eqnarray} ... 2 For any r\ge 0$$ \mathbb{E}(M_{\infty}-M_n)^2=\mathbb{E}(M_{\infty}-M_{n+r})^2+\mathbb{E}(M_{n+r}-M_n)^2 \\\overset{(d)}=\mathbb{E}(M_{\infty}-M_{n+r})^2+\sum_{k=n+1}^{n+r}\mathbb{E}(M_k-M_{k-1})^2. $$Taking r\to\infty, the first term converges to 0 by (f). 2 The uniform integrability of \{X_n\} implies that \mu is absolutely continuous with respect to \Bbb P on \mathscr A:=\cup_n\mathscr F_n, in the sense that for a given \epsilon>0 there exists \delta>0 such that if B\in\mathscr A and \Bbb P(B)<\delta, then |\mu(B)|<\epsilon. Now let \{A_k\} and A be as you describe. Fix ... 2 Note that$$ \left( \frac{N_t}{t}-\lambda \right)^2= \frac{1}{t^2} (N_t-t\lambda)^2 \leq \frac{1}{\sigma^2} (N_t-t\lambda)^2$$for all t \in [\sigma,\tau]. Hence,$$\mathbb{E} \left[ \sup_{t \in [\sigma,\tau]} \left( \frac{N_t}{t}-\lambda \right)^2 \right] \leq \frac{1}{\sigma^2} \mathbb{E} \left[ \sup_{t \in [\sigma,\tau]} (N_t-t\lambda)^2 \right].$$... 1 If random variable Y takes values in \mathbb R, you can find a measurable function f: \mathbb R\to \mathbb R, such that f(N) has the same distribution as Y, where N is a standard Gaussian random variable. Now, for a Brownian motion B_t, define$$ M_t = E(f(B_1)|\mathcal F^B_t)$$1 The proof is correct, but notice that you get something stronger than what you state in the last line of your question, indeed X_{\infty} is finite almost surely, which is the same as saying -\infty < X_{\infty} < \infty (and not only X_{\infty} < \infty). To go from the fact that E(|X_{\infty}|) < \infty to the fact that X_{\infty} \in ... 1 You've shown that there is a c such that P(T=\infty)>0. So:$$E[A_{T\wedge n}]=E[A_{T\wedge n}|T<\infty]P(T<\infty)+A_n P(T=\infty).$$Both terms terms on the r.h.s are non-negative and you've also shown that E[A_{T\wedge n}] is finite and you can take n\rightarrow\infty, so at the very least A_\infty<\infty since P(T=\infty)>0 1 Because N is a modification of M, E(M_t)=E(N_t) for all t\ge 0. What do you know about the right continuity of t\mapsto E(N_t)? 1 The compensator of Y is the (\mathcal F_s^Y)-adapted process t\mapsto\lambda\min(t,\tau), which has expected value 1-e^{-\lambda t}. That is, M_t:=1_{\{\tau\le t\}}-\lambda\min(t,\tau) is a martingale. Added detail: The idea is that the process to subtract from Y to make it a martingale (the compensator of Y) should be the integral from 0 to ... 1 The partial sums of \sum X_k are bounded by constant c iff |M_r|\le c for every r. This last occurs iff T_c=\infty (where I write T_c instead of T to make the dependence on c explicit). Therefore$$ \{\text{partial sums of $\sum X_k$ are bounded}\} = \bigcup_{c=1}^\infty \{T_c=\infty\}.\tag1 $$So if the LHS has positive probability then ... 1 Let's work it out (also let's finish the calculation):$$ \begin{aligned} E[B_s(B_t^2-t)]&=E[E[B_s (B_t^2-t) \mid \mathcal{F}_s]] \\ &=E[B_s E[(B_t^2-t) \mid \mathcal{F}_s]] \\ &=E[B_s (B_s^2-s)] \\ &=E[B_s^3]-E[sB_s] \\ &=0. \end{aligned} $$The first equality is the tower property. The second equality is "factoring out what is known". ... 1 The function f(x)=\min\{x,7\} is concave, hence by Jensen's inequality for conditional expectation$$ \mathbb{E}[Y_{n+1}|\mathcal{F}_n]=\mathbb{E}[f(X_{n+1})|\mathcal{F}_n]\leq f(\mathbb{E}[X_{n+1}|\mathcal{F}_n])=f(X_n)=Y_n$$1 The function u(t,x)=e^{t/2}\,\cos(x) satisfies the heat equation {d\over dt}u+{1\over 2}\Delta_x u=0, so that, by Ito's formula, u(t,W_t)=e^{t/2}\,\cos(W_t) is a martingale. 1 Indicating the dependence of T on c explicitly, you have$$\{T_c=\infty\}=\{\sup_n|M_n|\le c\}$$Therefore,$$\bigcup_{c>0,c\in\Bbb Q}\{T_c=\infty\}=\{\sup_n|M_n|<\infty\}$$1 The position of the ship after n space-hops is{}^* X_{n} = X_{n-1} + R_{n-1} U_n, where the vector V_n is independent of X_1,\dots,X_{n-1} and uniformly distributed on the unit sphere S_1. So R_n = R_{n-1}|e_{n-1}+V_n|, where e_{n-1} is a unit vector in the direction of X_{n-1}. Then, clearly, R_n= R_{n-2}|e+U_{n-1}||e+U_n| = ... 1 You should start as low as possible to reduce volatility given your situation. Then you can gradually move up towards the Kelly strategy:$$f^{*} = \frac{bp - q}{b}$$where f^* is how much to bet as a fraction of your bankroll b is the net odds, ("b to 1"). For a \1 bet how much you get back on top of the \1 you bet. p is the ... 1 OK, I think found (most of) the answer to this. According to Kallenberg (p. 451, Theorem 23.20), by the Rao Decomposition Theorem, is (up to a constant) a local quasimartingale if and only if it is a special semimartingale. (More information about quasimartingales and the Rao decomposition theorem: http://www.mscand.dk/article/viewFile/10921/8942 ... 1 I solved it myself. Actually \tau^{-1}\circ g^{-1}_k is measurable. Notice that obviously$$\{\omega|\tau< k-1\}\in \mathcal{F}_{k-2}\subseteq \mathcal{F}_{k-1}$$and since \mathcal{F}_{k-1} is a \sigma-field we get:$$\{\omega|\tau \geq k\}\in \mathcal{F}_{k-1}.

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