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4

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 ...


3

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 $0$, ...


3

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 ...


3

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} ...


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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} ...


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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 ...


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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 ...


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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 $P(T=\infty)>0$


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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 $t\mapsto E(N_t)$?


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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.


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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| = ...


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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 ...


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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 ...


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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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