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Let $\{t_n\}_{n=1}^\infty$ be a sequence of rational numbers in $(t, \infty)$, monotonically decreasing to $t\ge 0$ as $n\to\infty.$ We want to show that $\lim_{n\to\infty} \mathbb{E} [X_{t_n}]=\mathbb{E} [X_t].$ To this end, we use the theory of backward martingales: Definition: Let $\{\mathcal{F}_n\}_{n=1}^\infty$ be a decreasing sequence of sub-$\sigma$-...


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As @charlus said, for every $t \in \mathbb R^+$, $Z(t)$ is a random variable. So when we take the expectation $\mathbb E[Z(t)]$, we integrate with respect to $\omega$ and not $t$. In other words, $$\mathbb E[Z(t)] := \int_{\omega \in \Omega} Z(t,\omega) d\mathbb P(\omega) \quad \forall t\in\mathbb R^+ $$ So there isn't any notion of "taking expectation ...


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Martinagle property of $(X_t)$ is equivalent to the condition $E(X_t-X_s)|\mathcal F_s)=0$ for $s <t$. By definition of conditional expectation this is equivalent to the condition $\int_A (X_t-X_s)dP=0$ for all $A \in \mathcal F_s$. By the $\pi-\lambda$ theorem it is enough to verfy this for some class of sets which is closed under finite intersection and ...


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An observation that might prove fruitful for you, too much for a comment, but likely not the answer you're looking for. Anyway ... It seems your interest is in the long-time behavior of $p_t$. When your process starts, there is considerable flexibility in how $p_t$ evolves. However, that changes the first time you hit the "set $p_t = 1$" situation. ...


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$(\lim \sup \frac {S_n} {C_n}>a)= (\lim \sup \frac {X_k+X_{k+1}+...+X_n} {C_n}>a)$ for any $k$ since $\frac {X_1+X_2+..+X_{k-1}} {C_n} \to 0$. From this it is clear that $(\lim \sup \frac {S_n} {C_n}>a)$ belongs to $\sigma (X_k,X_{k+1},...)$ for each $k$ and hence it belongs to $\mathcal T$.


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Maybe an example will help. Let's take the sample space to be the unit interval $[0,1]$, and the probability measure $P$ thereon to be Lebesgue measure. If $\omega\in[0,1]$ let $$ \sum_{n=1}^\infty {\omega_n\over 2^2} $$ be the base 2 expansion of $\omega$. Now define $X_n(\omega) :=\omega_n$ to be the $n$th binary digit of $\omega$. [In the ambiguous case ...


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I hope the following is an answer in the way you wanted it to be. If there are any mistakes in it, please correct me. As far as I know, the quadratic variation is only defiened for semimartingales, so we just consider this type of process. As $[X]_0=X_0^2$ we need $|X_0|$ to be deterministic. For simplicity let's focus first on continous processes. Since ...


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If $\mathcal{F}_0$ is trivial, then the condition you are asking for is simply that $E[X_t] = X_0$ but $X_t$ is not a martingale. Take any integrable process $Y_t$ with $Y_0 = 0$, and a uniform $\pm1$ valued $U$ independent of $Y_t$. Then the process $X_t:=UY_t$ has this property, and is usually not a martingale. Indeed, $E[UY_t] = \frac12 E[Y_t] + \frac12 E[...


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If $f$ is step function it is easy to see that $\int f(s)dB_s$ has mean $0$. [The integral is a finite sum of terms of the type $c(B_v-B_u)$ where $c$ is a constant and $0 \leq u <v$]. By definition of the stochastic integral you can write the given intergal as a limit in mean square (hence also in mean) of integrals of this type. Hence $E\int_0^{t}sdB_s=...


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Observe that if $\tau=\{t\}$ and $\tau'=\{t'\}$ and $B,B' \in \mathcal{B}(\mathbb{R})$ with $t<t'$,then we have : $C(\tau,B)\cap C(\tau',B') = C(\{t,t'\},B\times B')\in \mathcal{A}$ and if $t=t'$ then : $C(\tau,B)\cap C(\tau',B') = C(\tau,B\cap B')\in \mathcal{A}$ From those two observations you can build any intersection of set in $\mathcal{A}$ and ...


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It's related with theory of queues from its definition if we read the Queueing Theory and Markov Property whats about:. You can think of a queue or a queue node as almost a black box. Jobs or "clients" arrive in the queue, possibly wait some time, take some time to process, and then exit the queue. However, the tail node is not a pure black box, ...


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The stochastic integral is usually constructed for adapted càglàd (continuous on the left, limit on the right) processes as integrands and semimartingales $\big($which are càdlàg (continuous on the right, limit on the left) processes and include the FV-processes$\big)$ as integrator by using predictable simple processes and the property that the predictable ...


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Because the numbers on these cards are rather arbitrary, I believe that it will be very hard to solve this problem by hand. Writing a piece of code to do this job is straightforward if you're used to it. I wrote mine in SageMath, since it has easy tools for combinatorics. This gave the following results: For the first part of the question: the distribution ...


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From the functional form of the montone class theorem, you can conclude immediately from the identity you have displayed that $$ \Bbb E[(X_t-X_s)\cdot F]=0 $$ for all bounded $\mathcal F_s:=\sigma(X_u: 0\le u\le s)$-measurable $F$. (See Theorem 1 here: https://almostsuremath.com/2019/10/27/the-functional-monotone-class-theorem/.)


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Let $y_n=E[X_n]$ then you have a linear recurrence relation with constant coefficients: $$y_{n+1}-2y_n+y_{n-1}=-2.$$ This system needs two boundary conditions. You only really have one, which is $y_0=0$. To fix that, you can artificially introduce $y_N=0$ (so that you are solving for the expected time to hit $0$ or $N$ from inside) and then send $N \to \...


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What specifically are you having trouble with in Ross's Stochastic Processes? I am familiar with this text and I would have to say it has its shortcomings. Although the preface states This text is a nonmeasure theoretic introduction to stochastic processes, and as such assumes a knowledge of calculus and elementary probability. The first chapter begins ...


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Apply the binomial theorem to one step of the random walk, $$ X_{n+1}^3=(X_n+\Delta X_n)^3=(M_n+A_n)+3X_n^2ΔX_n+3X_n +ΔX_n, $$ using $ΔX_n^2=1$. So with $M_{n+1}=M_n+(3X_n^2+1)ΔX_n$ and $A_{n+1}=A_n+3X_n$ you get a martingale and a predictable process.


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