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## Hot answers tagged martingales

3

You have $$\widetilde{W}_t=W_t+\int\Theta(u)du$$ which is in general not a Brownian motion, because it has a drift component. But 5.3.1 states $$M_t=M_0+\int \Gamma(u)dW_u\tag{5.3.1}$$ , which holds only for a Brownian motion $W$ (and $M_t$ martingale). So one cannot trivially replace $W_t$ and $W_t+\int\Theta(u)du=\widetilde{W}_t$ in 5.3.2 aswell by ...

3

Assuming you mean $\mathrm{E}(X|Y)$, then since the distribution of each die is identical and independent, if we are given that their sum is $y$, the expected value of each die would be the same: $y/2$. If you did mean $\mathrm{E}(X/Y)$, then in light of the preceding, this would be $\frac12$. Explicit Calculation of $\boldsymbol{\mathrm{E}(X|Y)}$ Given ...

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Hints Brownian motion: Apply Itô's formula to $f(t,x) := \exp(\beta x - t \beta^2/2).$ Poisson process: Use that $(N_t)_{t \geq 0}$ has independent increments, i.e. $N_t-N_s$ is independent of $F_s^N$. Start with the easier one: $$\mathbb{E}(N_t-\lambda t \mid F_s^N) = \mathbb{E}((N_t-N_s)+N_s \mid F_s^N) - \lambda t = \ldots$$

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We show how to handle the problem for one value of $y$, say $y=9$. Given that $Y=9$, $X$ takes on values $3$ to $6$ with equal probabilities. Thus $$E(X|Y=9)=\frac{3+4+5+6}{4}.$$ One value of $y$ done, $10$ more to do. Remark: The symmetry argument of robjohn is much better.

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If $x$ is not in $(0,r)$, then $\eta=0$ and $X_\eta$ is undefined, hence we assume that $x$ is in $(0,r)$. Since one knows that $\eta$ is integrable, Wald's theorem ensures that $S_\eta$ is integrable (and provides its expectation, which we will not need). Furthermore, $S_{\eta-1}$ is in $(0,r)$ by the definition of $\eta$ and ...

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At time $0$, $\xi^{Z_0}=\xi$. When $n\to\infty$, $Z_n\to+\infty$ on non-extinction hence $\xi^{Z_n}\to0$ on non-extinction and $Z_n\to0$ on extinction hence $\xi^{Z_n}\to1$ on extinction. Finally $|\xi^{Z_n}|\leqslant1$ uniformly, thus everything is in place for an application of martingale dominated convergence theorem.

2

Define a sequence of stopping times $(\tau_k)_k$ by $$\tau_k := \inf\{n \geq 0; X_n \geq k\}.$$ From $$M_{n \wedge \tau_k} = X_{n \wedge \tau_k}-X_0 - \underbrace{A_{n \wedge \tau_k}}_{\geq 0} \leq 2k$$ it follows that $(M_{n \wedge \tau_k})_{n \in \mathbb{N}}$ is a martingale which is bounded above. Consequently, by a standard convergence theorem, the ...

2

No, $X_{n \wedge N} \geq -K-M$ is correct. Note that, by definition, $X_k \geq -K$ for any $k < N$. Hence, in particular for $k := (N-1) \wedge n$. Consequently, $$X_{n \wedge N} = \underbrace{(X_{n \wedge N}-X_{(N-1) \wedge n})}_{\geq -M} + \underbrace{X_{(N-1) \wedge n}}_{\geq -K} \geq -K-M$$ where we have used that $|X_{k+1}-X_k| \leq M$ for all $k ... 1 Let us start from the answer of this question: Expectation of Stopping Time w.r.t a Brownian Motion. Use the martingale$B_t^3 - 3tB_t$to compute$E[T \mid B_T = a]$and$E[T \mid B_t = b]$. Deduce the value of$E[TB_t^2]$. Use the martingale$B_t^4 - 6tB_t^2 + 3t^2$to compute$E[T^2]$. 1 Take the collection of all partitions of$\Omega$and form the$\sigma$-algebra for each. The number of such partitions is called$B_n$, the Bell number (from Wiki, the source of all truth).$B_4 = 15$. To see that a$\sigma$-algebra${\cal A}$corresponds to a partition, choose$\omega \in \Omega$and let$A_\omega$be the intersection of all elements ... 1 The proof of Doob’s martingale maximal inequalities applies to the submartingale$−X$, see for example this one. 1 An indicator function has a value of$1$when the parameters are within the given range, and a value of$0$otherwise.$\operatorname{\bf 1}_{\{X_0<\lambda,\cdots,X_{n−1}<\lambda,X_n\ge\lambda\}} = \begin{cases} 1 & : X_0<\lambda,\cdots,X_{n−1}<\lambda;X_n\ge\lambda \\ 0 & :\text{elsewhere}\end{cases}$Let's look at a sample case. ... 1 Define the event$A_j:=\{X_j\lt\lambda\}$where$j\in\{0,\dots,m\}$and$B_n:=\bigcap_{j=0}^{n-1}A_j\cap A_n^c$. Then the sets are pairwise disjoint and$\bigcup_{n=0}^mB_n=\{\max_{0\leqslant n\leqslant m}X_n\geqslant \lambda\}$. 1 Hints Set$A_n := (0,a_n]$. Show that $$\sigma(X_1,\ldots,X_n) = \sigma(A_1,\ldots,A_n).$$ Recall that$\mathbb{E}(Y \mid \mathcal{F}) = X$if, and only if,$X$is$\mathcal{F}$-measurable and $$\int_G X \, d\mathbb{P} = \int_G Y \, d\mathbb{P} \tag{1}$$ for all$G \in \mathcal{G}$where$\mathcal{G}$is a$\cap$-stable generator of the$\sigma$-algebra ... 1 For every$k$, with the convention that$a_0=1$, consider the intervals $$I_k=(0,a_k],\qquad L_k=(a_k,a_{k-1}].$$ To compute $$Y_n=P(X \in I_{n+1}\mid\mathfrak{F}_n],$$ note first that$\mathfrak{F}_n$is generated by a partition of$(0,1]$, namely, $$\{I_n\}\cup\{L_k\mid1\leqslant k\leqslant n\}.$$ Second,$I_{n+1}\subseteq I_n$and$I_n$is a member of the ... 1 With the help given, I was able to solve the problem in a somewhat different way. Since we have a finite partition of$\Omega$and (without resctriction$a_0 = 1$)$\mathfrak{F}_n = \sigma(\{(0, a_n], (a_n, a_{n-1}], ..., (a_1, a_0]\})$. Then$\mathbb{E}[X_{n+1}|\mathfrak{F}_n] = 1_{(0, a_n]}\cdot\mathbb{E}[X_{n+1}|{(0, a_n]}] + \sum_{k=1}^n 1_{(a_k, ...

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The Gambler's Ruin example is probably helpful for intuition. Consider a Markov chain $X_n$ whose state space is the integers which jumps 1 unit to the left or to the right with equal probability. Now let $\tau = \inf \{ n : X_n = 0 \}$. Then the stopped process is the same as the original process until the gambler runs out of money, after which he ...

1

First of all, a filtration $( \mathscr{F}_t )_{t \geq 0 }$ is a "set" of sigma algebras indexed usually by time t that are increasing. That is, for every $t>0$, $\mathscr{F}_t$ is a sigma algebra and $\mathscr{F}_t \subseteq \mathscr{F}_T$ for all $0\leq t \leq T$. The canonical example, is the filtration generated by a process, say Brownian Motion $W$: ...

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This is a typo, one should read "Let $X$ be a simple random walk on $\mathbb Z$".

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