Tag Info

Hot answers tagged

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 ...


3

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: $$\mathbb{E}(N_t-\lambda t \mid F_s^N) = \mathbb{E}((N_t-N_s)+N_s \mid F_s^N) - \lambda t = \ldots$$


2

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.


2

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 ...


2

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, ...


1

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 $W$: ...


1

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



Only top voted, non community-wiki answers of a minimum length are eligible