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0

The condition that $E(S_{i+1}\mid S_i=n)\lt0$ for every $n\ne0$ is not sufficient to guarantee positive recurrence. Counterexamples are discrete Bessel processes of suitable indexes.


0

Actually the Markov property for Itô diffusions rather reads $E[f(X_{t+h})\mid F_t] = E[f(X_{t+h})\mid X_t]$. (In a second phase, the stationarity of the transitions of these processes then yields $E[f(X_{t+h})\mid X_t] = E^{X_t}[f(X_h)]$, that is, more explicitely, $E[f(X_{t+h})\mid X_t] = g(X_t)$ where, for every $x$, $g(x)=E^x[f(X_h)]$.) This holds for ...


0

Writing the definition of $S$ yields $S(s)=E(X(s)x(0)^*)=s\,E(X(s)X(s)^*)$.


1

The inequality $$|e^{\imath \, tu}-1 \leq |t| \cdot |u| \tag{1}$$ follows easily from the fact that $$e^{\imath \, tu}-1 = \frac{1}{\imath} \int_0^{tu} e^{\imath x} \, dx.$$ From $(1)$ we see that $$\begin{align*} \int |e^{\imath \, t u}-1| \, \nu(du) \leq \int \min\{|t|\cdot |u|,2\} \nu(du) = |t| \int_{-\frac{2}{t}}^{\frac{2}{t}} |u| \, \nu(du) +2 ...


1

If $a\in\mathbb{R}$ then: $$\left|e^{ia}-1\right|=\left|\cos a-1+i\sin a\right|=\sqrt{\left(\cos a-1\right)^{2}+\sin^{2}a}=\sqrt{2-2\cos a}=\sqrt{2-2\left(1-2\sin^{2}\left(\frac{1}{2}a\right)\right)}=2\left|\sin\left(\frac{1}{2}a\right)\right|\leq\left|a\right|$$ That could help for your first question. I don't see its functionality. ...


0

Hint: The invariant probability vector is a left-eigenvector associated with the maximal positive eigenvalue, which in this case is $\lambda = 1$. Your answer should be:


0

Consider \begin{align} S_{m} = \sum_{k=0}^{m} \binom{2x}{k} \binom{2m-2x}{m-k} \end{align} for which it can be seen as the following. \begin{align} S_{m} &= \sum_{k=0}^{m} \binom{2m-2x}{m} \frac{(-m)_{k}(-2x)_{k}}{k! (m-2x+1)_{k}} \\ &= \binom{2m-2x}{m} {}_{2}F_{1}(-m, -2x; m-2x+1; 1) \\ &= \binom{2m-2x}{m} \frac{(m-2x)! (2m)!}{(2m-2x)! m!} \\ ...


3

Here's a suggestion. Assume (a) and (b) hold. Let $\mu = E[X_0]$. Show that $P(X_0 > \mu) > 0$. (If not, then $X_0 \le E[X_0]$ almost surely. Conclude that $X_0$ is constant, contradicting the assumption in (b) that it has variance 1.) Then using continuity from below, show that there is some $\epsilon$ such that $P(X_0 > \mu + \epsilon) > ...


0

a is drift, sigma σ^2 is variance rate, and ν(x) is "jump measure": the arrival rate of jumps of size x.


2

Why is this equal to $\Pr(X_2=j\mid X_0=i)$? Because $$\Pr(X_1=j\mid X_0=k)=\Pr(X_2=j\mid X_1=k)=\Pr(X_2=j\mid X_1=k,X_0=i),$$ first by stationarity, then by the Markov property at time $1$, hence $$\Pr(X_1=j\mid X_0=k)\Pr(X_1=k\mid X_0=i)=\Pr(X_2=j,X_1=k\mid X_0=i),$$ in particular, $$\sum_k\Pr(X_1=j\mid X_0=k)\Pr(X_1=k\mid X_0=i)=\Pr(X_2=j\mid ...


2

Stochastic comes from Ancient Greek whereas random is an old French word. (fun fact: random has totally disappeared in modern French and was replaced by aléatoire which comes from... Latin) Otherwise there is no difference between them in the realm of Probability Theory.


0

This difference fall into context and perspective. In some contexts they are the same but in many others they are completely different things. The main difference (when exist) is related to the epistemology base or simply to the closed point of view from a subjectivity. A possible example: stochastic refers to deterministic processes that fall in a ...


1

Stochastic process and random process are just two different names for the same mathematical concept, which are used interchangeably, just like convergence in law and weak convergence, and there is no objective argument for or against one or the other. In analogy with the english language, you can interpret it as dialects (the British say rubbish, Americans ...


0

I used the following property from Wikipedia. Though it still seems a little bit confusing.


0

Remember that when you prove the Doob's decomposition theorem you define: $$A_n = \sum_{k=1}^n \mathbb{E}[X_k | \mathcal{F}_{k-1}]-X_{k-1}$$ and $$M_n =X_0+ \sum_{k=1}^n X_{k}-\mathbb{E}[X_k | \mathcal{F}_{k-1}]$$ Being $X_n$ a submartingale, you have $A_n \ge 0$ this implies $M_n \le X_n$ which implies $\sup_nM_n < + \infty$ if we are in the event ...


0

$A \in \sigma(X)$ means that we can decide for which $\omega \in \Omega$ the event $A$ happens if we know the value $X(\omega)$. For example, if we define $$A := \{\omega \in \Omega; X^2(\omega)=1\},$$ then $A \in \sigma(X)$, since we can check whether $X^2(\omega)=1$ if we know $X(\omega)$. In contrast, if $Y$ is another (independent) random variable and ...


0

Well one interpretation of a stochastic matrix is as a state transition matrix. So consider the state after the $n^\text{th}$ transition, $S=M^n$, and notice that $$\det(S) = \det(M^n) = \det(M)^n$$ Except in marginal cases when $\det(M) = 1$ or when the transitions don't converge, we expect that $\det(M^\infty) = \det(M)^\infty = 0$. So the closer the ...


1

Note that $$Z_k = f(X_1,\ldots,X_k)$$ for $$f(x_1,\ldots,x_k) := \prod_{j=1}^k (1+x_j).$$ Use that $X_1,\ldots,X_k$ are $\mathcal{F}_k$-measurable in order to conclude that $Z_k$ is $\mathcal{F}_k$-measurable.


0

Here's a tip. Look at the integration-by-parts formula for stochastic calculus: http://en.wikipedia.org/wiki/It%C5%8D_calculus#Integration_by_parts


0

Please refer to the following Thesis http://mspace.lib.umanitoba.ca/handle/1993/23349 Page 268.


1

There are very many applications of stochastic processes. To name a few, gambling, statistical sampling, insurance, communication networks, stock option pricing.


1

Quite generally, if the canonical filtration $(\mathcal F_t)$ is generated by the process $N=(N_t)$, then, for every stopping time $T$, the sigma-algebra $\mathcal F_T$ is generated by the process $N^T=(N^T_t)_t$ obtained by stopping $N$ at time $T$. That is, $N^T_t=N_{\min\{T,t\}}$ for every $t\geqslant0$. In your case $N^T$ is $\sigma(T)$-measurable since ...


2

It is exactly the other way round: We define $(1)$ to be the same thing as $(2)$; it's just a more convenient way to write $(2)$. That this notation makes sense can be seen by integrating both sides of $(1)$ from $0$ to $T$: $$\underbrace{\int_0^T dX_t}_{X_T-X_0 = X_T-x} = \int_0^T b(t,X_t) \, dt + \int_0^T \sigma(t,X_t) \,dW_t.$$ A similar situation pops ...


1

The second coordinate $Y$ of the hitting point has a symmetric Cauchy distribution. A notable feature of these distributions is that their mean do not exist. Recall that the symmetric Cauchy distribution with parameter $a\gt0$ has density on the real line $$y\mapsto\dfrac{a}{\pi(a^2+y^2)}.$$ Indeed it is quite different from a normal density since its ...


1

The statement $|X(t)|<M$ can be interpreted in two different ways: Either $$\exists M: |X(t,\omega)| \leq M \, \quad \text{for all} \, \omega \in \Omega \tag{1}$$ (i.e. the bound is uniform for all $\omega \in \Omega$) or $$\forall \omega \in \Omega \, \, \exists M=M(\omega): |X_t(\omega)| \leq M. \tag{2}$$ $(1)$ is in general not correct. To see ...


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 $W$: ...


0

To continue the hint: we have by Doob's inequality, for $c\gt 0$: $$\mathbb P\left(\max_{1\leqslant k\leqslant n}X_k\geqslant \lambda\right)=\mathbb P\left(\max_{1\leqslant k\leqslant n}(X_k+c)^2\geqslant (\lambda+c)^2\right)\leqslant \frac 1{(\lambda+c)^2}\mathbb E[(X_n+c)^2].$$ This is what we have to optimize over $c$ (the equality is true for each $c$, ...


0

Can you just show that since: $$P(Y_{n+1}|Y_n) \neq P(Y_{n+1}|Y_n,Y_{n-1}) $$ Then $Y_n$ cannot be a markov chain as it violates the markov property


0

Yes, this distribution can be described (although not always very explicitly). More generally, let $S$ be the total number of individuals of a Galton--Watson process started from $z_0$ individuals and with offspring distribution $\mu$. Let $(W_n, n \geq 0)$ be a random walk on $\mathbb{Z}$ with initial value $z_0$ and jump distribution $\nu(k) = \mu(k + ...


0

The problem is that you work with equalities in distribution, and that $\sum_{i=1}^n|B_{(i+1)/n}-B_{i/n}|^p$ is not even equal in distribution to $n|X_{1/n}|^p$. It is rather equal in distrubution to $\sum_{i=1}^nY_{i,n}$, where $(Y_{n,i})_{i=1}^n$ are i.i.d. and $Y_{n,i}$ is normally distributed, with mean $0$ and variance $1/n$. We can use the following ...


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 ...


1

A direct proof is to note that, if $(X_k)$ denotes the Markov process on $\{0,1,\ldots,N\}$, then the random process $(Y_k)$ defined by $$Y_k=u(X_k),\qquad u(x)=\max\{x,N-x\},$$ is a Markov chain on $\{N/2,N/2+1,\ldots,N\}$ with transition rates $2g_{N/2}=2r_{N/2}$ for $N/2\to N/2+1$, $g_n=r_{N-n}$ for $n\to n+1$ and $r_n=g_{N-n}$ for $n\to n-1$ for every ...


2

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 ...


1

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


0

For simplicity, consider a two-state Markov process $(X_t)$ on the states {$\texttt{alive}$,$\texttt{dead}$} with only one transition, from $\texttt{alive}$ to $\texttt{dead}$, with rate transition $r$. This means that, for every $t$, when $s\to0$, $$P(X_{t+s}=\texttt{dead}\mid X_t=\texttt{alive})=rs+o(s).$$ This implies that, for every $s$ and $t$, ...


4

First, simulate the paths of the process $(X_t)$ defined by $$X_t=\sigma W_t+\left(r-\tfrac12\sigma^2\right)t,$$ using Euler's scheme $X^\varepsilon_0=0$ and$$X_{n+1}^{\varepsilon}=X_n^{\varepsilon}+\sigma\sqrt\varepsilon Z_n+\left(r-\tfrac12\sigma^2\right)\varepsilon,$$ for every positive $\varepsilon$ and every $n$, where the process $(Z_n)$ is i.i.d. ...


2

Expand the inequality $\langle X-Y\rangle\geqslant0$.


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 ...


0

First, $S_\eta$ is a martingale whose increments have finite conditional expectation and $\mathbb{E}[\eta] < \infty$, so the optional stopping theorem implies that $S_\eta$ is integrable. Second, notice that $\{S_\eta \leq 0\}=\{X_\eta \leq 0\}$. Together, these imply that $$-\infty < \mathbb{E}[S_\eta I(S_\eta \leq 0)]$$ $$\quad\quad\quad\quad\quad= ...


0

By Proposition 5.22 on page 345 of Karatzas and Shreve, if $X_t$ is a weak solution to $dX_t = b(X_t)dt + \sigma(X_t)dW_t$ in an interval $I$ which in this case is $I = (0,\infty)$. If $p(x)$ is the scale function given by $p(x) = \int_0^x exp(-2 \int_0^y \frac{b(z)}{\sigma(z)^2}dz)dy = x$. We have $p(0^+) > -\infty$, and $p(\infty) = \infty$, then we ...


0

It's a counterintuitive, but provable, fact that the average waiting time for a customer that arrives randomly (and uniformly) will be $\geq \frac{T}{2}$. The intuitive explanation is that a passenger is more likely to arrive during a long interval than a short interval, so their waiting times will represent a biased sample of actual interarrival times (too ...


1

It follows indeed from the Markov property (...which is a consequence of the independence of the increments) that it suffices to show the claim for $i=1$. Let $t>0$, $\omega \in \Omega$ such that $|\Delta X_t(\omega)| > \varepsilon$. Since $X$ has (a.s.) càdlàg sample paths, we can find for any $k \in \mathbb{N}$ sufficiently large some $r_k, s_k \in ...


0

Repeated bets in roulette is a random walk on the wealthy variable. Since there is a dead end, when you can't continue the walk (when you're broke), the probability of loosing everything is greater than zero.


1

What you would need would be $f'_{LB}(x) \le f'(x) \le f'_{UB}(x)$, with $f'_{LB}(x) > 0$ for $x < x^*_{LB}$ and $f'_{UB}(x) < 0$ for $x > x^*_{UB}$.


2

Yes, they are indeed equivalent. Proof: Let $X_1,\ldots,X_n$ arbitrary random variables and denote by $$S_j := \sum_{k=1}^j X_k$$ the $j$-th partial sum. Since $$X_j = S_j - S_{j-1} \qquad \quad S_j = \sum_{k=1}^j X_k$$ it is not difficult to see that $$\sigma(X_1,\ldots,X_n) = \sigma(S_1,\ldots,S_n).$$ If we set $X_j := B_{t_j}-B_{t_{j-1}}$ for a ...


0

If $\mathcal{F} \subseteq \mathcal{A}$ is a sub-$\sigma$-algebra, then we can interpret $\mathcal{F}$ as our pool of information. For example for a stochastic process $(X_t)_{t \geq 0}$ the canonical filtration is given by $$\mathcal{F}_t := \sigma(X_s; s \leq t),$$ i.e. $\mathcal{F}_t$ contains the information about the process up to time $t$ ("the ...


0

Take $A=\text{diag}(a_1,...,a_n)$, $a_i>0$ and consider the vector SDE ($n$ components) \begin{equation} dX_t=-A X_t +\sigma dW_t, \end{equation} where $W_t$ is a vector of independent Wiener processes and $\sigma>0$ a scalar. This is a special case of the vector OU (Ornstein-Uhlenbeck) process. Its temporal autocovariance reads \begin{equation} ...


0

A process $f: [0,\infty) \times \Omega \to \mathbb{R}$ is $\mathcal{F}_s$-adapted if, and only if, $$(\Omega,\mathcal{F}_s) \ni \omega \mapsto f(s,\omega)$$ is measurable for any fixed $s \geq 0$, i.e. if $$\{\omega \in \Omega; f(s,\omega) \in B\} \in \mathcal{F}_s \quad \text{for any Borel set} \, B.$$ For fixed $ \geq 0$, the function $$\omega \mapsto ...


1

Regarding your first question: Let's just consider a simple example; namely, the (one-dimensional) Brownian motion. Then it follows from the homogenity of Brownian motion (in time and space) that $$\begin{align*} A(z,t) &:= \lim_{\varepsilon \to 0} \lim_{\Delta t \to 0} \frac{1}{\Delta t} \int_{|x-z|<\varepsilon} (x-z) p(x,t+ \Delta t \mid z,t) \, ...


0

The author did not say that the expectation is $0$: he rather meant that $M_{T(\omega)}^\theta=0$ if $T(\omega)=\infty$. This is justified because $0\lt \mathrm{sech}(x)\lt 1 $ for each $x$. We thus have $$\mathbb E[M_{T}^\theta]=\mathbb E[M_T^\theta\chi\{T<\infty\}].$$



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