# Tag Info

4

Once $k>2$ the answer to #1 depends on the graph; even in the limiting determinstic case of $p=1$, and taking $n_0=1$, the expected time till full zombification can range from about $C \log n$ (for an expander graph) to at least $n/k$ (when $k$ is even, the node set is ${\bf Z}/n{\bf Z}$, and each node is joined to its $k$ nearest neighbors). As to #2: ...

2

The condition means that, for every $n$, there exists some measurable set $A_n$ such that $$\tau\leqslant n\iff(X_1,\ldots,X_n)\in A_n.$$ Thus, every positive stopping time $\tau$ can be rewritten as $$\tau=\inf\{n\geqslant1\mid(X_1,\ldots,X_n)\in A_n\},$$ for some well-chosen sequence $(A_n)$.

2

Your definition of $\tau$ is flawed, actually, $$\tau=\int_0^\infty\mathbf 1_{W_t\gt t}\,\mathrm dt,$$ hence $$E(\tau)=\int_0^\infty P(W_t\gt t)\,\mathrm dt.$$ For every $t$, $W_t$ is normal centered with variance $t$ hence $$P(W_t\gt t)=P(Z\gt\sqrt{t}),$$ where $Z$ is standard normal. By symmetry, $P(Z\gt\sqrt{t})=P(Z\lt-\sqrt{t})$ hence $$... 2 For every s\lt t, E(B_s(B_t-B_s))=E(B_s)E(B_t-B_s)=0 because B_s and B_t-B_s are independent and B_t-B_s is centered. A consequence is that E(B_t(B_t-B_s))=E(B_t(B_t-B_s))-E(B_s(B_t-B_s))=E((B_t-B_s)^2). Finally, B_t-B_s is centered normal with variance t-s hence E((B_t-B_s)^2)=t-s. The hypothesis that one starts from B_0=0 is not ... 2 I will let you find it for yourself ;) Hint: Irreducible is a matrix (I wished to comment, but I couldn't since due to not enough points) 1 Because, if f(t) is a continuous function, then the existence of any t \in [0,1] such that f(t) > \epsilon implies the existence of a rational s \in [0,1] such that f(s) > \epsilon. In other words, for any n \in \mathbb{N}, the two sets$$\{\omega : \exists t \in [0,1] \text{ s.t. } |f(t,\omega)| > 1/n\} = \{\omega : \exists t \in [0,1] ...

1

Whenever we consider a process $(Y_t)_t$ of the form $Y_t = f(B_t)$ (where $f$ is a "nice" function and $(B_t)_t$ a Brownian motion), then by Itô's formula $$Y_t - Y_0 = \int_0^t f'(B_s) \, dB_s + \frac{1}{2} \int_0^t f''(B_s) \, ds. \tag{1}$$ Since the stochastic integral is a martingale, this yields in particular that $$\mathbb{E}Y_t = \mathbb{E} \left( ... 1 The proof is based on this standard fact from functional analysis: Theorem. Let H be a Hilbert space and let E \subset H be a linear subspace. Let E^\perp := \{ x \in H : \langle x,y \rangle = 0 \text{ for all } y \in E\}. Then E is dense in H iff E^\perp = \{0\}. So for this proof, H = L^2(\mathcal{F}_T, P) and E is the linear span ... 1 First of all,$$X_t := \int_0^t W_s \, ds \tag{1}$$is an Itô process, i.e. an process of the form$$Y_t - Y_0 = \int_0^t \sigma_s \, dW_s + \int_0^t b_s \, ds$$(here \sigma_t =0, b_t = W_t). For an Itô process (Y_t)_{t \geq 0} Itô's formula reads$$f(Y_t)-f(Y_0) = \int_0^t f'(Y_s) \, dY_s + \int_0^t f(Y_s) + \frac{1}{2} f''(Y_s) \, d \langle Y ...

1

The key point is to write $B_T^2$ in a clever way: \begin{align*} B_T^2 &= \sum_{j=1}^n (B_{t_j}^2-B_{t_{j-1}}^2) \\ &= \sum_{j=1}^n \big( (B_{t_j}-B_{t_j^{\ast}})+B_{t_j^{\ast}} \big)^2 - \big( (B_{t_{j-1}}-B_{t_j^{\ast}})+B_{t_j^{\ast}} \big)^2 \\ &= 2 \sum_{j=1}^n B_{t_j^{\ast}} (B_{t_j}-B_{t_{j-1}}) + \sum_{j=1}^n ... 1 There is a quote on this point from page 808 in J.L. Doob's Classical Potential Theory and Its Probabilistic Counterpart. Before martingales had been formally christened, Lévy [1, 1935; 2, 1937], Bernstein [1, 1937], and other mathematicians had analyzed some of their properties in special contexts; usually the martingales in question arose as ... 1 As far as I know, this naming has its origin in the connection between Brownian motion and harmonic functions. Let B be a d-dimensional Brownian motion. Let f:\mathbb{R}^d\to\mathbb{R} be twice continuously differentiable. By Ito's formula, we have f(B_t) = f(0) + \sum_{i=1}^d\int_0^t \frac{\partial f}{\partial x_i}(B_s) dB^i_s + ...

1

It can be written as a markov chain to obtain the probabilities. But there are 101 states and couldn't find a simple closed form like in the case of the original gambler's ruin. If the limit was 5 units instead, the markov chain is written as: \begin{align*} A &= \left(\begin{array}{rrrrrr} r & q & 0 & 0 & 0 & p \\ p & r ...

1

Let $u_n$ be the probability that $n$ Bernoulli trials result in an even number of successes. This occurs if an initial failure is followed by an even number of successes, or an initial success is followed by an odd number of successes. Therefore $u_0=1$ and for $n\geq 1$ $$u_n=q u_{n-1}+p(1-u_{n-1}).$$ Multiplying by $s^n$ and adding over $n$ we see that ...

1

Let $R_j$ denote the first return time to $j$, then (strong) Markov property at time $R_j$ yields$$P(X_n=j\mid X_0=j)=\sum_{m=1}^nP(R_j=m\mid X_0=j)P(X_{n-m}=j\mid X_0=j).$$ Nota: It might be time for you to get a good textbook since the proof above is in all those on the subject. You might try Markov chains by James Norris, available on the web.

1

Consider a trajectory of $n$ transitions, ending in state $j$. Either transition $n$ is the first return to state $j$, which has by definition has probability $f_j^{(n)} = f_j^{(n)} p_{jj}^{(0)}$, or there was a return at some earlier transition(s), the first of which we can label transition $m$. If first return was at $m \neq n$ (which itself has ...

1

Hint: First, note that $\psi$ is defined to be continuous (see top of proof). Then, if you define $<f, g> := \int fg$, and derive a norm from it, it is not too hard to show that the integral of the product of a continuous and a bounded function (which is the case here) is continuous (use Cauchy-Schwartz). Note also that this is a "for each $\omega$" ...

1

It all depends on what you want the set $\mathcal{V}^{m\times n}$ to be used for. In Oksendal's monograph, what it denotes is the following. Each set $\mathcal{V}^{m\times n}(0,T)$ denotes the set of $m\times n$ matrix processes with entries integrable over $(0,T]$. With Oksensdal's definition of $\mathcal{V}^{m\times n}$, this set becomes the set of ...

1

The proof is in two steps: it is true for elementary functions: you just have to prove that for every $Y_i$ mesurable with respect to $F_{t_{i-1}}$, $$E\left[\sum_{i=0}^{N-1} Y_i (B_{t_i} - B_{t_{i-1}})\right] =0$$(it is quite easy). then, write an element $f$ as an $L^2$ limit of such elementary processes $f_n$. Then,  E \left[\int_S^T f_n dB - ...

1

Since $g(\cdot, \omega)$ is continuous (and hence uniformly continuous on $[S,T]$), you can check that $\phi_n(\cdot, \omega) \to g(\cdot, \omega)$ uniformly on $[S,T]$. Therefore $(g(\cdot, \omega) - \phi_n(\cdot, \omega))^2 \to 0$ uniformly and this gives you the convergence of the integral. Continuity is not strictly necessary here; piecewise continuity ...

1

For Polish $X$: Optimal measure $M^*$ exists: Topics in Optimal Transportation (Theorem 1.3 on page 19) and Optimal Transport (Theorem 4.1 on page 32) Equality $K_d=W_d$ holds: Topics in Optimal Transportation (Theorem 1.14 on page 34) and Optimal Transport (Remark 6.5 on page 95). The first of two books is more analysis-oriented, and may be easier to ...

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