Tag Info

Hot answers tagged

3

Set $$M_t := f(X_t) - \int_0^t Af(X_s) \, ds$$ and write $P_t f(x) := \mathbb{E}^x f(X_t)$ for the semigroup associated with $(X_t)_{t \geq 0}$. Using the definition of the generator, it is not difficult to see that $$\frac{d}{dt} (\mathbb{E}^x f(X_t)) = \frac{d}{dt} P_t f(x) = P_t Af(x).$$ By the fundamental theorem of calculus, this implies $$P_t ...


3

$\nu \circ f^{-1}$ denotes the push-forward measure (or image measure) of $f$ with respect to $\nu$, i.e. $$\nu_1(B) = (\nu \circ f^{-1})(B) = \int 1_B(f(x)) \, \nu(dx)$$ for any Borel set $B$. This implies $$\int g(x) \, \nu_1(dx) = \int g(f(x)) \, \nu(dx)$$ for any $g \in L^1(\nu_1)$.


3

Let me give you first a non-rigorous "proof" of the statement; afterwards I'll explain how to justify it. If we define $$f_n(x) := n x 1_{[0,1/n]}(x) + 1_{[1/n,\infty)}(x),$$ then $f_n \uparrow 1_{(0,\infty)}$. Obviously, $f_n$ is not differentiable at $x=0$ at $x=1/n$, but for the moment we forget about this (that's the non-rigorous part) and apply ...


3

You need to show independence of increments, i.e. if $0\le a<b<c<d$ then $(N_1(d)+N_2(d)) - (N_1(c)+N_2(c))$ is independent of $(N_1(b)+N_2(b)) - (N_1(a)+N_2(a))$, and similarly for more than two intervals. You can prove that by using independence of increments of each of the two processes separately plus independence of $N_1$ and $N_2$. You also ...


3

Hmm. I don't know how you can directly find $A$, $B$, $C$ but I can suggest a way to go about the problem. Now that you have found the eigenvalues $\mu_i$, solve for the corresponding normalized eigenvectors $e_i$. Then $$P^n = E^{-1}\Lambda^n E$$ where $\Lambda$ is the diagonal matrix of eigenvalues and $E = \sum_{i=0}^2 {e_ie_i^T}$. Then you can read ...


2

To help hone your intuition, it might help to think about the following simple example, which shows that the Blumenthal law is using much more than just the continuity of Brownian motion. Let $Z$ be a random variable with $P(Z=1) = P(Z=-1) = 1/2$, i.e. a coin flip. Set $X_t = tZ$, so that the process $X_t$ just moves with constant speed $\pm 1$ depending ...


2

Why not just computing $P^2$, then interpolating? We have: $$P=\begin{bmatrix}\frac{1}{3}&0&\frac{2}{3}\\\frac{1}{3}&\frac{2}{3}&0\\\frac{1}{3}&\frac{1}{3}&\frac{1}{3}\end{bmatrix},\qquad ...


2

The result is false in this generality. Fix some $z\in S$ and define $O\mu=\mu_c+\mu_d(S)\delta_z$, where $\mu_d$ and $\mu_c=\mu-\mu_d$ are the discrete and continuous parts of $\mu$, respectively. Then $O$ satisfies your conditions, but cannot possibly be given by a kernel, if there is a non-discrete probability measure $\nu$ on $S$. Why not? If ...


2

No. The disjoint events $\{B_n\}$ are not independent. They are mutually exclusive. Hint: Begin with the Law of Total Probability: $\mathsf P(A) = \sum\limits_{n=1}^\infty \mathsf P(A\mid B_n)\;\mathsf P(B_n)$ Yes. (But watch your shift key; you have a $P(X\mid y=y)$ typo. Additionally you need $X=x$ there not $X$ by itself).


2

For sake of compact notation, we let $\Delta N_i \mathop{:=} N_i(t+\Delta t)-N_i(t)$, denote the change in event count over a period $\Delta t$, and make note of the memoryless property of Poisson distribution's waiting times.   Meaning, $\mathsf P(\Delta N_i=n)$ depends only on $\Delta t, \lambda_i,$ and $n$; it is irrespective of $t$.   Indeed: ...


2

Define an array as a sequence of random variables on a probability space $(\Omega, \mathcal{F},P)$. Introduce doubly infinite arrays of random variables $X_{n,j},~\mathcal{F}_{n,j}$ for $j,n\geq 1$, and set sub $\sigma -$algebras of a $\sigma -$algebra. Adapting the array to the filtration, then $\{X_{n,j}\}$ is a MDA if the relation ...


1

We know the distribution of the vector $V:=(X(7)-X(5), X(5)-X(1), X(1))$, since the increments are independent. Defining the function $f\colon (x,y,z)\mapsto (x+y+z)(y+z)z$, the wanted expectation is $\mathbb E\left[f(V)\right]$.


1

First, $N_1(0) + N_2(0) = 0 + 0 = 0$. Second, we need to show $\{N_1(t) + N_2(t)\}$ has independent increments, namely, for $0 < t_1 < t_2 < \cdots < t_k$, $N_1(t_1) + N_2(t_1), (N_1(t_2) + N_2(t_2)) - (N_1(t_1) + N_2(t_1)), \ldots, (N_1(t_k) + N_2(t_k)) - (N_1(t_{k - 1}) + N_2(t_{k - 1}))$ are independent, which follows from that $\{N_1(t)\}$ ...


1

The only requirement is that the random variables $X_t$ are defined on the same probability space.


1

Stochastic process is a collection of random variables — nothing more and nothing less. That means your collection of iid standard normal random variables is an example of stochastic process that even has a special name — it's called a white noise.


1

Hint: You know that for $n=0$ we get $$ A + B = I, $$ and for $n=1$ we get $$ A + B/3 = P. $$ So you can express $A$ and $B$ in terms of $I$ and $P$...


1

Some related summary comments on ergodic Markov chains. (There are several fleas hopping around triangles elsewhere on this site, each according to the same transition matrix, but with varying questions about the associated process; search 'Markov flea vertex'.) With help from @Did, essentially by manipulating difference equations, I believe you have ...


1

The closed communicating class might be the whole set. For the second part, you will need an infinite state-space.


1

In some sense, there is an $x$ on the right-hand side because the equality $$\mathbb{E}_x(Y \circ \theta_s \mid \mathcal{F}_s^+)(\omega) = \mathbb{E}_{B_s(\omega)}(Y)$$ holds only $\mathbb{P}_x$-almost surely. For a function $f: \mathbb{R} \to \mathbb{R}$ the Markov property reads $$\mathbb{E}_x (f(B_{t+s}) \mid \mathcal{F}_s^+) = \mathbb{E}_{B_s} f(B_{t}), ...


1

Since $P$ is symmetric, we know from the spectral theorem that $P$ has real eigenvalues and is diagonalizable. Since $P$ is a doubly stochastic matrix, the stationary distribution is uniform, i.e. $$\pi=\left(\frac16,\frac16,\frac16,\frac16,\frac16,\frac16\right).$$ Since $P$ has a stationary distribution, we know that $1$ is an eigenvalue. In particular, ...


1

That is mistaken. Let's look at the probability distribution of $X_2$ given $X_0=6$. \begin{align} & \Pr(X_2 = 6\mid X_0=6) \\[10pt] = {} & \Pr\Big( \underbrace{(X_1 = 1\ \&\ X_2 = 6)\text{ or }(X_1=2\ \&\ X_2=6)\text{ or }\cdots}_{\text{five disjuncts}}\mid X_0=6\Big) \\[10pt] = {} & \underbrace{\Pr\Big(X_1 = 1\ \&\ X_2 = 6 \mid ...



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