New answers tagged

1

The equation $Lu=0$ where $L$ is the generator of your process (the formal adjoint of the Fokker-Planck operator) does indeed have significance, but its significance depends on the boundary conditions that you impose. In the case of the stationary FPE, the natural "boundary condition" is just the normalization and positivity conditions. About the only other ...


2

Your actual stochastic integral term (in your analytical solution) is huge. (Exercise: use the Ito isometry to compute its variance.) Its contribution to the overall quantity is not huge, because it is multiplied by a tiny factor $e^{-\theta t}$ on the outside. But that doesn't matter in computer arithmetic: to a computer, your stochastic integral is so huge ...


2

Here is a trick to map an instance of your problem to the framework of the paper "Distributed Stochastic Optimization via Correlated Scheduling" (IEEE Trans Netw. April 2016): http://ee.usc.edu/stochastic-nets/docs/distributed-optimization-ton.pdf Let's perform a notation shift: \begin{align} (Y_1,Y_2) &\leftrightarrow (\omega_1, \omega_2) \quad \mbox{[...


3

If the time vector is t = 0:dt:500; there is an overflow in exp(th*t(i+1)) in the for loop, as t gets large. Perhaps something like this: th = 2; mu = 1.2; sig = 0.3; dt = 1e-2; t = 0:dt:500; % Time vector x0 = 0; % Set initial condition rng(1); % Set random seed W = zeros(1,length(t)); % Allocate integrated W ...


1

I'm guessing that what you have in mind is a Poisson process. Two essential facts about the process described in your question are the following. Let $X$ be the number of sites falling within a particular region whose area is $a$. Then $\operatorname{E}(X) = a.$ If two regions do not intersect each other, then the numbers of sites falling within them are ...


0

To my understanding, Skorohod works with $\Omega$ being the Skorohod space. Then canonical projection $X$ is $X_t(\omega) = \omega(t)$ You see càdlàg processes as random variables on the skorohod space. You just care about law. That is very useful to prove existence in law of processes defined as limit of processes.


2

If, for all $T$, $P(|X_T-X_0|\leq 2c)=1$, then there is nothing to prove, the process is bounded so it is integrable. If for some $T$ the above probability $\rho$ is less than $1$, then \begin{align}P(\tau_1=\infty)\leq P(\tau_1\geq NT)&\leq P(|X_T-X_0|\leq2c,|X_{2T}-X_T|\leq 2c,\cdots,|X_{NT}-X_{(N-1)T}|\leq 2c) \\&=\prod_{i=1}^N P(|X_{iT}-X_{(i-1)...


1

By Fubini's Theorem, (2) is equivalent to $$\int_0^t E[X_s^2] ds < \infty$$ If $$E[X_s^2] = \infty$$ then $$\int_0^t E[X_s^2] ds = \infty \ ↯$$ Hence $$E[X_s^2] < \infty$$ Some specifics with $s$ and $t$ but I think (2) implies (1)


0

Just a summary of my comments that may clarify: You "should never" consider the sigma algebra generated by an event (since it has nothing to do with expressions like $E[X|Y=y]$). You "should" consider the sigma algebra generated by a random variable, and this has lots to do with expressions like $E[X|Y=y]$. Specifically... Let $S$ be a sample space. Let $(...


1

If $\{B_t:t\in\mathbb R_+\}$ is a standard Brownian motion, then $$X_t := X_0\exp\left(\left(\mu-\frac12\sigma^2\right)t+\sigma B_t\right) $$ (where $X_0$, $\mu$, and $\sigma$ are constants) defines a geometric Brownian motion. It is clear that if $X_0=0$ then $X_t$ is identically zero, and similarly if $\sigma=0$ then $X_t = X_0 e^{\mu t}$ with probability ...


1

Sasha's answer is so nice. I want to offer other way. Other way Let $\alpha\in \mathbb{R}^+$. It is well known that $X_t=\exp\left(\alpha W_t-\frac{1}{2}\alpha^2 t\right)$ is a martingale, therefore $$\mathbb{E}[X_{T_{a,b}}]=\mathbb{E}[X_0]=1$$ so $$\mathbb{E}[X_{T_{a,b}}]=\mathbb{E}\left[\exp\left(\alpha W_{T_{a,b}}-\frac{1}{2}\alpha^2 T_{a,b}\right)\...


-1

I would make a distinction: for example in a queueing system the arrival times (or interval times) might be modelled by a Poisson process which would be time independent and would not be bound my initial conditions. This would be an example of a random process which outputs random variables. Service time in the queue would be dependent on the previous state(...


4

If $W_t:=\rho W^{(1)}_t+\sqrt{1-\rho^2} W^{(2)}_t$ then we can show that, $W_t$ is a Brownian motion. Proof Let $(\Omega, \mathcal{F},\mathbb{P},\{\mathcal{F_t}\})$ be a probability space . Clearly, $W_t$ has continuous sample paths and $W_0=0$. $$\mathbb{E}[W_t|\mathcal{F_s}]=\rho\,\mathbb{E}[W^{(1)}_t|\mathcal{F_s}]+\sqrt{1-\rho^2}\,\mathbb{E}[W^{(2)...


3

For any $t>0$, we have $W_t^{(1)},W_t^{(2)}\stackrel{\mathrm{i.i.d.}}\sim \mathcal N(0,t)$, so $\rho W_t^{(1)}\sim\mathcal N\left(0,\rho^2 t\right)$ and $\sqrt{1-\rho^2}W_t^{(2)}\sim\mathcal N\left(0,(1-\rho^2)t \right)$, from which $$W_t \sim \mathcal N\left(0, t \right). $$ Therefore $\{W_t:t\in\mathbb R_+\}$ is a Gaussian process, and from independence ...


1

Following your second attempt using Kolmogorov's zero one law: Define for each $n \in \mathbb{N}$ $$\mathcal{F}_n := \sigma(X_t-X_s; n \leq s \leq t <n+1)$$ then the so-defined $\sigma$-algebras are independent since the Lévy process has independent increments. Now set $$\mathcal{G}_n := \sigma \left( \bigcup_{k=n}^{\infty} \mathcal{F}_k \right) = \...


1

A Markov chain can have stationary distribution $\pi(x)$ and not satisfy the detailed balance equation $\pi(x) \pi(x_p | x) = \pi(x_p) \pi(x | x_p)$. If an aperiodic and ergodic Markov chain doesn't require that $\pi(x_p | x) > 0$ for all $x_p, x$ then you can construct a counter example by having some $\pi(x_p | x) = 0$ and another $\pi(x | x_p) > 0$. ...


1

No, recall that the distribution of $X_t$ converges, but the process is ergodic - the measure defined by $X$ can be interpreted as the proportion of time spent by $X_t$ in a set as $t$ gets large. In particular, $\lim_t X_t$ does not exist.


0

Indeed $f(n, t_1, \ldots, t_n) = \dfrac{\lambda^n \mathsf e^{-\lambda t}}{t^n}$ . $$\begin{align}f(n, t_1, \cdots, t_n) =&~\Big(\prod_{i=1}^n{t^{-1}\lambda e^{-\lambda (t_i - t_{i-1})}}\Big) \times e^{-\lambda (t - t_n)}\\[1ex] =&~ t^{-n} \lambda^n\exp({-\lambda \sum_{i=1}^n(t_i-t_{i-1})}-\lambda (t-t_n)) \\[1ex] =&~ t^{-n} \lambda^n\exp(-\...


4

Actually, the answer to this lies in quite a bit more advanced topic called rough path theory (beware: PDF). A rough path is a way of "enhancing" a $\alpha$-Hölder continuous path with some extra information. A rough path is an ordered pair, $\textbf{X}=(X, \mathbb{X})$ where $X\colon [0,T]\to V$ where $V$ is some Banach space (typically $\Bbb{R}$) and a ...


0

Set $f(x,t)=xe^{\lambda t}$. By application of Ito's lemma, we have $$df(t,X_t)=\frac{\partial f}{\partial t}(t,X_t)dt+\frac{\partial f}{\partial x}(t,X_t)dX_t+\frac 12\frac{\partial^2 f}{\partial x^2}(t,X_t)d[X_t,X_t]$$ therefore $$df(X_te^{\lambda t})=\lambda X_te^{\lambda t}dt+e^{\lambda t}(-\lambda X_t dt+dW_t)$$ as a result $$d(X_te^{\lambda t})=e^{\...


0

$\mathfrak{U}(C)$ is by definition the minimal $\sigma$-Algebra such that the functions $p_t: x\mapsto x_t$ are $\mathfrak{U}(C)$-$\mathscr{B}(\mathbb{R})$-measurable for $t \in [0, \infty)$. Now, it can easily be shown that the system of sets \begin{align*} \mathscr{A}:=\left\{ A\in \mathscr{B}(\mathbb{R}): p_t^{-1}(A) \in \sigma(\left\{ p_t^{-1}(B):B\in ...


0

I think the question should have been put differently: Under the assumption that there is a fixed but unknown probability of heads $p$ can one predict or estimate its value with some degree of accuracy based on the results of the previous tosses(our data). Now, the LLN says that under specific conditions when an experiment is repeated ad ...


0

So, not that I want to answer my own question, but I seem to have been able to work out the first two moments at least. Lemmas/Facts Let $\{B^i\}$ and $\{W^i\}$ be standard independent Brownian motions. $\mathbb{E}[B_t^i] = 0$, $\;\mathbb{E}[(B_t^i)^2] = t$, $\;(B_{t+k}^i-B_{t}^i)\sim\mathcal{N}(0,k)$. [1] If $\theta\sim\mathcal{N}(0,\sigma^2)$, $\;\...


0

Section 5.4 on p. 13 of this document, seems to confirm the notion that we can extend the class of allowable integrands to locally square-integrable processes (or at least when we restrict to predictable processes, instead of general progressively measurable processes). Not only that, but apparently we can also expand the class of integrators to include ...


3

This is a great question that digs at the heart of how we use probabilities. There are several great answers, but I'd like to offer one which is more intuitive in nature. The key to understanding the issue you are dealing with is understanding your assumptions. Let us say that you have a coin which, when flipped, lands heads 50% of the time and tails 50% ...


1

If a coin came up heads 1000 times in a row, I'd suspect that it was NOT a fair coin. Maybe both sides are heads, or the tails side is heavier, or there are hidden magnets involved, or some other magic trick. If I saw a coin come up heads 1000 times in a row, I'd bet that the next toss will be heads again, not tails. But assuming that we somehow know that ...


-1

Also, to play devil's advocate, remember there is also the Gambler's Fallacy theorem. "I will wager a bet on either red or black. There is a 1 in 2 chance of either outcome. There has been 49 red, so the next outcome must be black!" Each incidence is a specific, discrete event that is unrelated to the other (all other factors considered, in relation to the ...


1

A given process has a certain probability distribution. This distribution does not change due to the fact that you have performed that process finite many times and got a sequence of results that is very unlikely (unless you alter the process in that way). In principle you can only get a knowledge of this probability distribution if you perform the process ...


2

While we know that the proportion of heads we get from flipping a fair coin should be about half, it's worth looking at the expected amount of deviation: If you flip a coin a million times, it's fairly rare to get exactly half a million heads: that happens with less than 0.1% probability! Even having the error be 100 or less is uncommon: that happens with ...


0

There is a remote possibility that a coin could be made where the result of one throw would influence the result of the next throw. Well, just make a coin that weighs 50kg and it will probably come up the same way all the time :-) Ignoring that possibility, the probability for that coin for throwing heads or tails doesn't change. However, each throw gives ...


3

As Did mentioned, $[W_1,W_2](t)=0$. Let $\{t_i\}_{i=1}^{n}$ be a partition of $[0,t]$. We want to consider $$A_n=\sum_{i=0}^{n-1}(\,W_1(t_{i+1})-W_1(t_{i})\,)(\,W_2(t_{i+1})-W_2(t_{i})\,)$$ Using independent of $W_1$ and $W_2$ , thus $\mathbb{E}[A_n]=0$. Since increment of Wiener process are independent, the variance of sum is sum of variance , and we have ...


2

Basic understanding of probability suggests that the probability of flipping heads is .5 and tails is .5 If that is what you believe, then I have a coin here of uncertain provenance and a wager you might be interested to undertake. I will of course call heads or tails, all you are required to do is put down your $100. Suddenly you aren't so confident it'...


4

Probability is a mathematical model. You need to separate the modeling process from the application of the model. We can model a coin toss by a two-element probability space {H,T} with a probability of 0.5 assigned to each element. If you sample 100 times from this probability space, and get 100 heads, then the probability of a head on the next sampling is ...


50

If you don't know whether it is a fair coin to start with, then it isn't a dumb question at all. (EDIT) You ask if the coin will be biased towards Tails to account for the all of the heads. If the coin was fair, then the answer from tilper addresses this well, with the overall answer being "No". Without that assumption of fairness, the overall answer ...


32

You have wandered into the realm of Bayesian versus frequentist statistics. The Bayesian philosophy is most attractive to me in understanding this question. Your "basic understanding of probability" can be interpreted as a prior expectation for what the distribution of heads and tails should be. But that prior expectation actually is itself a ...


20

Is the probability of flipping heads/tails still .5 each? If you already had the assumption that the coin is fair, then yes. Although statistically it's highly unlikely to flip heads a "very large number of times" in a row, the probability will not be changed by past outcomes. This is because each coin toss is independent from all other coin tosses. ...


1

Since $X(0) = Y(0) = 1$, the process obviously cannot live on a unit circle. But it does live on a circle of radius $\sqrt{2}$. In fact, it is easy to see that the solution to your system is given by $X(t) = \sqrt{2}\cos (W(t)+\frac\pi4)$, $Y(t) = \sqrt{2}\sin (W(t)+\frac\pi4)$. Indeed, $X(0)=Y(0)=1$ and, by Itô's lemma, $dX(t) = -Y(t) dW(t) - \frac12 ...


0

To show that the system reaches an equilibrium distribution for all $\mu, \lambda$ you must show that $P_i(t) \to P_i(\infty)$ for $i$. This has to be done from the set of differential equations for the $P_i$. Start with $P_0$ and show that it satisfies $P_0' = \mu(1-P_0) - \lambda P_0$. Find the solution and show that it has a limit. Then look at $P_1$. ...


1

Let me try to indicate an answer to what I think is your real question ("Is there a direct way to show that a continuous Lévy process is square integrable?") A detailed discussion of this and much more, using stochastic calculus, can be found on Geo. Lowther's blog, specifically https://almostsure.wordpress.com/2010/09/15/processes-with-independent-...


0

I personally like Finite Difference (FD) approaches, as they are quite suitable for this problems where the domain is generally rectangular, and you do not need too much work for evaluation of the Boundary Conditions. What you wrote is a perfectly reasonable Explicit Euler (EE) scheme, and you can expect it to converge first order in time and second order ...


1

The only continuous Lévy processes are Brownian motion (with or without drift). There are Lévy processes that are not square-integrable; viz. the Cauchy processes.


0

I would use an ADI scheme to solve such a problem, as you can isolate each variable, and ultimately , in your case, solve tridiagonal matrix equations, while profiting from the Crank-Nicholson accuracy. I googled some scheme , and I found that link here. Steps : Set up your 2D grid of state variables $(S_t,\sigma_t)$, and your time grid. Fill your ...


0

Not sure if I can answer satisfactory, but I'll try. This is not $\sqrt 3$. The claim is that the average walking distance for n steps approaches $\sqrt n$ for $n \rightarrow \infty$, not exact equality. I visualize this by thinking of the random walk as the sum of $n$ Bernoulli experiments with the outcomes ${-1,+1}$, meaning go left or right. ...


2

First of all, what does $o(h)$ (it is a class of functions, called Landau or Big- O notation) actually mean? Generally it means for a function $f\in o(g)$ that we have $$ {\displaystyle \lim _{x\to a}\left|{\frac {f(x)}{g(x)}}\right|=0} $$ As an example, take a differentiable function $f$, then we have for $h\to0$ the following $$ f(x+h)=f(x)+hf'(x)+o(h) $$ ...


0

I think I have figured out a way to prove that $v(x)$ is differentiable. The idea is to use the sort of techniques used in viscosity solution for optimal control. I will give a sketch of the argument below. My proof hardly makes use of the structure given above. In fact, I think, for a piecewise Lipschitz, bounded $\mu,\sigma$ and $f$, the same reasoning ...


1

It isn't $1$, it is $1+o(h)$. The $o(h)$ term is combined with other $o(h)$ terms so it looks like it disappears.


3

The inequality which both ziT and I used is a direct consequence of the following inequality. Lemma 1: Let $X_1,\ldots,X_n$ be independent random variables and $S_k := \sum_{j=1}^k X_j$, $k=1,\ldots,n$. Then for any $a,b \geq 0$ $$\mathbb{P} \left( \max_{1 \leq j \leq n} |S_j|>a+b \right) \leq \frac{\mathbb{P}(|S_n|>a)}{\mathbb{P} \left( \max_{1 \...


1

You have to make further assumptions on $w$, since, for instance, the solution of $$ \ddot x + kx = \sin(t\sqrt{k}) $$ is not bounded, even if $\sin(t\sqrt{k})$ is smooth and bounded. You have to avoid resonance, i.e. completely avoid a particular frequency in the spectrum of your brownian noise.


0

Note that \begin{align*} X_t = X_0 e^{\alpha t} + \sigma\int_0^t e^{\alpha(t-s)} dW_s. \end{align*} Therefore \begin{align*} E(X_t) = X_0 e^{\alpha t}, \end{align*} and \begin{align*} E\left(X_t^2\right) &= E\left(\left(X_t-E(X_t)\right)^2\right) + \left(E\left(X_t\right) \right)^2\\ &=\sigma^2 \int_0^te^{2\alpha(t-s)} ds + X_0^2 e^{2\alpha t}\\ &...


0

Let $f(x) = \ln(1+x^2)$. Then \begin{align*} |f'(x)| = \frac{2|x|}{1+x^2} \le 1. \end{align*} Therefore the Lipschitz condition is satisfied, and there is a unique strong solution.



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