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4 votes
1 answer
622 views

Why is every Gaussian process a linear process?

In Section 4.2.4 of [1], the authors write In this section we consider a causal linear process $$ X_t = \sum_{j = 0}^\infty a_j \varepsilon_{t - j}, \quad t \in \mathbb{N}, $$ where, without loss of ...
AlbertRapp's user avatar
4 votes
1 answer
337 views

Support of bivariate joint distribution of stationary and ergodic sequence

Let $\{X_t\}_{t\in \mathbb{N}}$ be a strictly stationary and ergodic sequence of real valued random variables and let the support of $X_1$ equal $[-1,1]$. Can the support of $(X_1,X_2)$ equal the unit ...
user424747's user avatar
3 votes
0 answers
98 views

Probability measure on $\mathbb{R}^n$ with given marginals and given correlation matrix

In all what follows, let $\mathcal{P}(\mathbb{R}^n)$ denote the set of probability measures on $(\mathbb{R}^n, \mathcal{B}(\mathbb{R}^n))$ and $\mathcal{C}_n$ the set of $n \times n$ correlation ...
Tom's user avatar
  • 279
2 votes
1 answer
120 views

Approximation of a stationary process by a sequence of ergodic and stationary sequence of stochastic processes

Let $X = [X_t : t \in \mathbb{Z}] \sim P$ and $Y = [Y_t : t \in \mathbb{Z}]\sim Q$ be two stochastic processes. Let's define the Mallows metric. Let $\mathcal{M}_m$ be the random vectors $(X,Y)$ ...
Fam's user avatar
  • 135
2 votes
1 answer
65 views

On the stationarity of Gaussian processes

I am trying to understand and prove the statement: The normal (or Gaussian) process is stationary in the wide sense if and only if it is strictly stationary. I know the following: A strictly ...
MathematicalMind1618's user avatar
2 votes
0 answers
432 views

Cointegration of (multiple) time series

Some time ago I stumbled upon the notion of cointegration of time series (see the wiki for some basic fact). Unfortunately, my knowledge of time series is a bit sketchy, and moreover I was able to ...
Mirco A. Mannucci's user avatar
1 vote
2 answers
236 views

Counterexample for absolute summability of autocovariances of strictly stationary strongly mixing sequence

Suppose $(X_i)_{i\in\mathbb{Z}}$ is a strictly stationary, strongly (i.e. $\alpha-$)mixing sequence of real random variables. If we have $\mathbb{E}[|X_1|^{2+\epsilon}]<\infty$ for some $\epsilon&...
Dasherman's user avatar
  • 203
1 vote
1 answer
77 views

Why shocks are independent with weighted sum of normal process

I am doing a problem and got stuck by the definition of "normal process". The problem is stated as follows: Suppose $e_t = \sum_{j}^{\infty}\theta^j Y_{t - j} $ and assume that $Y_t$ is a ...
tobinz's user avatar
  • 119
1 vote
1 answer
372 views

Calculate Average and Correlation of WSS Random Processes

Given two stochastic processes, $X[n]$ and $Y[n]$, both being WSS (wide state stationary) and independents. What would be the Average and Autocorrelation function of $Z[n] = Y[n] X[n]$? Is the ...
Sergio's user avatar
  • 11
1 vote
0 answers
81 views

Urn model with delayed replacement

Suppose I have x red and y blue balls. At each timestep I draw a ball with probability $$P(\text{red ball}) = (x/(x+y))^z, P(\text{blue ball}) = 1-P(\text{red ball})$$ where z is fixed. Each ball is ...
timbuktu's user avatar
0 votes
0 answers
30 views

Distribution of bivariate vectors for strictly stationary processes

Consider a strictly stationary process $X_t$, $t\in\mathbb{Z}_{\geq 1}$. Could you help me to disprove the following statement: "For $t, s > 0$, the bivariate vectors $(X_s, X_t)$ and $(X_t, ...
iom10's user avatar
  • 23
0 votes
1 answer
66 views

Estimating operators of functional linear processes

Let $H= L^2[0,1]$ be the space of measurable and square integrable functions from $[0,1]$ to $\mathbb{R},$ let $(\varepsilon_k)_{k \in \mathbb{Z}}$ denote the iid (or strict stationary) $H$-valued ...
Obriareos's user avatar
  • 195
-4 votes
1 answer
303 views

Reference request in optimal stopping [closed]

I am given the following task. Distributed over a trading day, I am supposed to buy a certain quantity of a good. The price of this good changes during the day. The goal is to buy the required ...
Bettina Kraus's user avatar