MHB Prove Weakly Stationary Process w/$E(X^2_t ) < ∞$

AI Thread Summary
A strictly stationary process with a finite second moment, $E(X^2_t) < ∞$, is proven to be weakly stationary by demonstrating that its mean value function does not depend on time and that the autocovariance is independent of time for any given lag. The mean value function $E[X_t]$ is constant over time due to the independence of the distribution of $X_t$ from $t$. The second moment's finiteness ensures that the joint distribution of $X_t$ and $X_{t+h}$ is well-defined and independent of $t$. Thus, both points in the solution are valid, clarifying that without the assumption of a finite second moment, the necessary quantities for weak stationarity may not exist. Understanding these principles is crucial for grasping weak stationarity in stochastic processes.
nacho-man
Messages
166
Reaction score
0
Not 100% sure if this is the right board.

My question is to
Show that a strictly stationary process with $E(X^2_t ) < ∞$ is weakly stationary.

So weakly stationary implies two things:

- the mean value function $u_t$ does not depend on time $t$
and
- the autocovariance $\gamma_x(t+h,t)$ is independent of $t$ for each $h$

The first point is fairly intuitive, but just to clarify, the second point is saying that the ACF does not depend on the time $t$ at any point when finding the autocovariance between two different points in time for any given $h$ ?Anyway, continuing on to the question,

the solutions say:

$E[X_t]$ is independent of $t$ since the distribution of ${X_t}$ is independent of $t$ and $E[X_t]$ exists.
So that proves the first point, but isn't this just an obvious re-iteration of the first definition of a weakly stationary set?
How does the 2nd moment being finite allow us to conclude that the distribution of ${X_t}$ is independent of $t$?

The 2nd point of the solution says: the joint distribution of $X_t , X_{t+h}$ is independent of t, hence ${X_t}$ is weakly stationary.

Are these both valid solutions? It seems as if just the definition of a weakly stationary set has been provided. I feel a little confused as to why these conclusions cannot be made if $E^2[X] \nless \infty$

Any help with understanding this is very appreciated.
 
Physics news on Phys.org
Hello,

The assumption of finiteness of $\mathbb E(X_t^2)$ is done because otherwise the quantities $\mathbb E(X_tX_{t+h})$ involved in the definition of weak stationarity do not necessarily make sense.

The solution is valid.
 
Back
Top