Covariogram estimation for the process contaminated with linear trend

AI Thread Summary
The discussion focuses on estimating the covariogram R(h) for a process Y(t) that is contaminated by a linear trend, contrasting it with the covariogram Rs(h) of a stationary process S(t). It highlights that Y(t) is not second-order stationary due to the linear trend introduced by the contamination factor k. The goal is to demonstrate that the estimate of R(h) converges in probability to Rs(h) plus a specific term involving k and n. A participant clarifies the definition of a covariogram, emphasizing its role in measuring covariance in spatial processes. The thread seeks assistance in overcoming challenges related to this estimation.
New_Galatea
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Let {S(t), t=1,2,...} be a zero-mean, unit variance, second-order stationary process in R^1,
and define Y(t)=S(t)+k(t-(n+1)/2), t=1,2,...,n.
Then the process Y(t) is not second-order stationary process since it is contaminated with linear trend, k – degree of contamination.

Define R(h) – covariogram for Y(t) process and
Define Rs(h) - covariogram for S(t) process.

Could you help me to show that estimate of R(h) converges in probability to estimate of Rs(h) + ((k^2) * (n^2))/12

Thank in advance
 
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What is a covariogram?
 
As I know “Covariogram” is synonym of “Covariance”.
A strict definition is following:
Let x(t) be a spatial process. Covariogram for spatial process x(t) is a function
R(t1,t2)= M[(x(t1)-Mx(t1))(x(t2)-Mx(t2))].
Here M – symbol of mean.
 
Have you attempted a solution? Is there a specific obstacle you cannot get around?
 
I was reading a Bachelor thesis on Peano Arithmetic (PA). PA has the following axioms (not including the induction schema): $$\begin{align} & (A1) ~~~~ \forall x \neg (x + 1 = 0) \nonumber \\ & (A2) ~~~~ \forall xy (x + 1 =y + 1 \to x = y) \nonumber \\ & (A3) ~~~~ \forall x (x + 0 = x) \nonumber \\ & (A4) ~~~~ \forall xy (x + (y +1) = (x + y ) + 1) \nonumber \\ & (A5) ~~~~ \forall x (x \cdot 0 = 0) \nonumber \\ & (A6) ~~~~ \forall xy (x \cdot (y + 1) = (x \cdot y) + x) \nonumber...
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