Covariogram estimation for the process contaminated with linear trend

[New_Galatea]

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

 

[lippyka]

!Solution: We can prove the result by using the law of large numbers. Let Y_1, Y_2, ..., Y_n be a sequence of independent and identically distributed random variables with finite mean μ and variance σ^2. By the law of large numbers, lim n -> ∞ (1/n) * ∑_(i=1)^n▒Y_i = μ and lim n -> ∞ (1/n) * ∑_(i=1)^n▒Y_i^2 = σ^2 with probability 1. For the process Y(t), we have Y(t) = S(t) + k(t-(n+1)/2) so (1/n) * ∑_(i=1)^n▒Y(t) = (1/n) * ∑_(i=1)^n▒S(t) + k * (1/n) * ∑_(i=1)^n▒(t-(n+1)/2) The first term on the right-hand side converges to the mean of S(t), μ_s, with probability 1. The second term on the right-hand side converges to 0 since the summand is a constant, k. Therefore, lim n -> ∞ (1/n) * ∑_(i=1)^n▒Y(t) = μ_s with probability 1. Similarly, for the squared values, (1/n) * ∑_(i=1)^n▒Y(t)^2 = (1/n) * ∑_(i=1)^n▒S(t)^2 + k^2 * (1/n) * ∑_(i=1)^n▒(t-(n+1)/2)^2 The first term on the right-hand side converges to the variance of S(t), σ_s^2, with probability 1. The second term on the right-hand side