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I have the following argument in my Time Series class notes:

Let ${u_t}$ be a mean zero covariance stationary process.

Define $\gamma(j) = \mathbb{E}u_tu_{t-j}$ and $Y_t = \mu +C(L)u_t$ where $C(L)=\sum_{j=0}^{\infty}c_jL^j$

The objective is that by using Chebyshev's inequality I can show that $P(|X - \mathbb{E}X| >\epsilon) \le Var(X)/\epsilon^2$ and therefore show that $n^{-1}\sum_{t=1}^n Y_t \rightarrow_p \mu$.

The answer to this question goes as follow:

$P(|X - \mathbb{E}X| >\epsilon) \le \frac{Var(n^{-1/2}\sum_{t=1}^{n}(Y_t - \mu))}{n\epsilon^2}$ by Chebyshev.

$\frac{Var(n^{-1/2}\sum_{t=1}^{n}(Y_t - \mu))}{n\epsilon^2} = \frac{2\pi fy(0)}{{n\epsilon^2}}$ where $fy(\lambda)$ denotes the spectral density of $Y$. If we denote the spectral density function of $u$ we have that

$fy(\lambda) = f_u(\lambda)\mid \sum_{j=0}^{\infty}c_je^{-i\lambda j}\mid^2$

and therefore $2\pi fy(0) = (\sum_{j=- \infty}^\infty \gamma(j)) (\sum_{j=0}^\infty c_j)^2$

My question is, where does the spectral density $2\pi fy(0)$ comes from? I would like to have an explanation as for why we can use the spectral density. Quite frankly, I don't really understand the implication of spectral density and its usage in this problem.

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I guess the correction meant an inequality. Write that $$\operatorname{Var}\left(\sum_{t=1}^n(Y_t-\mu)\right)=C(L)\sum_{k=1}^n(n-k)\gamma(k),$$ which gives $$\operatorname{Var}\left(n^{-1/2}\sum_{t=1}^n(Y_t-\mu)\right)\leqslant C(L)\sum_{k=1}^n\gamma(k).$$ Now we can conclude if $\sum_k\gamma(k)$ is finite, which seems to be assumed.

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  • $\begingroup$ Yes your right, this is basically the point of this question is to prove both $c_j$ and $\gamma_j$ are finite but in my answer sheet I wish to understand why we must go through the spectral density. Thanks $\endgroup$
    – ChuckM
    Apr 6, 2013 at 17:42
  • $\begingroup$ Because when you compute the variance, you write is a double sum of expectations for indexes $i,j$ from $1$ to $n$. By weak stationarity, the terms depend only on the distance $|i-j|$. $\endgroup$ Apr 6, 2013 at 17:46
  • $\begingroup$ ok thanks for the help $\endgroup$
    – ChuckM
    Apr 6, 2013 at 18:15
  • $\begingroup$ You are welcome. $\endgroup$ Apr 6, 2013 at 18:51

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