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Say that we have strong mixing coefficients that satisfy the following: $\alpha(m) = O(m^{-a-\epsilon})$ for some $\epsilon > 0$. If we have $h\in {\mathbb N}$ that is finite and $h>m$, I have in my course note the following: $\alpha(m-h) = O(m^{-a-\epsilon})$. Why is that? What's the proof that shows that $\alpha(m-h) = O(m^{-a-\epsilon})$?

Can we say that $\alpha(m-h)=O(m^{-a-\epsilon}) \equiv \alpha(m)=O(m^{-a})$? That's what I seem to understand from the proof in Davidson's book "Stochastic Limit Theory" (1994, p.211)

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  • $\begingroup$ @DavideGiraudo yes these tags are more appropriate. thanks $\endgroup$
    – ChuckM
    Mar 31, 2013 at 23:24

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Let $K$ be a constant such that $\alpha(m)\leqslant K\cdot m^{-a-\varepsilon}$ for each $m$. Then we have $$\alpha(m-h)\leqslant K(m-h)^{-a-\varepsilon}=Km^{-a-\varepsilon}\left(1-\frac hm\right)^{-a-\varepsilon}.$$ We conclude using the fact that the sequence $\left(\left(1-\frac hm\right)^{-a-\varepsilon},m\geqslant 1\right)$ is bounded.

Note that we didn't use features of mixing coefficients, but only properties of sequences.

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  • $\begingroup$ great! thanks! that's clear! I always have some kind of struggle with $Op(1)$ $op(1)$ and all that $\endgroup$
    – ChuckM
    Mar 31, 2013 at 22:14

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