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Proving asymptotic covariance of Cross-Spectrum Estimation I came across this mathematical problem:

Consider two constants $M,L$ such that $M\to \infty, L \to \infty \hbox{ and } \frac{L}{M} \to 0$, And $\ell,s,r \in \mathbb{Z}$ $\nu_1,\nu_1 \in \mathbb{R}$

$$ \lim_{M\to +\infty} \left\{\frac{1}{L}\sum_{\ell=-M}^{M} U_M(r,\ell,s) w\left(\frac{\ell+s}{L}\right)\ w\left(\frac{\ell}{L}\right) e^{-2\pi i\ell(\nu_1-\nu_2)}\right\}$$

where $r$, $s$ are constants and:

$$ U_M(r,\ell,s)= \left\{\begin{matrix} 0 & r<-M+\ell+s\\ 1-\frac{\ell+s+r}{M}& -M+\ell+s\leq r \leq 0 \\ 1-\frac{\ell+s}{M}& 0 < r \leq s\\ 1-\frac{\ell+r}{M}& s<r \leq M-\ell\\ 0 & r>M-\ell \end{matrix}\right.$$ And $w(x)$ is a function which is even, bounded, square-integrable and such that $w(0)=1$.

It has been proven [1] that this limit exists and it is equal to 0 for $\nu_1\neq\nu_2$ and : $$=\int_{-\infty}^{\infty}w^2(x)dx $$ for $\nu_1=\nu_2$. I am trying to prove that replacing in the limit $U_M(r,\ell,s)$ by

$$V_M(r)= \left(1-\frac{|r|}{M}\right)$$

Produces the same result but I don't have the background to understand the mentioned proof, is there a simple way to prove this claim? I am only interested in the case of $V_M(r)$.

[1] Hannan, E. J. Multiple time series, vol. 38. John Wiley & Sons, 2009. Proof of Theorem 9 chapter V, page 314.

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    $\begingroup$ Would math.SE be a better fit for this question? What is the conexion to signal processing? $\endgroup$ – MBaz Jul 17 '17 at 18:21
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    $\begingroup$ Ya, but Ted Hannan always thought of himself as a statistician, so Math.SE might be better. Or crossvalidated.SE. Let me know and I can migrate it. $\endgroup$ – Peter K. Jul 17 '17 at 18:47
  • $\begingroup$ I'm studiying asymptotic covariance for a specific Blackman-Tukey type correlogram. Here, w(x) is the tappering window. Sorry for not making it clear. $\endgroup$ – Augusto Zebadúa Jul 17 '17 at 18:54
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    $\begingroup$ The triangle window was kind of apparent. Some notations in your question are uneven, and don't help the reading (like arguments to $U_M$), and the constants that tend to infinity, well. I believe you could further simplify the question to forster answers $\endgroup$ – Laurent Duval Jul 17 '17 at 18:59
  • $\begingroup$ The definition of function U is problematic. If r and s are fixed integers, then according to the second piece-wise definition condition r must be less than zero (hence it's a negative number) but then according to the third and fifth definitions, r is greater than zero ? The same also happens for the s variable. The conditions to be satisfied by those fixed integers seem to be inconsistent... $\endgroup$ – Fat32 Jul 17 '17 at 19:27
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A first thing to try is to reduce the unwanted variables. What happens if you set $\ell=s=0$?

$$ U_M(r,0,0)= \left\{\begin{matrix} 0 & r<-M\\ 1-\frac{r}{M}& -M\leq r \leq 0 \\ 1-\frac{0}{M}& 0 < r \leq 0\\ 1-\frac{r}{M}& 0<r \leq M\\ 0 & r>M \end{matrix}\right.$$

Luckily, this definition seems to fit with $V_M$, since the third inequality never happens. So your $V_M(r)$ is just a special case. Nothing more to do.

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    $\begingroup$ The definition of function U is problematic. If r and s are fixed integers, then according to the second piece-wise definition condition r must be less than zero (hence it's a negative number) but then according to the third and fifth definitions, r is greater than zero ? The same also happens for the s variable. The conditions to be satisfied by those fixed integers seems to be inconsistent... $\endgroup$ – Fat32 Jul 17 '17 at 19:22
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    $\begingroup$ where, r,s are constants and just look at the line below the limit... $\endgroup$ – Fat32 Jul 17 '17 at 19:29
  • $\begingroup$ Now M,L are positive integers then the first condition says r is negative, while the last condition says r is positive. Only one of them can be satisfied...Probably he is making some typos. $\endgroup$ – Fat32 Jul 17 '17 at 19:30
  • $\begingroup$ sure there's a typo, but not sure which one :-)) $\endgroup$ – Fat32 Jul 17 '17 at 19:32
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    $\begingroup$ There were in fact many typos. I think everything is consistent now. $\endgroup$ – Augusto Zebadúa Jul 17 '17 at 19:58

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