2
$\begingroup$

I want to know how the authors of this arxiv paper (p. 10) solved the equation \begin{align} g\left(\lambda\right) ={}& \frac{1}{2\pi}\sum_{\omega\in\left[0,2\pi\right]:f\left(\omega\right)=\lambda} \frac{1}{\left|f'\left(\omega\right)\right|} \tag{1a} \\ \overset{\text{?}}{=}{}& \frac{1}{\pi(\vartheta+\varphi\lambda)\sqrt{\left[(1+\vartheta)^{2}-\lambda(1-\varphi)^{2}\right]\left[\lambda(1+\varphi)^{2}-(1-\vartheta)^{2}\right]}}\mathbf{1}_{(\lambda_{-},\lambda_{+})}(\lambda) \text{,} \tag{1b} \end{align} with \begin{equation} \lambda,\vartheta,\varphi\in\mathbb{R}\text{,}\;\left|\varphi\right|<1\text{,} \quad \lambda_{-} = \min{(\lambda^{-},\lambda^{+})}\text{,} \quad \lambda_{+} = \max{(\lambda^{-},\lambda^{+})}\text{,} \quad \lambda^{\pm} = \frac{(1\pm\vartheta)^{2}}{(1\mp\varphi)^{2}} \text{.} \end{equation} $f$ is the Fourier transform of the autocovariance function of a $\operatorname{ARMA}\left(1,1\right)$ process with $\operatorname{MA}\left(1\right)$ polynomial $a$ and $\operatorname{AR}\left(1\right)$ polynomial $b$. It is given by \begin{equation} f\left(\omega\right) = \left|\frac{b\left(\operatorname{e}^{\mathsf{i}\omega}\right)}{a\left(\operatorname{e}^{\mathsf{i}\omega}\right)}\right|^{2} = \frac{1+\vartheta^{2}+2\vartheta\cos\left(\omega\right)}{1+\varphi^{2}-2\varphi\cos\left(\omega\right)} \text{,} \tag{2} \end{equation} for $\omega\in\left[0,2\pi\right]$. The derivative of $f$ is given by \begin{equation} f'\left(\omega\right) = -\frac{2\left(\varphi+\vartheta\right)\left(1+\varphi\vartheta\right)\sin\left(\omega\right)}{\left(1+\varphi^{2}-2 \varphi \cos (\omega )\right)^2} \text{.} \tag{3} \end{equation}

Any ideas on how to solve equation (1)? The solution looks as if it was calculated by use of the residue theorem, but I do not know how to start (or how to transform the implicit sum to an integral).

Any help is much appreciated.

$\endgroup$
  • $\begingroup$ it reminds me at the identify $\delta(f(x))=\sum_{x_i,f(x_i)=0} \delta(x-x_i)$ but i'm not 100% sure if this is the right direction $\endgroup$ – tired Aug 21 '15 at 16:41
0
$\begingroup$

I just visited this problem again and made the effort to calculate the function $g\left(\lambda\right)$ by hand.

As a result, I noticed that the authors omited a factor $\left(\vartheta+\varphi\right)\left(1+\vartheta\varphi\right)$ in equation (1b). The following equation (3.2) in their paper is right again, so it seems the authors simply forgot the mentioned factor.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.