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I have a regression model: $y_i=\exp(a \sin(\frac{2 \pi i}{n}) + b \cos(\frac{2 \pi i} {n})+\varepsilon_i)$ where a, b are the regression parameters. Let ${\varepsilon}_i = {\varepsilon}_i(t)$ be independent identically distributed random processes. I want to evaluate an accuracy.

Let $\hat{y} = \exp(\hat{a} \sin(\frac{2 \pi i}{n}) + \hat{b} \cos(\frac{2 \pi i} {n}))$ be the model with estimated parameters $\hat{a}, \hat{b}$.

Then, $\hat{\varepsilon}_i = \ln{y_i} - \ln{\hat{y_i}}$ is a residual on ith point.

$$Z_n(t)=\frac{1}{\sigma \sqrt{n}} \sum_{i=1}^{[nt]} \hat{\varepsilon_i}.$$

I need to find $\lim_{n\to\infty} Z_n(t)$ in distribution.

I tried to proceed as follows:

$$Z_n(t)=\frac{1}{\sigma \sqrt{n}} \sum_{i=1}^{[nt]} \hat{\varepsilon_i}=\frac{1}{\sigma \sqrt{n}} \sum_{i=1}^{[nt]}[\ln{y_i} - \ln{\hat{y_i}}]$$

$$Z_n(t)=\frac{1}{\sigma \sqrt{n}} \sum_{i=1}^{[nt]} [(a-\hat{a}) \sin(\frac{2 \pi i}{n}) + (b-\hat{b}) \cos(\frac{2 \pi i} {n})+\varepsilon_i]$$

$\frac{1}{\sigma \sqrt{n}} \sum_{i=1}^{[nt]} \varepsilon_i \to_{n\to\infty} N(0, 1)$. Here $N(0,1)$ is a standart normal distribution.

$$Z_n(t) \to N(0,1) + \lim_{n\to\infty}\frac{1}{\sigma \sqrt{n}} \sum_{i=1}^{[nt]} [(a-\hat{a}) \sin(\frac{2 \pi i}{n}) + (b-\hat{b}) \cos(\frac{2 \pi i} {n})]$$

Now I think we can simplify the sum limit to something like $(a-\hat{a}) + (b-\hat{b})$ and can get $Z_n(t)\to N(a-\hat{a} + b-\hat{b}, 1)$. My question is how to get it and is it all correct with my calculations?

Thanks in advance.

EDIT: Well, now I have got a partial solution of this problem...

First of all, we should get the OLS-estimators of $\hat{a}$ and $\hat{b}$.

Let $\hat{u}_i = \ln(\hat{y}_i)$.

Then, our model is $U = X \theta$, where $\theta=\begin{pmatrix} a \\ b \end{pmatrix}$, $X = \begin{pmatrix} \sin(\frac{2 \pi 1}{n}) & \cos(\frac{2 \pi 1} {n}) \\ ... & ... \\ \sin(\frac{2 \pi n}{n}) & \cos(\frac{2 \pi n} {n})\end{pmatrix}$

An assessment can be found by using the formula: $\hat{\theta} = (X^T X)^{-1} X^T U $.

After some calculations, $\hat{\theta} = \begin{pmatrix} 2 \overline{u_i \cos(\frac{2 \pi i}{n}}) \\ 2 \overline{u_i \sin(\frac{2 \pi i}{n}}) \end{pmatrix}$.

So, $\hat{\varepsilon} = u - \hat{u} = -n\cdot \overline{U}$. Here $\overline{U}$ is a mean value of $U$, i.e. $\frac{1}{n}\Sigma_{i=1}^n u_i$.

Now we want to describe a random process $Z_n(t)=\frac{1}{\sigma \sqrt{n}} \sum_{i=1}^{[nt]} \hat{\varepsilon_i}$.

AFAIK, Ian B. MacNeill's theorem is just about it, but I can't yet understand it... Could you please help me to complete solution here?

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  • $\begingroup$ Is $\sigma$ some arbitrary constant? $\endgroup$ – DanZimm May 31 '13 at 3:40
  • $\begingroup$ Yes, $\sigma$ is just a distribution parameter for $\varepsilon$ $\endgroup$ – Alexander Mihailov May 31 '13 at 3:42
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    $\begingroup$ You omit to say whether the $\varepsilon_i$ are independent or maybe at least uncorrelated. $\endgroup$ – Michael Hardy May 31 '13 at 3:44
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    $\begingroup$ Is the limit you're trying to find supposed to be (1) a limit in distribution or (2) a limit in probability or (3) an almost-sure limit or (4) something else? (I'm guessing (1).) $\endgroup$ – Michael Hardy May 31 '13 at 3:45
  • $\begingroup$ @MichaelHardy what is an "almost sure limit"? xD $\endgroup$ – DanZimm May 31 '13 at 3:46
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Okay, I have got the solution for the case of the model:

$$u_i = a \sin \frac{2 \pi i}{n} + b + \varepsilon_i$$

The original problem is similar to this one.

According to the Ian B. MacNeill's theorem, $Z_n \to B_g \in C[0,1]$ (in distribution, weak convergence), where $B_g$ is the Gaussian process $$M = 0$$ $$K_g(t,u) =\min(t,u) - \int_0^t \int_0^u g(x)^T\cdot G^{-1} \cdot g(y) dx dy,$$

here: $g(x) = X(\frac{i}{n}\to x)$, $G_{ij}=\int_0^1 g_i(x) g_j(x) dx$

So, in our problem:

$$g(x)=\begin{pmatrix} \sin (2\pi x) \\ 1 \end{pmatrix}$$

$$G_{11}=\int_0^1 \sin^2(x) dx = 1/2;\ G_{22}=\int_0^1 dx = 1$$

$$G_{12}=G_{21}=\int_0^1 \sin(x) dx = 0$$

$$G = \begin{pmatrix} 0.5 & 0 \\ 0 & 1 \end{pmatrix}; G^{-1} = \begin{pmatrix} 2 & 0 \\ 0 & 1 \end{pmatrix}$$

$$\star = g(x)^T\cdot G^{-1} \cdot g(y) = 2 \sin(2\pi x) \sin(2\pi y) + 1$$

$$\int_0^t \int_0^u \star dx dy = \frac{1}{2 \pi^2} \cos 2\pi t \cos 2\pi u + tu$$

Finally, the answer is: $$Z_n \to_{n\to\infty} B_g (0, \min(t,u)-\frac{1}{2 \pi^2} \cos 2\pi t \cos 2\pi u - tu$$

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