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I want to find a 1-$\alpha$ confident region for a value $\left(\Gamma_1,\Gamma_2\right)$. (Which are not random variables)

I can obtain diferent $m_i$ and $b_i$ that satisfy $\Gamma_2=m_i\Gamma_1+b_i$.

With $\left(m_i,b_i\right)=f\left(Y_{1i},\ldots,Y_{ki},X_1,\ldots,X_l\right)$.

And $X_j\sim\mathcal{N}(\mu_j,\sigma^2)$ ; $Y_{ji}\sim\mathcal{N}(\theta_{ji},\sigma^2)$.

The equality holds when $\left(m_i,b_i\right)=f\left(\theta_{1i},\ldots,\theta_{ki},\mu_1,\ldots,\mu_l\right)$.

Then, I have a Linear System with Random Coefficients.

\begin{alignat*}{4} m_1\Gamma_1 & {}-{} & \Gamma_2 ={} & -b_1 \\ m_2\Gamma_1 & {}-{} & \Gamma_2 ={} & -b_2 \\ & {}\vdots{} & \vdots{} & \\ m_n\Gamma_1 & {}-{} & \Gamma_2 ={} & -b_n \\ \end{alignat*}

Formally, I want to find a confidence region for the solution. That means a region S so that $(1-\alpha)100\%$ times contains $\left(\Gamma_1,\Gamma_2\right)$. Is this a well know problem? I can't find anything on the web. Could someone point me to a book or paper where this is done? It would be of great help.

I have tried finding $(1-\alpha)100\%$ simultaneous confidence intervals (Bonferroni method) for all $m_i$ and $b_i$ . Then, for each pair of slope and y-intercept I got a region $S_i$ of possible values of $(\Gamma_1,\Gamma_2)$. I take as the region $S=\cap S_i$. Unfortunately the regions I obtain are bigger than one would expect. And they are extremely conservative, Over a 10000 samples simulation it held the value every time.

Thanks in advance for taking the time to read the question :)

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    $\begingroup$ There's much you still need to tell us in order to communicate your question. Do you know the parameters $\mu_j$, $\sigma$, $\theta_{ji}$? Do you know $f$? If you do, then does the form of $f$ guarantee that there exist unique $\Gamma_i$ for which $m_i\Gamma_1-\Gamma_2=-b_i$? What do you actually observe--the $m_i,b_i$, the $X_j$ and $Y_{ji}$, something else? $\endgroup$ – whuber Jun 10 '13 at 19:37
  • $\begingroup$ I observe $X_j$ and $Y_{ji}$. $f$ is a quotient of equally distributed polinomial operations with $X_j$ and $Y_{ji}$ . And with 3 measures $\Gamma_i$ exists and is unique. $\endgroup$ – Manuel Jun 10 '13 at 19:44
  • $\begingroup$ I'm sorry, I still can't make sense of this. You explicitly define $f$ has having two values, $(m,b)$, whereas a quotient of polynomials will be a single value. Do you know the coefficients of $f$ (whatever it may be)? $\endgroup$ – whuber Jun 10 '13 at 19:49
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    $\begingroup$ You are right. I missexplaiend myself. I am very sorry for my english too.$\left(m_i,b_i\right)=\left(f_1(\ldots,Y_{ji},\ldots,X_j,\ldots),f_2(\ldots,Y_{ji},\ldots,X_j,\ldots)\right)$. Both $f_1$ and $f_2$ are quite complex to express here. They are rational functions of polinomyals where no variable has a power grater than 2. The function form is known, i just evaluate it in the observations. $\endgroup$ – Manuel Jun 10 '13 at 19:59
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    $\begingroup$ OK, I think I get it now: in effect, you observe $\{(m_i,b_i)\}$ and assume a somewhat complicated model for their joint distribution. The $\Gamma_i$ are just regression coefficients, which becomes clear if we change to a more usual notation where you observe $(x_i,y_i)$ and wish to estimate $\beta_0$ and $\beta_1$ for which $y_i=\beta_0+\beta_1x_i$; the translation is $x_i=m_i$, $y_i=b_i$, $\beta_0=\Gamma_2$, and $\beta_1=-\Gamma_1$. (A) Is this a correct interpretation? (B) If so, is there any obstacle to applying the usual Maximum Likelihood machinery? $\endgroup$ – whuber Jun 10 '13 at 20:08

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