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We have a following inequality:

$$ \sum(x_i - \bar{x})^2 (\beta - \hat{\beta})^2 + 2\sum(x_i-\bar{x})(z_i-\bar{z})(\beta-\hat{\beta})(\gamma-\hat{\gamma})+\sum(z_i-\bar{z})^2(\gamma-\hat{\gamma})^2 \le 2{\hat{\sigma}}^2F $$

Only $\beta$ and $\gamma$ are unknown terms, whereas all other values are known ($x_i$, $\bar{x}$, $\hat{\beta}$, $z_i$, $\bar{z}$, $\hat{\gamma}$, ${\hat{\sigma}}^2$, $F$).

This is supposed to denote an elliptical confidence region around some parameters.

I was trying to play around with the formula and I first moved the expression $2{\hat{\sigma}}^2F$ to the left and then changing the inequality sign to the equality sign so that I have:

$$ \sum(x_i - \bar{x})^2 (\beta - \hat{\beta})^2 + 2\sum(x_i-\bar{x})(z_i-\bar{z})(\beta-\hat{\beta})(\gamma-\hat{\gamma})+\sum(z_i-\bar{z})^2(\gamma-\hat{\gamma}) -2{\hat{\sigma}}^2F = 0 $$

Additionaly from the previous literature I know that:

$$ \gamma = -A\sqrt{1-\frac{\beta^2}{A^2}}$$

where the value of $A$ is also known. Thus the previous equation can be rewritten as:

$$ \sum(x_i - \bar{x})^2 (\beta - \hat{\beta})^2 + 2\sum(x_i-\bar{x})(z_i-\bar{z})(\beta-\hat{\beta})(-A\sqrt{1-\frac{\beta^2}{A^2}}-\hat{\gamma})+\sum(z_i-\bar{z})^2(-A\sqrt{1-\frac{\beta^2}{A^2}}-\hat{\gamma}) -2{\hat{\sigma}}^2F = 0 $$

Trying to solve this with some help from online sources has led me to the fact that this will be a quartic equation with four solutions, $\beta_1$, $\beta_2$, $\beta_3$ and $\beta_4$. I will therefore also have $\gamma_1$, $\gamma_2$, $\gamma_3$ and $\gamma_4$.

I am wondering what will these numbers tell me. Remember: I took the inequality defining the which points fall into an elliptical region on the graph and then changed it to the equality and rearranged the equation so that the right-hand term is zero.

Four values per $\beta$ and $\gamma$ would seem to form four points in the coordinate system. Would I be correct to assume that these four points would be the four points where the minor and the major axis touch the ellipsis? Or would these points represent something else?

KReiser
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J. Doe
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    should be a final square $ \sum(x_i - \bar{x})^2 (\beta - \hat{\beta})^2 + 2\sum(x_i-\bar{x})(z_i-\bar{z})(\beta-\hat{\beta})(\gamma-\hat{\gamma})+\sum(z_i-\bar{z})^2(\gamma-\hat{\gamma})^2 \le 2{\hat{\sigma}}^2F $I think letters $u,v$ are unused, let $u = \beta - \hat{\beta} $ and $v = \gamma-\hat{\gamma},$ pull those outside of each summation, you get summations $P,Q,R$ and right hand side $S$ in $Pu^2 + Q uv + Rv^2 \leq S $ which is a $u,v$ ellipse when $P,R \geq 0$ and $Q^2 - 4PR < 0 $ – Will Jagy Apr 29 '23 at 21:17
  • Alright, $Q^2 - 4 PR < 0$ follows from Cauchy-Schwarz – Will Jagy Apr 29 '23 at 21:19
  • @WillJagy By "final square", do you mean the square/rectangle into which the ellipse would be inscribed? So then the four points would be the four corners of that rectangle and taking half the distance between points would give me the points where the major and the minor axis of the ellipsis touch the ellipsis (and also touch the rectangle)? – J. Doe Apr 29 '23 at 21:28
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    No, I believe you made a typo. Just before the inequality sign you have $(\gamma-\hat{\gamma})$ when it should read $(\gamma-\hat{\gamma})^2$ – Will Jagy Apr 29 '23 at 21:31
  • @WillJagy You're right! Thanks, I corrected it! – J. Doe Apr 29 '23 at 21:32
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    Those four points are the intersections between the ellipse and a circle (defined by the equation for $\gamma$). – Intelligenti pauca Apr 29 '23 at 21:44
  • @Intelligentipauca Hm, I see. Well, I was trying to somehow get to the equation of that ellipse. If I knew these four points, would that allow me to do that? – J. Doe Apr 29 '23 at 23:36
  • I don't understand: you already have the equation of that ellipse, it's the second formula. Just replace $\beta$ and $\gamma$ with $x$ and $y$. – Intelligenti pauca Apr 30 '23 at 06:44
  • You might be looking for this post to turn your formula into center, radii and rotation. Or this post for more details on intersecting an ellipse with a circle. – MvG Apr 30 '23 at 19:27

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