How can I prove monotonicity of slope MLE in EIV regression model?

Question

I'm trying to figure out Casella and Berger Exercise 12.4(c), regarding monotonicity of the maximum likelihood estimator of the slope of an errors-in-variables regression model. The goal is to show that $$ \begin{align*} \hat{\beta}(\lambda) &= \frac{-(S_{xx} - \lambda S_{yy}) + \sqrt{(S_{xx} - \lambda S_{yy})^{2} + 4 \lambda S_{xy}^{2}}}{2 \lambda S_{xy}} \end{align*} $$

is monotonic in $\lambda$, increasing if $S_{xy} > 0$ and decreasing if $S_{xy} < 0$.

The derivative is $$ \begin{align*} \hat{\beta}'(\lambda) &= \left(\frac{S_{xx}\sqrt{(S_{xx} - \lambda S_{yy})^{2} + 4\lambda S_{xy}^{2}} + \lambda S_{xx}S_{yy} - 2\lambda S_{xy}^{2} - S_{xx}^{2}}{\sqrt{(S_{xx} - \lambda S_{yy})^{2} + 4\lambda S_{xy}^{2}}}\right) \left(\frac{1}{2\lambda^{2} S_{xy}}\right) \end{align*} $$

If I can show that the numerator ($S_{xx}\sqrt{(S_{xx} - \lambda S_{yy})^{2} + 4S_{xy}^{2}\lambda} + \lambda S_{xx}S_{yy} - 2S_{xy}^{2}\lambda - S_{xx}^{2}$) of the left-hand term is positive, then I'm finished.

We have the constraints $S_{xx} > 0$, $S_{yy} > 0$, $S_{xy}^{2} \le S_{xx}S_{yy}$.

I've been working on this for far more hours than I care to admit, and I'm no closer to the goal than when I started. Every attempt to use the $S_{xy}^{2} \le S_{xx}S_{yy}$ constraint to derive a useful "initial term greater than 0" inequality hits a dead end.

I would appreciate any pointers.

Answer 1

I think I finally got it(?). This shows that the numerator is nonnegative, rather than positive as I'd hoped. For ease of notation, let $a = S_{xx}$, $b = S_{yy}$, and $c = S_{xy}$.

$$ \begin{align*} &a\sqrt{(a - b\lambda)^{2} + 4c^{2}\lambda} + b^{2}\lambda^{2} - ab\lambda + 2c^{2}\lambda - (a - b\lambda)^{2} - 4c^{2}\lambda \\ \qquad &= a\sqrt{(a - b\lambda)^{2} + 4c^{2}\lambda} + b^{2}\lambda^{2} - ab\lambda - 2c^{2}\lambda - (a - b\lambda)^{2} \\ &= a\sqrt{(a - b\lambda)^{2} + 4c^{2}\lambda} - ab\lambda - 2c^{2}\lambda - a^{2} + 2ab\lambda \\ &= a\sqrt{a^{2} - 2(ab - 2c^{2})\lambda + b^{2}\lambda^{2}} + (ab - 2c^{2})\lambda - a^{2} \\ &\ge a\sqrt{a^{2} - 2(ab - 2c^{2})\lambda + \left(b^{2} - \frac{4c^{2}}{a^{2}}\left(ab - c^{2}\right)\right)\lambda^{2}} + (ab - 2c^{2})\lambda - a^{2} \\ &= a\sqrt{a^{2} - 2(ab - 2c^{2})\lambda + \left(\frac{a^{2}b^{2} - 4abc^{2} + 4c^{4}}{a^{2}}\right)\lambda^{2}} + (ab - 2c^{2})\lambda - a^{2} \\ &= a\sqrt{a^{2} - 2(ab - 2c^{2})\lambda + \left(\frac{ab - 2c^{2}}{a}\right)^{2}\lambda^{2}} + (ab - 2c^{2})\lambda - a^{2} \\ &= a\sqrt{\left(a - \left(\frac{ab - 2c^{2}}{a}\right)\lambda\right)^{2}} + (ab - 2c^{2})\lambda - a^{2} \\ &= a\left(a - \left(\frac{ab - 2c^{2}}{a}\right)\lambda\right) + (ab - 2c^{2})\lambda - a^{2} \\ &= a^{2} - (ab - 2c^{2})\lambda + (ab - 2c^{2})\lambda - a^{2} \\ &= 0 \end{align*} $$