I have the following situation:
I have two estimators of $\alpha$, both via maximum likelihood of the density:
$$ f(x,y\mid \alpha,\beta) = f(y \mid x,\alpha,\beta)f(x \mid \alpha) $$
One uses only $f(x \mid \alpha)$ and has variance $V_{x,\alpha \alpha}^{-1}$ via Fisher Information.
The other has variance $ (V_{x,\alpha \alpha} + V_{y,\alpha \alpha} - V_{y,\alpha \beta} V_{y,\beta \beta}^{-1} V_{y,\beta \alpha})^{-1}$ (once again via Fisher Information) by using the full likelihood.
Now I am supposed to show that the first estimator is less efficient:
$$ V_{x,\alpha \alpha}^{-1} - (V_{x,\alpha \alpha} + V_{y,\alpha \alpha} - V_{y,\alpha \beta} V_{y,\beta \beta}^{-1} V_{y,\beta \alpha})^{-1} \geq 0 $$
This is the same as:
$$ 1 \geq \frac{V_{y,\alpha \beta} V_{y,\beta \alpha}}{V_{y,\beta \beta} V_{y,\alpha \alpha}} $$
Which reminds of the Cauchy-Schwarz inequaliy, although I am not so sure how to apply this to the case at hand, since $V$'s are derivatives of the likelihood.
(If notation wasnt clear, $V_{w,\delta\nu}=\frac{\partial^2 f(w)}{\partial \delta \nu}$).
Can anybody provide me any insight? Maybe define some distance function using the derivatives and use Cauchy-Schwarz?
Thanks.