Pagina:Teoria degli errori e fondamenti di statistica.djvu/301

285

da cui, dividendo e moltiplicando l’integrando per ${\mathcal {L}}$, risulta

$\int _{-\infty }^{+\infty }\,\mathrm {d} x_{1}\int _{-\infty }^{+\infty }\,\mathrm {d} x_{2}$	$\cdots \int _{-\infty }^{+\infty }\,\mathrm {d} x_{N}\,{\mathcal {L}}\left({\frac {1}{\mathcal {L}}}\,{\frac {\partial {\mathcal {L}}}{\partial \theta }}\right)\;=$
	$=\int _{-\infty }^{+\infty }\,\mathrm {d} x_{1}\int _{-\infty }^{+\infty }\,\mathrm {d} x_{2}\cdots \int _{-\infty }^{+\infty }\,\mathrm {d} x_{N}\,{\mathcal {L}}\,{\frac {\partial \left(\ln {\mathcal {L}}\right)}{\partial \theta }}$
	$=\;\int _{-\infty }^{+\infty }\,\mathrm {d} x_{1}\,f_{1}(x_{1};\theta )\cdots \int _{-\infty }^{+\infty }\,\mathrm {d} x_{N}f_{N}(x_{N};\theta )\,{\frac {\partial \left(\ln {\mathcal {L}}\right)}{\partial \theta }}$
	$=\;0$

ossia

$E\left\{{\dfrac {\partial \left(\ln {\mathcal {L}}\right)}{\partial \theta }}\right\}=0$

(E.1)

Se $t$ è imparziale

$E(t)\;=\;\int _{-\infty }^{+\infty }\,\mathrm {d} x_{1}\cdots \int _{-\infty }^{+\infty }\,\mathrm {d} x_{N}\,t(x_{1},x_{2},\ldots ,x_{N})\,{\mathcal {L}}(x_{1},x_{2},\ldots ,x_{N};\theta )\;=\;\tau (\theta )$

da cui, derivando ambo i membri rispetto a $\theta$,

$\int _{-\infty }^{+\infty }\,\mathrm {d} x_{1}\int _{-\infty }^{+\infty }\,\mathrm {d} x_{2}\cdots \int _{-\infty }^{+\infty }\,\mathrm {d} x_{N}\,t\,{\frac {\partial {\mathcal {L}}}{\partial \theta }}=\tau '(\theta )$.

Dividendo e moltiplicando poi l’integrando per la verosimiglianza ${\mathcal {L}}$, risulta

$\int _{-\infty }^{+\infty }\,\mathrm {d} x_{1}\int _{-\infty }^{+\infty }\,\mathrm {d} x_{2}$	$\cdots \int _{-\infty }^{+\infty }\,\mathrm {d} x_{N}\,t\,{\frac {\partial {\mathcal {L}}}{\partial \theta }}\;=$
	$=\;\int _{-\infty }^{+\infty }\,\mathrm {d} x_{1}\cdots \int _{-\infty }^{+\infty }\,\mathrm {d} x_{N}\,t\,{\mathcal {L}}\left({\frac {1}{\mathcal {L}}}\,{\frac {\partial {\mathcal {L}}}{\partial \theta }}\right)$
	$=\;\int _{-\infty }^{+\infty }\,\mathrm {d} x_{1}\,f_{1}(x_{1};\theta )\cdots \int _{-\infty }^{+\infty }\,\mathrm {d} x_{N}\,f_{N}(x_{N};\theta )\,t\,{\frac {\partial \left(\ln {\mathcal {L}}\right)}{\partial \theta }}$
	$=\;E\left\{t\,{\frac {\partial \left(\ln {\mathcal {L}}\right)}{\partial \theta }}\right\}$

e, in definitiva,

$E\left\{t\,{\frac {\partial \left(\ln {\mathcal {L}}\right)}{\partial \theta }}\right\}=\tau '(\theta )$