288 Appendice E - La funzione di verosimiglianza
Nel caso particolare che tutte le
x
i
{\displaystyle x_{i}}
f
(
x
;
θ
)
{\displaystyle f(x;\theta )}
∂
(
ln
L
)
∂
θ
=
∂
∂
θ
∑
i
=
1
N
ln
f
(
x
i
;
θ
)
=
∑
i
=
1
N
∂
∂
θ
ln
f
(
x
i
;
θ
)
{\displaystyle {\frac {\partial (\ln {\mathcal {L}})}{\partial \theta }}\;=\;{\frac {\partial }{\partial \theta }}\sum _{i=1}^{N}\ln f(x_{i};\theta )\;=\;\sum _{i=1}^{N}{\frac {\partial }{\partial \theta }}\ln f(x_{i};\theta )}
E
{
∂
(
ln
L
)
∂
θ
}
=
∑
i
=
1
N
E
{
∂
∂
θ
ln
f
(
x
i
;
θ
)
}
=
N
⋅
E
{
∂
∂
θ
ln
f
(
x
;
θ
)
}
{\displaystyle E\left\{{\frac {\partial (\ln {\mathcal {L}})}{\partial \theta }}\right\}\;=\;\sum _{i=1}^{N}E\left\{{\frac {\partial }{\partial \theta }}\ln f(x_{i};\theta )\right\}\;=\;N\cdot E\left\{{\frac {\partial }{\partial \theta }}\ln f(x;\theta )\right\}}
e, tenuto conto della (E.1) , questo implica che
Ora
E
{
[
∂
(
ln
L
)
∂
θ
]
2
}
=
E
{
[
∑
i
=
1
N
∂
∂
θ
ln
f
(
x
i
;
θ
)
]
[
∑
k
=
1
N
∂
∂
θ
ln
f
(
x
k
;
θ
)
]
}
{\displaystyle E\left\{\left[{\frac {\partial (\ln {\mathcal {L}})}{\partial \theta }}\right]^{2}\right\}=E\left\{\left[\sum _{i=1}^{N}{\frac {\partial }{\partial \theta }}\ln f(x_{i};\theta )\right]\left[\sum _{k=1}^{N}{\frac {\partial }{\partial \theta }}\ln f(x_{k};\theta )\right]\right\}}
=
∑
i
=
1
N
E
{
[
∂
∂
θ
ln
f
(
x
i
;
θ
)
]
2
}
+
∑
i
,
k
i
≠
k
E
{
∂
∂
θ
ln
f
(
x
i
;
θ
)
⋅
∂
∂
θ
ln
f
(
x
k
;
θ
)
}
{\displaystyle =\sum _{i=1}^{N}E\left\{\left[{\frac {\partial }{\partial \theta }}\ln f(x_{i};\theta )\right]^{2}\right\}+\sum _{\begin{array}{c}i,k\\i\neq k\end{array}}E\left\{{\frac {\partial }{\partial \theta }}\ln f(x_{i};\theta )\cdot {\frac {\partial }{\partial \theta }}\ln f(x_{k};\theta )\right\}}
=
N
⋅
E
{
[
∂
∂
θ
ln
f
(
x
;
θ
)
]
2
}
+
∑
i
,
k
i
≠
k
E
{
∂
∂
θ
ln
f
(
x
i
;
θ
)
}
⋅
E
{
∂
∂
θ
ln
f
(
x
k
;
θ
)
}
{\displaystyle =N\cdot E\left\{\left[{\frac {\partial }{\partial \theta }}\ln f(x;\theta )\right]^{2}\right\}+\sum _{\begin{array}{c}i,k\\i\neq k\end{array}}E\left\{{\frac {\partial }{\partial \theta }}\ln f(x_{i};\theta )\right\}\cdot E\left\{{\frac {\partial }{\partial \theta }}\ln f(x_{k};\theta )\right\}}
(tenendo conto del fatto che le
x
i
{\displaystyle x_{i}}
(E.3) , ed infine, in questo caso, il minorante della varianza della stima si può scrivere
Col metodo della massima verosimiglianza si assume, come stima del valore vero
θ
∗
{\displaystyle \theta ^{*}}
θ
{\displaystyle \theta }
θ
^
{\displaystyle {\widehat {\theta }}}
L
{\displaystyle {\mathcal {L}}}
x
1
,
x
2
,
…
,
x
N
{\displaystyle x_{1},x_{2},\ldots ,x_{N}}
Ora, nel caso esista una stima di minima varianza
t
{\displaystyle t}
τ
(
θ
)
{\displaystyle \tau (\theta )}
(E.2) la condizione perché la funzione di verosimiglianza abbia un estremante diviene
∂
(
ln
L
)
∂
θ
=
t
−
τ
(
θ
)
R
(
θ
)
=
0
{\displaystyle {\frac {\partial \left(\ln {\mathcal {L}}\right)}{\partial \theta }}\;=\;{\frac {t-\tau (\theta )}{R(\theta )}}\;=\;0}
![{\displaystyle E\left\{\left[{\frac {\partial (\ln {\mathcal {L}})}{\partial \theta }}\right]^{2}\right\}=E\left\{\left[\sum _{i=1}^{N}{\frac {\partial }{\partial \theta }}\ln f(x_{i};\theta )\right]\left[\sum _{k=1}^{N}{\frac {\partial }{\partial \theta }}\ln f(x_{k};\theta )\right]\right\}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/f0ffee3afe86246457ae4cc92a70338489d522db)
![{\displaystyle =\sum _{i=1}^{N}E\left\{\left[{\frac {\partial }{\partial \theta }}\ln f(x_{i};\theta )\right]^{2}\right\}+\sum _{\begin{array}{c}i,k\\i\neq k\end{array}}E\left\{{\frac {\partial }{\partial \theta }}\ln f(x_{i};\theta )\cdot {\frac {\partial }{\partial \theta }}\ln f(x_{k};\theta )\right\}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/a1d86be752858e50873fcc25cc6da5554d4d2b6c)
![{\displaystyle =N\cdot E\left\{\left[{\frac {\partial }{\partial \theta }}\ln f(x;\theta )\right]^{2}\right\}+\sum _{\begin{array}{c}i,k\\i\neq k\end{array}}E\left\{{\frac {\partial }{\partial \theta }}\ln f(x_{i};\theta )\right\}\cdot E\left\{{\frac {\partial }{\partial \theta }}\ln f(x_{k};\theta )\right\}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/aa69642c59b6ae0520e49272fb7e8687cd98016d)
![{\displaystyle {\sigma _{t}}^{2}\;\geq \;{\frac {[\tau '(\theta )]^{2}}{N\cdot E\left\{\left[{\dfrac {\partial }{\partial \theta }}\ln f(x;\theta )\right]^{2}\right\}}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/20c1988999461e514990dbb73217acbab09579ce)
