What/when does one need $E$ in expected Fisher information for?
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What/when does one need $E$ in expected Fisher information for?
Since I read an example which merely calculated the second derivatives, put a minus on them and then wrote them in matrix form. It seems that $E bigg( -partial^2 log(f) bigg)=-partial^2 log(f)$.
Normally I think that $E$ merely applies to random variables, since it treats constants as $E(c)=c$. But is it then just that in this case $E bigg( -partial^2 log(f) bigg)=-partial^2 log(f)$, but not in general.
fisher-information
asked Nov 21 at 9:46
mavavilj
2,6501932
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up vote
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What/when does one need $E$ in expected Fisher information for?
Since I read an example which merely calculated the second derivatives, put a minus on them and then wrote them in matrix form. It seems that $E bigg( -partial^2 log(f) bigg)=-partial^2 log(f)$.
Normally I think that $E$ merely applies to random variables, since it treats constants as $E(c)=c$. But is it then just that in this case $E bigg( -partial^2 log(f) bigg)=-partial^2 log(f)$, but not in general.
fisher-information
asked Nov 21 at 9:46
mavavilj
2,6501932
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
What/when does one need $E$ in expected Fisher information for?
Since I read an example which merely calculated the second derivatives, put a minus on them and then wrote them in matrix form. It seems that $E bigg( -partial^2 log(f) bigg)=-partial^2 log(f)$.
Normally I think that $E$ merely applies to random variables, since it treats constants as $E(c)=c$. But is it then just that in this case $E bigg( -partial^2 log(f) bigg)=-partial^2 log(f)$, but not in general.
fisher-information
asked Nov 21 at 9:46
mavavilj
2,6501932
What/when does one need $E$ in expected Fisher information for?
Since I read an example which merely calculated the second derivatives, put a minus on them and then wrote them in matrix form. It seems that $E bigg( -partial^2 log(f) bigg)=-partial^2 log(f)$.
Normally I think that $E$ merely applies to random variables, since it treats constants as $E(c)=c$. But is it then just that in this case $E bigg( -partial^2 log(f) bigg)=-partial^2 log(f)$, but not in general.
fisher-information
fisher-information
asked Nov 21 at 9:46
mavavilj
2,6501932
asked Nov 21 at 9:46
mavavilj
2,6501932
asked Nov 21 at 9:46
mavavilj
2,6501932
asked Nov 21 at 9:46
mavavilj
2,6501932
asked Nov 21 at 9:46
mavavilj
2,6501932
2,6501932
add a comment |
add a comment |
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