How to prove alternative form of Fisher information
$begingroup$
Assuming the FI regularity conditions hold. The Fisher information matrix $I(theta;X)$ about $theta$ based on $X$ is defined as the matrix with elements
$$
I_{i,j}(theta;X)=Cov_{theta}Big(frac{partial}{partialtheta_{i}}log f_{X}(X|theta),frac{partial}{partialtheta_{j}}log f_{X}(X|theta)Big)
$$
However, an alternative form is given in my studybook
$$
I_{i,j}=-E_{theta}(frac{partial^{2}}{partialtheta_{i}partialtheta_{j}}log f_{X}(x|theta))
$$
I am not able to proof the alternative form, since I am not able to find why the following equality holds
$$
-frac{partial^{2}}{partialtheta_{i}partialtheta_{j}}log f_{X}(x|theta)=frac{partial}{partialtheta_{i}}log f_{X}(x|theta)frac{partial}{partialtheta_{j}}log f_{X}(x|theta)
$$
statistics fisher-information
$endgroup$
add a comment |
$begingroup$
Assuming the FI regularity conditions hold. The Fisher information matrix $I(theta;X)$ about $theta$ based on $X$ is defined as the matrix with elements
$$
I_{i,j}(theta;X)=Cov_{theta}Big(frac{partial}{partialtheta_{i}}log f_{X}(X|theta),frac{partial}{partialtheta_{j}}log f_{X}(X|theta)Big)
$$
However, an alternative form is given in my studybook
$$
I_{i,j}=-E_{theta}(frac{partial^{2}}{partialtheta_{i}partialtheta_{j}}log f_{X}(x|theta))
$$
I am not able to proof the alternative form, since I am not able to find why the following equality holds
$$
-frac{partial^{2}}{partialtheta_{i}partialtheta_{j}}log f_{X}(x|theta)=frac{partial}{partialtheta_{i}}log f_{X}(x|theta)frac{partial}{partialtheta_{j}}log f_{X}(x|theta)
$$
statistics fisher-information
$endgroup$
add a comment |
$begingroup$
Assuming the FI regularity conditions hold. The Fisher information matrix $I(theta;X)$ about $theta$ based on $X$ is defined as the matrix with elements
$$
I_{i,j}(theta;X)=Cov_{theta}Big(frac{partial}{partialtheta_{i}}log f_{X}(X|theta),frac{partial}{partialtheta_{j}}log f_{X}(X|theta)Big)
$$
However, an alternative form is given in my studybook
$$
I_{i,j}=-E_{theta}(frac{partial^{2}}{partialtheta_{i}partialtheta_{j}}log f_{X}(x|theta))
$$
I am not able to proof the alternative form, since I am not able to find why the following equality holds
$$
-frac{partial^{2}}{partialtheta_{i}partialtheta_{j}}log f_{X}(x|theta)=frac{partial}{partialtheta_{i}}log f_{X}(x|theta)frac{partial}{partialtheta_{j}}log f_{X}(x|theta)
$$
statistics fisher-information
$endgroup$
Assuming the FI regularity conditions hold. The Fisher information matrix $I(theta;X)$ about $theta$ based on $X$ is defined as the matrix with elements
$$
I_{i,j}(theta;X)=Cov_{theta}Big(frac{partial}{partialtheta_{i}}log f_{X}(X|theta),frac{partial}{partialtheta_{j}}log f_{X}(X|theta)Big)
$$
However, an alternative form is given in my studybook
$$
I_{i,j}=-E_{theta}(frac{partial^{2}}{partialtheta_{i}partialtheta_{j}}log f_{X}(x|theta))
$$
I am not able to proof the alternative form, since I am not able to find why the following equality holds
$$
-frac{partial^{2}}{partialtheta_{i}partialtheta_{j}}log f_{X}(x|theta)=frac{partial}{partialtheta_{i}}log f_{X}(x|theta)frac{partial}{partialtheta_{j}}log f_{X}(x|theta)
$$
statistics fisher-information
statistics fisher-information
edited Jan 10 at 12:41
rs4rs35
asked Jan 10 at 11:57
rs4rs35rs4rs35
237
237
add a comment |
add a comment |
1 Answer
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$begingroup$
We have that:
$-frac{partial^{2}}{partialtheta_{i}partialtheta_{j}}log f_{X}(x|theta)=-frac{frac{partial^{2}}{partialtheta_{i}partialtheta_{j}}f_{X}(x|theta)}{f_{X}(x|theta)}+frac{frac{partial}{partialtheta_{i}}f_{X}(x|theta)frac{partial}{partialtheta_{j}} f_{X}(x|theta)}{f^{2}_{X}(x|theta)}$
Now taking the expectation left and right of this relation, we find that:
$-mathbb{E}_{theta}[frac{partial^{2}}{partialtheta_{i}partialtheta_{j}}log f_{X}(x|theta)]=mathbb{E}[(frac{partial}{partialtheta_{i}}log f_{X}(x|theta)(frac{partial}{partialtheta_{j}}log f_{X}(x|theta))]$
This is because of the fact that:
$-mathbb{E}_{theta}[frac{partial}{partialtheta_{i}} log f_X(x;theta)]=frac{partial}{partialtheta_{i}}int f_X(x|theta)text{dx}$
So the alternative form follows.
$endgroup$
add a comment |
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1 Answer
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1 Answer
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$begingroup$
We have that:
$-frac{partial^{2}}{partialtheta_{i}partialtheta_{j}}log f_{X}(x|theta)=-frac{frac{partial^{2}}{partialtheta_{i}partialtheta_{j}}f_{X}(x|theta)}{f_{X}(x|theta)}+frac{frac{partial}{partialtheta_{i}}f_{X}(x|theta)frac{partial}{partialtheta_{j}} f_{X}(x|theta)}{f^{2}_{X}(x|theta)}$
Now taking the expectation left and right of this relation, we find that:
$-mathbb{E}_{theta}[frac{partial^{2}}{partialtheta_{i}partialtheta_{j}}log f_{X}(x|theta)]=mathbb{E}[(frac{partial}{partialtheta_{i}}log f_{X}(x|theta)(frac{partial}{partialtheta_{j}}log f_{X}(x|theta))]$
This is because of the fact that:
$-mathbb{E}_{theta}[frac{partial}{partialtheta_{i}} log f_X(x;theta)]=frac{partial}{partialtheta_{i}}int f_X(x|theta)text{dx}$
So the alternative form follows.
$endgroup$
add a comment |
$begingroup$
We have that:
$-frac{partial^{2}}{partialtheta_{i}partialtheta_{j}}log f_{X}(x|theta)=-frac{frac{partial^{2}}{partialtheta_{i}partialtheta_{j}}f_{X}(x|theta)}{f_{X}(x|theta)}+frac{frac{partial}{partialtheta_{i}}f_{X}(x|theta)frac{partial}{partialtheta_{j}} f_{X}(x|theta)}{f^{2}_{X}(x|theta)}$
Now taking the expectation left and right of this relation, we find that:
$-mathbb{E}_{theta}[frac{partial^{2}}{partialtheta_{i}partialtheta_{j}}log f_{X}(x|theta)]=mathbb{E}[(frac{partial}{partialtheta_{i}}log f_{X}(x|theta)(frac{partial}{partialtheta_{j}}log f_{X}(x|theta))]$
This is because of the fact that:
$-mathbb{E}_{theta}[frac{partial}{partialtheta_{i}} log f_X(x;theta)]=frac{partial}{partialtheta_{i}}int f_X(x|theta)text{dx}$
So the alternative form follows.
$endgroup$
add a comment |
$begingroup$
We have that:
$-frac{partial^{2}}{partialtheta_{i}partialtheta_{j}}log f_{X}(x|theta)=-frac{frac{partial^{2}}{partialtheta_{i}partialtheta_{j}}f_{X}(x|theta)}{f_{X}(x|theta)}+frac{frac{partial}{partialtheta_{i}}f_{X}(x|theta)frac{partial}{partialtheta_{j}} f_{X}(x|theta)}{f^{2}_{X}(x|theta)}$
Now taking the expectation left and right of this relation, we find that:
$-mathbb{E}_{theta}[frac{partial^{2}}{partialtheta_{i}partialtheta_{j}}log f_{X}(x|theta)]=mathbb{E}[(frac{partial}{partialtheta_{i}}log f_{X}(x|theta)(frac{partial}{partialtheta_{j}}log f_{X}(x|theta))]$
This is because of the fact that:
$-mathbb{E}_{theta}[frac{partial}{partialtheta_{i}} log f_X(x;theta)]=frac{partial}{partialtheta_{i}}int f_X(x|theta)text{dx}$
So the alternative form follows.
$endgroup$
We have that:
$-frac{partial^{2}}{partialtheta_{i}partialtheta_{j}}log f_{X}(x|theta)=-frac{frac{partial^{2}}{partialtheta_{i}partialtheta_{j}}f_{X}(x|theta)}{f_{X}(x|theta)}+frac{frac{partial}{partialtheta_{i}}f_{X}(x|theta)frac{partial}{partialtheta_{j}} f_{X}(x|theta)}{f^{2}_{X}(x|theta)}$
Now taking the expectation left and right of this relation, we find that:
$-mathbb{E}_{theta}[frac{partial^{2}}{partialtheta_{i}partialtheta_{j}}log f_{X}(x|theta)]=mathbb{E}[(frac{partial}{partialtheta_{i}}log f_{X}(x|theta)(frac{partial}{partialtheta_{j}}log f_{X}(x|theta))]$
This is because of the fact that:
$-mathbb{E}_{theta}[frac{partial}{partialtheta_{i}} log f_X(x;theta)]=frac{partial}{partialtheta_{i}}int f_X(x|theta)text{dx}$
So the alternative form follows.
edited Jan 10 at 13:03
answered Jan 10 at 12:53
S. CrimS. Crim
363112
363112
add a comment |
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