Fisher information of normal distribution with unknown mean and variance?Convergence rate of empirical Fisher information matrixFisher Information and minimum variance estimatorsSufficient Statistics, MLE and Unbiased Estimators of Uniform Type DistributionCalculating a Fisher expected informationFisher information matrix of MLE'sFor a Fisher Information, $mathcalI(theta)$, why does $mathcalI(theta) = nmathcalI_1(theta)$ not hold for multiple dimensions?Fisher's information of normal distributionFisher information comparisonComparing Fisher Information of sample to that of statisticCalculating Fisher Information for Bernoulli rv
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Fisher information of normal distribution with unknown mean and variance?
Convergence rate of empirical Fisher information matrixFisher Information and minimum variance estimatorsSufficient Statistics, MLE and Unbiased Estimators of Uniform Type DistributionCalculating a Fisher expected informationFisher information matrix of MLE'sFor a Fisher Information, $mathcalI(theta)$, why does $mathcalI(theta) = nmathcalI_1(theta)$ not hold for multiple dimensions?Fisher's information of normal distributionFisher information comparisonComparing Fisher Information of sample to that of statisticCalculating Fisher Information for Bernoulli rv
$begingroup$
I am asked to find the fisher information contained in $X_1 sim N(theta_1, theta_2)$ (ie: two unknown parameters, only one observation). How would I find the Fisher information here?
I know that with a sample $X_1,X_2,ldots,X_n $~$N(mu,sigma^2)$ and $sigma^2=1$, Fisher's information is given by :
$$
-E(fracd^2dmu^2 ln f(x))=1/sigma^2.
$$
Though this is the case with one paramter and I am not sure how it would map on to the case with two parameters. I imagine there is some use of a Hessian but I am not sure what to do.
probability statistics expected-value fisher-information
asked 2 days ago
j doej doe
11
$endgroup$
add a comment |
$begingroup$
I am asked to find the fisher information contained in $X_1 sim N(theta_1, theta_2)$ (ie: two unknown parameters, only one observation). How would I find the Fisher information here?
I know that with a sample $X_1,X_2,ldots,X_n $~$N(mu,sigma^2)$ and $sigma^2=1$, Fisher's information is given by :
$$
-E(fracd^2dmu^2 ln f(x))=1/sigma^2.
$$
Though this is the case with one paramter and I am not sure how it would map on to the case with two parameters. I imagine there is some use of a Hessian but I am not sure what to do.
probability statistics expected-value fisher-information
asked 2 days ago
j doej doe
11
$endgroup$
add a comment |
$begingroup$
I am asked to find the fisher information contained in $X_1 sim N(theta_1, theta_2)$ (ie: two unknown parameters, only one observation). How would I find the Fisher information here?
I know that with a sample $X_1,X_2,ldots,X_n $~$N(mu,sigma^2)$ and $sigma^2=1$, Fisher's information is given by :
$$
-E(fracd^2dmu^2 ln f(x))=1/sigma^2.
$$
Though this is the case with one paramter and I am not sure how it would map on to the case with two parameters. I imagine there is some use of a Hessian but I am not sure what to do.
probability statistics expected-value fisher-information
asked 2 days ago
j doej doe
11
$endgroup$
I am asked to find the fisher information contained in $X_1 sim N(theta_1, theta_2)$ (ie: two unknown parameters, only one observation). How would I find the Fisher information here?
I know that with a sample $X_1,X_2,ldots,X_n $~$N(mu,sigma^2)$ and $sigma^2=1$, Fisher's information is given by :
$$
-E(fracd^2dmu^2 ln f(x))=1/sigma^2.
$$
Though this is the case with one paramter and I am not sure how it would map on to the case with two parameters. I imagine there is some use of a Hessian but I am not sure what to do.
probability statistics expected-value fisher-information
probability statistics expected-value fisher-information
asked 2 days ago
j doej doe
11
asked 2 days ago
j doej doe
11
edited 2 days ago
j doe
asked 2 days ago
j doej doe
11
asked 2 days ago
j doej doe
11
asked 2 days ago
j doej doe
11
11
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
It will be the expected value of the Hessian matrix of $ln f(x;mu, sigma^2)$. Specifically for the normal distribution, you can check that it will a diagonal matrix. The $mathcalI_11$ you have already calculated. For the second diagonal term
$$
ln f(x;mu, sigma)=-frac12ln(2 sigma^2)+frac12sigma^2(x-mu)^2,
$$
$$
l'_sigma^2 = - frac12sigma^2 - frac12sigma^4(x-mu)^2,
$$
hence
$$
mathcalI_22= -mathbbE[l''_sigma^2]
=
- mathbbE [ frac12sigma^4 - frac1sigma^6(x-mu)^2]
=
-frac12sigma^4 + frac2sigma^4 = frac12sigma^4 .
$$
And for the non-diagonal terms
$$
mathcalI_22= -mathbbE[l''_sigma^2,mu] = - mathbbEfrac2(x-mu)2sigma^4 = 0.
$$
answered 2 days ago
V. VancakV. Vancak
11.3k3926
$endgroup$
$begingroup$
Can you say a bit more about that? What would be the entries in the Hessian?
$endgroup$
– j doe
2 days ago
$begingroup$
If you let $l$ be the log-likelihood function and write $vequiv sigma^2$ for simplicity, the Hessian of the log-likelihood function will be equal to $$colorbluebeginbmatrixfracpartial^2 lpartial mu^2 & fracpartial^2 lpartial mu partial v \ fracpartial^2 lpartial mu partial v & fracpartial^2 lpartial v^2endbmatrix.$$Note that the Hessian is a symmetric matrix, so once you calculate the top right entry, you can copy it into the bottom left entry.
$endgroup$
– Minus One-Twelfth
2 days ago
$begingroup$
@jdoe pls see the edited answer
$endgroup$
– V. Vancak
yesterday
add a comment |
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1 Answer
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active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
It will be the expected value of the Hessian matrix of $ln f(x;mu, sigma^2)$. Specifically for the normal distribution, you can check that it will a diagonal matrix. The $mathcalI_11$ you have already calculated. For the second diagonal term
$$
ln f(x;mu, sigma)=-frac12ln(2 sigma^2)+frac12sigma^2(x-mu)^2,
$$
$$
l'_sigma^2 = - frac12sigma^2 - frac12sigma^4(x-mu)^2,
$$
hence
$$
mathcalI_22= -mathbbE[l''_sigma^2]
=
- mathbbE [ frac12sigma^4 - frac1sigma^6(x-mu)^2]
=
-frac12sigma^4 + frac2sigma^4 = frac12sigma^4 .
$$
And for the non-diagonal terms
$$
mathcalI_22= -mathbbE[l''_sigma^2,mu] = - mathbbEfrac2(x-mu)2sigma^4 = 0.
$$
answered 2 days ago
V. VancakV. Vancak
11.3k3926
$endgroup$
$begingroup$
Can you say a bit more about that? What would be the entries in the Hessian?
$endgroup$
– j doe
2 days ago
$begingroup$
If you let $l$ be the log-likelihood function and write $vequiv sigma^2$ for simplicity, the Hessian of the log-likelihood function will be equal to $$colorbluebeginbmatrixfracpartial^2 lpartial mu^2 & fracpartial^2 lpartial mu partial v \ fracpartial^2 lpartial mu partial v & fracpartial^2 lpartial v^2endbmatrix.$$Note that the Hessian is a symmetric matrix, so once you calculate the top right entry, you can copy it into the bottom left entry.
$endgroup$
– Minus One-Twelfth
2 days ago
$begingroup$
@jdoe pls see the edited answer
$endgroup$
– V. Vancak
yesterday
add a comment |
$begingroup$
It will be the expected value of the Hessian matrix of $ln f(x;mu, sigma^2)$. Specifically for the normal distribution, you can check that it will a diagonal matrix. The $mathcalI_11$ you have already calculated. For the second diagonal term
$$
ln f(x;mu, sigma)=-frac12ln(2 sigma^2)+frac12sigma^2(x-mu)^2,
$$
$$
l'_sigma^2 = - frac12sigma^2 - frac12sigma^4(x-mu)^2,
$$
hence
$$
mathcalI_22= -mathbbE[l''_sigma^2]
=
- mathbbE [ frac12sigma^4 - frac1sigma^6(x-mu)^2]
=
-frac12sigma^4 + frac2sigma^4 = frac12sigma^4 .
$$
And for the non-diagonal terms
$$
mathcalI_22= -mathbbE[l''_sigma^2,mu] = - mathbbEfrac2(x-mu)2sigma^4 = 0.
$$
answered 2 days ago
V. VancakV. Vancak
11.3k3926
$endgroup$
$begingroup$
Can you say a bit more about that? What would be the entries in the Hessian?
$endgroup$
– j doe
2 days ago
$begingroup$
If you let $l$ be the log-likelihood function and write $vequiv sigma^2$ for simplicity, the Hessian of the log-likelihood function will be equal to $$colorbluebeginbmatrixfracpartial^2 lpartial mu^2 & fracpartial^2 lpartial mu partial v \ fracpartial^2 lpartial mu partial v & fracpartial^2 lpartial v^2endbmatrix.$$Note that the Hessian is a symmetric matrix, so once you calculate the top right entry, you can copy it into the bottom left entry.
$endgroup$
– Minus One-Twelfth
2 days ago
$begingroup$
@jdoe pls see the edited answer
$endgroup$
– V. Vancak
yesterday
add a comment |
$begingroup$
It will be the expected value of the Hessian matrix of $ln f(x;mu, sigma^2)$. Specifically for the normal distribution, you can check that it will a diagonal matrix. The $mathcalI_11$ you have already calculated. For the second diagonal term
$$
ln f(x;mu, sigma)=-frac12ln(2 sigma^2)+frac12sigma^2(x-mu)^2,
$$
$$
l'_sigma^2 = - frac12sigma^2 - frac12sigma^4(x-mu)^2,
$$
hence
$$
mathcalI_22= -mathbbE[l''_sigma^2]
=
- mathbbE [ frac12sigma^4 - frac1sigma^6(x-mu)^2]
=
-frac12sigma^4 + frac2sigma^4 = frac12sigma^4 .
$$
And for the non-diagonal terms
$$
mathcalI_22= -mathbbE[l''_sigma^2,mu] = - mathbbEfrac2(x-mu)2sigma^4 = 0.
$$
answered 2 days ago
V. VancakV. Vancak
11.3k3926
$endgroup$
It will be the expected value of the Hessian matrix of $ln f(x;mu, sigma^2)$. Specifically for the normal distribution, you can check that it will a diagonal matrix. The $mathcalI_11$ you have already calculated. For the second diagonal term
$$
ln f(x;mu, sigma)=-frac12ln(2 sigma^2)+frac12sigma^2(x-mu)^2,
$$
$$
l'_sigma^2 = - frac12sigma^2 - frac12sigma^4(x-mu)^2,
$$
hence
$$
mathcalI_22= -mathbbE[l''_sigma^2]
=
- mathbbE [ frac12sigma^4 - frac1sigma^6(x-mu)^2]
=
-frac12sigma^4 + frac2sigma^4 = frac12sigma^4 .
$$
And for the non-diagonal terms
$$
mathcalI_22= -mathbbE[l''_sigma^2,mu] = - mathbbEfrac2(x-mu)2sigma^4 = 0.
$$
answered 2 days ago
V. VancakV. Vancak
11.3k3926
edited yesterday
answered 2 days ago
V. VancakV. Vancak
11.3k3926
answered 2 days ago
V. VancakV. Vancak
11.3k3926
answered 2 days ago
V. VancakV. Vancak
11.3k3926
11.3k3926
$begingroup$
Can you say a bit more about that? What would be the entries in the Hessian?
$endgroup$
– j doe
2 days ago
$begingroup$
If you let $l$ be the log-likelihood function and write $vequiv sigma^2$ for simplicity, the Hessian of the log-likelihood function will be equal to $$colorbluebeginbmatrixfracpartial^2 lpartial mu^2 & fracpartial^2 lpartial mu partial v \ fracpartial^2 lpartial mu partial v & fracpartial^2 lpartial v^2endbmatrix.$$Note that the Hessian is a symmetric matrix, so once you calculate the top right entry, you can copy it into the bottom left entry.
$endgroup$
– Minus One-Twelfth
2 days ago
$begingroup$
@jdoe pls see the edited answer
$endgroup$
– V. Vancak
yesterday
add a comment |
$begingroup$
Can you say a bit more about that? What would be the entries in the Hessian?
$endgroup$
– j doe
2 days ago
$begingroup$
If you let $l$ be the log-likelihood function and write $vequiv sigma^2$ for simplicity, the Hessian of the log-likelihood function will be equal to $$colorbluebeginbmatrixfracpartial^2 lpartial mu^2 & fracpartial^2 lpartial mu partial v \ fracpartial^2 lpartial mu partial v & fracpartial^2 lpartial v^2endbmatrix.$$Note that the Hessian is a symmetric matrix, so once you calculate the top right entry, you can copy it into the bottom left entry.
$endgroup$
– Minus One-Twelfth
2 days ago
$begingroup$
@jdoe pls see the edited answer
$endgroup$
– V. Vancak
yesterday
$begingroup$
Can you say a bit more about that? What would be the entries in the Hessian?
$endgroup$
– j doe
2 days ago
$begingroup$
Can you say a bit more about that? What would be the entries in the Hessian?
$endgroup$
– j doe
2 days ago
$begingroup$
If you let $l$ be the log-likelihood function and write $vequiv sigma^2$ for simplicity, the Hessian of the log-likelihood function will be equal to $$colorbluebeginbmatrixfracpartial^2 lpartial mu^2 & fracpartial^2 lpartial mu partial v \ fracpartial^2 lpartial mu partial v & fracpartial^2 lpartial v^2endbmatrix.$$Note that the Hessian is a symmetric matrix, so once you calculate the top right entry, you can copy it into the bottom left entry.
$endgroup$
– Minus One-Twelfth
2 days ago
$begingroup$
If you let $l$ be the log-likelihood function and write $vequiv sigma^2$ for simplicity, the Hessian of the log-likelihood function will be equal to $$colorbluebeginbmatrixfracpartial^2 lpartial mu^2 & fracpartial^2 lpartial mu partial v \ fracpartial^2 lpartial mu partial v & fracpartial^2 lpartial v^2endbmatrix.$$Note that the Hessian is a symmetric matrix, so once you calculate the top right entry, you can copy it into the bottom left entry.
$endgroup$
– Minus One-Twelfth
2 days ago
$begingroup$
@jdoe pls see the edited answer
$endgroup$
– V. Vancak
yesterday
$begingroup$
@jdoe pls see the edited answer
$endgroup$
– V. Vancak
yesterday
add a comment |
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