Fisher information of the Rayleigh distribution
Multi tool use
$begingroup$
Problem description: Find the Fisher information of the Rayleigh distribution. I was satisfied with my solution until I saw that it disagreed with the solution obtained in one of the problem sets from Princeton. http://www.princeton.edu/~cuff/ele530/files/hw4_sn.pdf p2. I calculate the Fisher information by the following Thm: $$I(theta) = -E(frac{partial^2 log L}{partial theta^2}) $$ where L is the likelihood function of the pdf. The princeton problem set uses another argument I am not familiar with, where they obtain $I(theta) = frac{n}{theta^2}$. I will show my calculations below, maybe someone can spot the error(if there is one). $$log L = prod log f(x_{i})$$ where $X_{i}$'s are iid Rayleigh distributed for $i=1,2,...n$. I end up with $$log L = sum_{i}(log x_{i} -frac{2}{theta} - frac{x_{i}^2}{2theta^2}) $$ Then I take the partial derivative with respect to $theta$ two times and obtain $$frac{partial^2 log L}{partial theta^2} = frac{2ntheta^2-3sum_{i}x_{i}^2}{theta^4} $$ So $I(theta) = -E(frac{2ntheta^2-3sum_{i}x_{i}^2}{theta^4} )$. This is where I am confused, should this expected value be evauated with respect to the $x_{i}^2$'s or $theta$?
probability-distributions expectation
$endgroup$
add a comment |
$begingroup$
Problem description: Find the Fisher information of the Rayleigh distribution. I was satisfied with my solution until I saw that it disagreed with the solution obtained in one of the problem sets from Princeton. http://www.princeton.edu/~cuff/ele530/files/hw4_sn.pdf p2. I calculate the Fisher information by the following Thm: $$I(theta) = -E(frac{partial^2 log L}{partial theta^2}) $$ where L is the likelihood function of the pdf. The princeton problem set uses another argument I am not familiar with, where they obtain $I(theta) = frac{n}{theta^2}$. I will show my calculations below, maybe someone can spot the error(if there is one). $$log L = prod log f(x_{i})$$ where $X_{i}$'s are iid Rayleigh distributed for $i=1,2,...n$. I end up with $$log L = sum_{i}(log x_{i} -frac{2}{theta} - frac{x_{i}^2}{2theta^2}) $$ Then I take the partial derivative with respect to $theta$ two times and obtain $$frac{partial^2 log L}{partial theta^2} = frac{2ntheta^2-3sum_{i}x_{i}^2}{theta^4} $$ So $I(theta) = -E(frac{2ntheta^2-3sum_{i}x_{i}^2}{theta^4} )$. This is where I am confused, should this expected value be evauated with respect to the $x_{i}^2$'s or $theta$?
probability-distributions expectation
$endgroup$
$begingroup$
how to take an expected value with respect to the $x_i^2$?
$endgroup$
– Karl
Nov 16 '14 at 13:26
$begingroup$
The expectation is always respected to $x_i$'s, how can you take the expectation of a parameter $theta$?
$endgroup$
– Sayan
Nov 16 '14 at 13:44
$begingroup$
I did take them with respect to $x_{i}$'s. I will post the result asap.
$endgroup$
– user29163
Nov 16 '14 at 13:54
add a comment |
$begingroup$
Problem description: Find the Fisher information of the Rayleigh distribution. I was satisfied with my solution until I saw that it disagreed with the solution obtained in one of the problem sets from Princeton. http://www.princeton.edu/~cuff/ele530/files/hw4_sn.pdf p2. I calculate the Fisher information by the following Thm: $$I(theta) = -E(frac{partial^2 log L}{partial theta^2}) $$ where L is the likelihood function of the pdf. The princeton problem set uses another argument I am not familiar with, where they obtain $I(theta) = frac{n}{theta^2}$. I will show my calculations below, maybe someone can spot the error(if there is one). $$log L = prod log f(x_{i})$$ where $X_{i}$'s are iid Rayleigh distributed for $i=1,2,...n$. I end up with $$log L = sum_{i}(log x_{i} -frac{2}{theta} - frac{x_{i}^2}{2theta^2}) $$ Then I take the partial derivative with respect to $theta$ two times and obtain $$frac{partial^2 log L}{partial theta^2} = frac{2ntheta^2-3sum_{i}x_{i}^2}{theta^4} $$ So $I(theta) = -E(frac{2ntheta^2-3sum_{i}x_{i}^2}{theta^4} )$. This is where I am confused, should this expected value be evauated with respect to the $x_{i}^2$'s or $theta$?
probability-distributions expectation
$endgroup$
Problem description: Find the Fisher information of the Rayleigh distribution. I was satisfied with my solution until I saw that it disagreed with the solution obtained in one of the problem sets from Princeton. http://www.princeton.edu/~cuff/ele530/files/hw4_sn.pdf p2. I calculate the Fisher information by the following Thm: $$I(theta) = -E(frac{partial^2 log L}{partial theta^2}) $$ where L is the likelihood function of the pdf. The princeton problem set uses another argument I am not familiar with, where they obtain $I(theta) = frac{n}{theta^2}$. I will show my calculations below, maybe someone can spot the error(if there is one). $$log L = prod log f(x_{i})$$ where $X_{i}$'s are iid Rayleigh distributed for $i=1,2,...n$. I end up with $$log L = sum_{i}(log x_{i} -frac{2}{theta} - frac{x_{i}^2}{2theta^2}) $$ Then I take the partial derivative with respect to $theta$ two times and obtain $$frac{partial^2 log L}{partial theta^2} = frac{2ntheta^2-3sum_{i}x_{i}^2}{theta^4} $$ So $I(theta) = -E(frac{2ntheta^2-3sum_{i}x_{i}^2}{theta^4} )$. This is where I am confused, should this expected value be evauated with respect to the $x_{i}^2$'s or $theta$?
probability-distributions expectation
probability-distributions expectation
asked Nov 16 '14 at 13:06
user29163user29163
360313
360313
$begingroup$
how to take an expected value with respect to the $x_i^2$?
$endgroup$
– Karl
Nov 16 '14 at 13:26
$begingroup$
The expectation is always respected to $x_i$'s, how can you take the expectation of a parameter $theta$?
$endgroup$
– Sayan
Nov 16 '14 at 13:44
$begingroup$
I did take them with respect to $x_{i}$'s. I will post the result asap.
$endgroup$
– user29163
Nov 16 '14 at 13:54
add a comment |
$begingroup$
how to take an expected value with respect to the $x_i^2$?
$endgroup$
– Karl
Nov 16 '14 at 13:26
$begingroup$
The expectation is always respected to $x_i$'s, how can you take the expectation of a parameter $theta$?
$endgroup$
– Sayan
Nov 16 '14 at 13:44
$begingroup$
I did take them with respect to $x_{i}$'s. I will post the result asap.
$endgroup$
– user29163
Nov 16 '14 at 13:54
$begingroup$
how to take an expected value with respect to the $x_i^2$?
$endgroup$
– Karl
Nov 16 '14 at 13:26
$begingroup$
how to take an expected value with respect to the $x_i^2$?
$endgroup$
– Karl
Nov 16 '14 at 13:26
$begingroup$
The expectation is always respected to $x_i$'s, how can you take the expectation of a parameter $theta$?
$endgroup$
– Sayan
Nov 16 '14 at 13:44
$begingroup$
The expectation is always respected to $x_i$'s, how can you take the expectation of a parameter $theta$?
$endgroup$
– Sayan
Nov 16 '14 at 13:44
$begingroup$
I did take them with respect to $x_{i}$'s. I will post the result asap.
$endgroup$
– user29163
Nov 16 '14 at 13:54
$begingroup$
I did take them with respect to $x_{i}$'s. I will post the result asap.
$endgroup$
– user29163
Nov 16 '14 at 13:54
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Continuing the above discussion, I find the following when computing $-E(frac{2ntheta^2-3sum_{i}x_{i}^2}{theta^4} )$.
$$-E(frac{2ntheta^2-3sum_{i}x_{i}^2}{theta^4}) = -E(frac{2ntheta^2+3sum_{i}-x_{i}^2}{theta^4})$$
now since I found that $$E(T(x)) = E(-x_{j}^2) = -2theta^2 $$ for arbitrary $x_{j}$. $$E(frac{2ntheta^2+3sum_{i}-x_{i}^2}{theta^4}) = frac{2n}{theta^2} + frac{3nE(T(x))}{theta^4} = frac{-4n}{theta^2} $$ since E is linear and $x_{i}$'s are iid, hence $$-E(frac{2ntheta^2+3sum_{i}-x_{i}^2}{theta^4}) = frac{4n}{theta^2}$$
$endgroup$
add a comment |
$begingroup$
The Princeton version is correct. When you compute the Fisher Information for a Rayleigh you have to exploit the fact that if a r.v. X~Rayleigh with parameter k then a r.v. Y=X^2 has a negative exponential distribution with parameter 1/k. The property would change depending on the definition that you use of both distribution, but it must work in any case.
$endgroup$
add a comment |
Your Answer
StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
});
});
}, "mathjax-editing");
StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "69"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);
StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});
function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});
}
});
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f1024198%2ffisher-information-of-the-rayleigh-distribution%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Continuing the above discussion, I find the following when computing $-E(frac{2ntheta^2-3sum_{i}x_{i}^2}{theta^4} )$.
$$-E(frac{2ntheta^2-3sum_{i}x_{i}^2}{theta^4}) = -E(frac{2ntheta^2+3sum_{i}-x_{i}^2}{theta^4})$$
now since I found that $$E(T(x)) = E(-x_{j}^2) = -2theta^2 $$ for arbitrary $x_{j}$. $$E(frac{2ntheta^2+3sum_{i}-x_{i}^2}{theta^4}) = frac{2n}{theta^2} + frac{3nE(T(x))}{theta^4} = frac{-4n}{theta^2} $$ since E is linear and $x_{i}$'s are iid, hence $$-E(frac{2ntheta^2+3sum_{i}-x_{i}^2}{theta^4}) = frac{4n}{theta^2}$$
$endgroup$
add a comment |
$begingroup$
Continuing the above discussion, I find the following when computing $-E(frac{2ntheta^2-3sum_{i}x_{i}^2}{theta^4} )$.
$$-E(frac{2ntheta^2-3sum_{i}x_{i}^2}{theta^4}) = -E(frac{2ntheta^2+3sum_{i}-x_{i}^2}{theta^4})$$
now since I found that $$E(T(x)) = E(-x_{j}^2) = -2theta^2 $$ for arbitrary $x_{j}$. $$E(frac{2ntheta^2+3sum_{i}-x_{i}^2}{theta^4}) = frac{2n}{theta^2} + frac{3nE(T(x))}{theta^4} = frac{-4n}{theta^2} $$ since E is linear and $x_{i}$'s are iid, hence $$-E(frac{2ntheta^2+3sum_{i}-x_{i}^2}{theta^4}) = frac{4n}{theta^2}$$
$endgroup$
add a comment |
$begingroup$
Continuing the above discussion, I find the following when computing $-E(frac{2ntheta^2-3sum_{i}x_{i}^2}{theta^4} )$.
$$-E(frac{2ntheta^2-3sum_{i}x_{i}^2}{theta^4}) = -E(frac{2ntheta^2+3sum_{i}-x_{i}^2}{theta^4})$$
now since I found that $$E(T(x)) = E(-x_{j}^2) = -2theta^2 $$ for arbitrary $x_{j}$. $$E(frac{2ntheta^2+3sum_{i}-x_{i}^2}{theta^4}) = frac{2n}{theta^2} + frac{3nE(T(x))}{theta^4} = frac{-4n}{theta^2} $$ since E is linear and $x_{i}$'s are iid, hence $$-E(frac{2ntheta^2+3sum_{i}-x_{i}^2}{theta^4}) = frac{4n}{theta^2}$$
$endgroup$
Continuing the above discussion, I find the following when computing $-E(frac{2ntheta^2-3sum_{i}x_{i}^2}{theta^4} )$.
$$-E(frac{2ntheta^2-3sum_{i}x_{i}^2}{theta^4}) = -E(frac{2ntheta^2+3sum_{i}-x_{i}^2}{theta^4})$$
now since I found that $$E(T(x)) = E(-x_{j}^2) = -2theta^2 $$ for arbitrary $x_{j}$. $$E(frac{2ntheta^2+3sum_{i}-x_{i}^2}{theta^4}) = frac{2n}{theta^2} + frac{3nE(T(x))}{theta^4} = frac{-4n}{theta^2} $$ since E is linear and $x_{i}$'s are iid, hence $$-E(frac{2ntheta^2+3sum_{i}-x_{i}^2}{theta^4}) = frac{4n}{theta^2}$$
answered Nov 16 '14 at 19:12
user29163user29163
360313
360313
add a comment |
add a comment |
$begingroup$
The Princeton version is correct. When you compute the Fisher Information for a Rayleigh you have to exploit the fact that if a r.v. X~Rayleigh with parameter k then a r.v. Y=X^2 has a negative exponential distribution with parameter 1/k. The property would change depending on the definition that you use of both distribution, but it must work in any case.
$endgroup$
add a comment |
$begingroup$
The Princeton version is correct. When you compute the Fisher Information for a Rayleigh you have to exploit the fact that if a r.v. X~Rayleigh with parameter k then a r.v. Y=X^2 has a negative exponential distribution with parameter 1/k. The property would change depending on the definition that you use of both distribution, but it must work in any case.
$endgroup$
add a comment |
$begingroup$
The Princeton version is correct. When you compute the Fisher Information for a Rayleigh you have to exploit the fact that if a r.v. X~Rayleigh with parameter k then a r.v. Y=X^2 has a negative exponential distribution with parameter 1/k. The property would change depending on the definition that you use of both distribution, but it must work in any case.
$endgroup$
The Princeton version is correct. When you compute the Fisher Information for a Rayleigh you have to exploit the fact that if a r.v. X~Rayleigh with parameter k then a r.v. Y=X^2 has a negative exponential distribution with parameter 1/k. The property would change depending on the definition that you use of both distribution, but it must work in any case.
answered Oct 14 '16 at 23:16
Filippo PalombaFilippo Palomba
11
11
add a comment |
add a comment |
Thanks for contributing an answer to Mathematics Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f1024198%2ffisher-information-of-the-rayleigh-distribution%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
c75CjF8u6yj3K 9pFQ74udjWYi,zGROM7lRWtcT,HFW3scOSnR,4KRWpwz hnLfl,a AJaLF9jR qr,EdZBm25 yQuowYUT3f3MJKJOoV
$begingroup$
how to take an expected value with respect to the $x_i^2$?
$endgroup$
– Karl
Nov 16 '14 at 13:26
$begingroup$
The expectation is always respected to $x_i$'s, how can you take the expectation of a parameter $theta$?
$endgroup$
– Sayan
Nov 16 '14 at 13:44
$begingroup$
I did take them with respect to $x_{i}$'s. I will post the result asap.
$endgroup$
– user29163
Nov 16 '14 at 13:54