limiting distribution and continuous transformation
up vote
0
down vote
favorite
Assume that $sqrt{n}(b-beta)$ converges in distribution to a normal variable with zero mean and some variance S (here $b$ is a random variable and $beta$ is a constant, both are scalar valued). I want to find the limiting distribution of $sqrt{n}(e^{b}-e^beta)$.
I know the conversation property of convergence for continuous transformations, thus if I can rewrite $sqrt{n}(e^{b}-e^beta)$ in terms of $sqrt{n}(b-beta)$ then the limiting distribution is easy to find. But I couldn't write it in terms of $sqrt{n}(b-beta)$. Does anybody have any idea about how to proceed?
edit: I guess the delta method suits well into this case, I will proceed using the delta method.
statistics probability-distributions convergence
asked Nov 19 at 19:16
Osman Bulut
556
add a comment |
up vote
0
down vote
favorite
Assume that $sqrt{n}(b-beta)$ converges in distribution to a normal variable with zero mean and some variance S (here $b$ is a random variable and $beta$ is a constant, both are scalar valued). I want to find the limiting distribution of $sqrt{n}(e^{b}-e^beta)$.
I know the conversation property of convergence for continuous transformations, thus if I can rewrite $sqrt{n}(e^{b}-e^beta)$ in terms of $sqrt{n}(b-beta)$ then the limiting distribution is easy to find. But I couldn't write it in terms of $sqrt{n}(b-beta)$. Does anybody have any idea about how to proceed?
edit: I guess the delta method suits well into this case, I will proceed using the delta method.
statistics probability-distributions convergence
asked Nov 19 at 19:16
Osman Bulut
556
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
Assume that $sqrt{n}(b-beta)$ converges in distribution to a normal variable with zero mean and some variance S (here $b$ is a random variable and $beta$ is a constant, both are scalar valued). I want to find the limiting distribution of $sqrt{n}(e^{b}-e^beta)$.
I know the conversation property of convergence for continuous transformations, thus if I can rewrite $sqrt{n}(e^{b}-e^beta)$ in terms of $sqrt{n}(b-beta)$ then the limiting distribution is easy to find. But I couldn't write it in terms of $sqrt{n}(b-beta)$. Does anybody have any idea about how to proceed?
edit: I guess the delta method suits well into this case, I will proceed using the delta method.
statistics probability-distributions convergence
asked Nov 19 at 19:16
Osman Bulut
556
Assume that $sqrt{n}(b-beta)$ converges in distribution to a normal variable with zero mean and some variance S (here $b$ is a random variable and $beta$ is a constant, both are scalar valued). I want to find the limiting distribution of $sqrt{n}(e^{b}-e^beta)$.
I know the conversation property of convergence for continuous transformations, thus if I can rewrite $sqrt{n}(e^{b}-e^beta)$ in terms of $sqrt{n}(b-beta)$ then the limiting distribution is easy to find. But I couldn't write it in terms of $sqrt{n}(b-beta)$. Does anybody have any idea about how to proceed?
edit: I guess the delta method suits well into this case, I will proceed using the delta method.
statistics probability-distributions convergence
statistics probability-distributions convergence
asked Nov 19 at 19:16
Osman Bulut
556
asked Nov 19 at 19:16
Osman Bulut
556
edited Nov 19 at 19:53
asked Nov 19 at 19:16
Osman Bulut
556
asked Nov 19 at 19:16
Osman Bulut
556
asked Nov 19 at 19:16
Osman Bulut
556
556
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
up vote
1
down vote
accepted
Yes, delta method applies here.
$$sqrt{n}(X_n - theta) xrightarrow{D} N(0, sigma^2)$$
then we have
$$sqrt{n}(g(X_n) - g(theta)) xrightarrow{D} N(0, sigma^2[g'(theta)]^2)$$
if $g'(theta)$ exists and non-zero.
Here, $sigma^2 = S, theta = beta$, $g(x)=exp(x)=g'(x)$.
Hence
$$sqrt{n}(e^{b_n} - e^beta) xrightarrow{D} N(0, Sexp(2beta)).$$
answered Nov 20 at 1:47
Siong Thye Goh
97.4k1463116
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',
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%2f3005384%2flimiting-distribution-and-continuous-transformation%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
Yes, delta method applies here.
$$sqrt{n}(X_n - theta) xrightarrow{D} N(0, sigma^2)$$
then we have
$$sqrt{n}(g(X_n) - g(theta)) xrightarrow{D} N(0, sigma^2[g'(theta)]^2)$$
if $g'(theta)$ exists and non-zero.
Here, $sigma^2 = S, theta = beta$, $g(x)=exp(x)=g'(x)$.
Hence
$$sqrt{n}(e^{b_n} - e^beta) xrightarrow{D} N(0, Sexp(2beta)).$$
answered Nov 20 at 1:47
Siong Thye Goh
97.4k1463116
add a comment |
up vote
1
down vote
accepted
Yes, delta method applies here.
$$sqrt{n}(X_n - theta) xrightarrow{D} N(0, sigma^2)$$
then we have
$$sqrt{n}(g(X_n) - g(theta)) xrightarrow{D} N(0, sigma^2[g'(theta)]^2)$$
if $g'(theta)$ exists and non-zero.
Here, $sigma^2 = S, theta = beta$, $g(x)=exp(x)=g'(x)$.
Hence
$$sqrt{n}(e^{b_n} - e^beta) xrightarrow{D} N(0, Sexp(2beta)).$$
answered Nov 20 at 1:47
Siong Thye Goh
97.4k1463116
add a comment |
up vote
1
down vote
accepted
up vote
1
down vote
accepted
Yes, delta method applies here.
$$sqrt{n}(X_n - theta) xrightarrow{D} N(0, sigma^2)$$
then we have
$$sqrt{n}(g(X_n) - g(theta)) xrightarrow{D} N(0, sigma^2[g'(theta)]^2)$$
if $g'(theta)$ exists and non-zero.
Here, $sigma^2 = S, theta = beta$, $g(x)=exp(x)=g'(x)$.
Hence
$$sqrt{n}(e^{b_n} - e^beta) xrightarrow{D} N(0, Sexp(2beta)).$$
answered Nov 20 at 1:47
Siong Thye Goh
97.4k1463116
Yes, delta method applies here.
$$sqrt{n}(X_n - theta) xrightarrow{D} N(0, sigma^2)$$
then we have
$$sqrt{n}(g(X_n) - g(theta)) xrightarrow{D} N(0, sigma^2[g'(theta)]^2)$$
if $g'(theta)$ exists and non-zero.
Here, $sigma^2 = S, theta = beta$, $g(x)=exp(x)=g'(x)$.
Hence
$$sqrt{n}(e^{b_n} - e^beta) xrightarrow{D} N(0, Sexp(2beta)).$$
answered Nov 20 at 1:47
Siong Thye Goh
97.4k1463116
answered Nov 20 at 1:47
Siong Thye Goh
97.4k1463116
answered Nov 20 at 1:47
Siong Thye Goh
97.4k1463116
answered Nov 20 at 1:47
Siong Thye Goh
97.4k1463116
97.4k1463116
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.
Some of your past answers have not been well-received, and you're in danger of being blocked from answering.
Please pay close attention to the following guidance:
- 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.
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%2f3005384%2flimiting-distribution-and-continuous-transformation%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
