Proving a statistic is ancillary
Suppose $X_1$ and $X_2$ are iid observations from the pdf $f(x|alpha)=alpha x^{alpha-1}e^{-x^alpha}$, $x>0$, $alpha>0$. Show that $frac{log X_1}{log X_2}$ is an ancillary statistic.
I guess I need to show that the distribution of this statistic is independent of $alpha$, i.e. it is the same as if $alpha = 1$. However, this concept is a little bit confusing to me and I am not sure how to approach this problem.
statistics statistical-inference
asked Dec 28 '18 at 14:49
Analysis801Analysis801
1175
add a comment |
Suppose $X_1$ and $X_2$ are iid observations from the pdf $f(x|alpha)=alpha x^{alpha-1}e^{-x^alpha}$, $x>0$, $alpha>0$. Show that $frac{log X_1}{log X_2}$ is an ancillary statistic.
I guess I need to show that the distribution of this statistic is independent of $alpha$, i.e. it is the same as if $alpha = 1$. However, this concept is a little bit confusing to me and I am not sure how to approach this problem.
statistics statistical-inference
asked Dec 28 '18 at 14:49
Analysis801Analysis801
1175
So did you actually try finding the distribution of $log X_1/log X_2$? Where are you stuck?
– StubbornAtom
Dec 28 '18 at 15:09
add a comment |
Suppose $X_1$ and $X_2$ are iid observations from the pdf $f(x|alpha)=alpha x^{alpha-1}e^{-x^alpha}$, $x>0$, $alpha>0$. Show that $frac{log X_1}{log X_2}$ is an ancillary statistic.
I guess I need to show that the distribution of this statistic is independent of $alpha$, i.e. it is the same as if $alpha = 1$. However, this concept is a little bit confusing to me and I am not sure how to approach this problem.
statistics statistical-inference
asked Dec 28 '18 at 14:49
Analysis801Analysis801
1175
Suppose $X_1$ and $X_2$ are iid observations from the pdf $f(x|alpha)=alpha x^{alpha-1}e^{-x^alpha}$, $x>0$, $alpha>0$. Show that $frac{log X_1}{log X_2}$ is an ancillary statistic.
I guess I need to show that the distribution of this statistic is independent of $alpha$, i.e. it is the same as if $alpha = 1$. However, this concept is a little bit confusing to me and I am not sure how to approach this problem.
statistics statistical-inference
statistics statistical-inference
asked Dec 28 '18 at 14:49
Analysis801Analysis801
1175
asked Dec 28 '18 at 14:49
Analysis801Analysis801
1175
asked Dec 28 '18 at 14:49
Analysis801Analysis801
1175
asked Dec 28 '18 at 14:49
Analysis801Analysis801
1175
asked Dec 28 '18 at 14:49
Analysis801Analysis801
1175
1175
So did you actually try finding the distribution of $log X_1/log X_2$? Where are you stuck?
– StubbornAtom
Dec 28 '18 at 15:09
add a comment |
So did you actually try finding the distribution of $log X_1/log X_2$? Where are you stuck?
– StubbornAtom
Dec 28 '18 at 15:09
So did you actually try finding the distribution of $log X_1/log X_2$? Where are you stuck?
– StubbornAtom
Dec 28 '18 at 15:09
So did you actually try finding the distribution of $log X_1/log X_2$? Where are you stuck?
– StubbornAtom
Dec 28 '18 at 15:09
add a comment |
1 Answer
1
active
oldest
votes
If $Uapproxmathsf{Exp}(1)$ then:$$P(U^{frac1{alpha}}>x)=P(U>x^{alpha})=e^{-x^{alpha}}$$ indicating that $U^{frac1{alpha}}$ has the distribution of $X_1,X_2$.
So $U_{i}:=X_{i}^{alpha}$ are iid and have standard exponential distribution.
Then it follows easily that: $$frac{ln X_{1}}{ln X_{2}}=frac{ln U_{1}}{ln U_{2}}$$and distribution of RHS does not depend on $alpha$ because the distribution of $U_1$ and $U_2$ does not depend on $alpha$.
answered Dec 28 '18 at 15:14
drhabdrhab
98.4k544129
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%2f3054960%2fproving-a-statistic-is-ancillary%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
If $Uapproxmathsf{Exp}(1)$ then:$$P(U^{frac1{alpha}}>x)=P(U>x^{alpha})=e^{-x^{alpha}}$$ indicating that $U^{frac1{alpha}}$ has the distribution of $X_1,X_2$.
So $U_{i}:=X_{i}^{alpha}$ are iid and have standard exponential distribution.
Then it follows easily that: $$frac{ln X_{1}}{ln X_{2}}=frac{ln U_{1}}{ln U_{2}}$$and distribution of RHS does not depend on $alpha$ because the distribution of $U_1$ and $U_2$ does not depend on $alpha$.
answered Dec 28 '18 at 15:14
drhabdrhab
98.4k544129
add a comment |
If $Uapproxmathsf{Exp}(1)$ then:$$P(U^{frac1{alpha}}>x)=P(U>x^{alpha})=e^{-x^{alpha}}$$ indicating that $U^{frac1{alpha}}$ has the distribution of $X_1,X_2$.
So $U_{i}:=X_{i}^{alpha}$ are iid and have standard exponential distribution.
Then it follows easily that: $$frac{ln X_{1}}{ln X_{2}}=frac{ln U_{1}}{ln U_{2}}$$and distribution of RHS does not depend on $alpha$ because the distribution of $U_1$ and $U_2$ does not depend on $alpha$.
answered Dec 28 '18 at 15:14
drhabdrhab
98.4k544129
add a comment |
If $Uapproxmathsf{Exp}(1)$ then:$$P(U^{frac1{alpha}}>x)=P(U>x^{alpha})=e^{-x^{alpha}}$$ indicating that $U^{frac1{alpha}}$ has the distribution of $X_1,X_2$.
So $U_{i}:=X_{i}^{alpha}$ are iid and have standard exponential distribution.
Then it follows easily that: $$frac{ln X_{1}}{ln X_{2}}=frac{ln U_{1}}{ln U_{2}}$$and distribution of RHS does not depend on $alpha$ because the distribution of $U_1$ and $U_2$ does not depend on $alpha$.
answered Dec 28 '18 at 15:14
drhabdrhab
98.4k544129
If $Uapproxmathsf{Exp}(1)$ then:$$P(U^{frac1{alpha}}>x)=P(U>x^{alpha})=e^{-x^{alpha}}$$ indicating that $U^{frac1{alpha}}$ has the distribution of $X_1,X_2$.
So $U_{i}:=X_{i}^{alpha}$ are iid and have standard exponential distribution.
Then it follows easily that: $$frac{ln X_{1}}{ln X_{2}}=frac{ln U_{1}}{ln U_{2}}$$and distribution of RHS does not depend on $alpha$ because the distribution of $U_1$ and $U_2$ does not depend on $alpha$.
answered Dec 28 '18 at 15:14
drhabdrhab
98.4k544129
edited Dec 28 '18 at 15:23
answered Dec 28 '18 at 15:14
drhabdrhab
98.4k544129
answered Dec 28 '18 at 15:14
drhabdrhab
98.4k544129
answered Dec 28 '18 at 15:14
drhabdrhab
98.4k544129
98.4k544129
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%2f3054960%2fproving-a-statistic-is-ancillary%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
So did you actually try finding the distribution of $log X_1/log X_2$? Where are you stuck?
– StubbornAtom
Dec 28 '18 at 15:09