Mixing
Power function exponential distribution
$begingroup$
I am trying to find the power function for a test.
I know that the power function is calculated by $beta(0) = P_0(x in R)$ where $R$ is the rejection region. What I know about this test is that $X sim Exp(beta) \H_0: beta = beta_0 quad H_1: beta > beta_0 \ Y_n = min(X_1,..,X_n) quad Y sim Exp(frac{beta}{n})$
What I have tried until now, is that I think that $P_0(x in R) = P_0(Y_n > c_n) $ since in this scenario the null hypothesis is rejected if all values for $X$ are higher than $C_n$. However I am unsure how to apply the properties of the exponential distribution to $Y_n$ and $c_n$
probability statistics
asked May 5 '16 at 16:01
StevenSteven
2217
$endgroup$
add a comment |
$begingroup$
I am trying to find the power function for a test.
I know that the power function is calculated by $beta(0) = P_0(x in R)$ where $R$ is the rejection region. What I know about this test is that $X sim Exp(beta) \H_0: beta = beta_0 quad H_1: beta > beta_0 \ Y_n = min(X_1,..,X_n) quad Y sim Exp(frac{beta}{n})$
What I have tried until now, is that I think that $P_0(x in R) = P_0(Y_n > c_n) $ since in this scenario the null hypothesis is rejected if all values for $X$ are higher than $C_n$. However I am unsure how to apply the properties of the exponential distribution to $Y_n$ and $c_n$
probability statistics
asked May 5 '16 at 16:01
StevenSteven
2217
$endgroup$
$begingroup$
I'm not saying it is wrong, but how do you conclude that the power function is $mathbb{P}_0(Y_n < c_n)$? There is no description of what kind of test.
$endgroup$
– Siron
May 5 '16 at 16:26
$begingroup$
@Siron I think I have specified in my question that $X_1,..,X_n sim Exp(beta)$ I concluded this because for this scenario $Y_n = min(X_1,..,X_n)$ thus I thought that .. although I don't have proof for it yet
$endgroup$
– Steven
May 5 '16 at 16:35
$begingroup$
Yes, that is the reason that I asked my question. Why do you think $mathbb{P}(Y_n > c_n)$? If this is indeed true then it is not hard to compute the power function.
$endgroup$
– Siron
May 5 '16 at 21:09
$begingroup$
Ok. Yes I gained more insight. For now I know that this is the region for which $H_0$ will be rejected. Because $H_0$ whill be rejected when the value for the test statistic of the hypothesis's distribution is smaller than the alpha level. So thus I should calculate this region using the information I know about this is distributed? But will I need to calculate test statistics of the exponential distribution? I know what the region means but I don"t get how to get further..
$endgroup$
– Steven
May 5 '16 at 21:23
add a comment |
$begingroup$
I am trying to find the power function for a test.
I know that the power function is calculated by $beta(0) = P_0(x in R)$ where $R$ is the rejection region. What I know about this test is that $X sim Exp(beta) \H_0: beta = beta_0 quad H_1: beta > beta_0 \ Y_n = min(X_1,..,X_n) quad Y sim Exp(frac{beta}{n})$
What I have tried until now, is that I think that $P_0(x in R) = P_0(Y_n > c_n) $ since in this scenario the null hypothesis is rejected if all values for $X$ are higher than $C_n$. However I am unsure how to apply the properties of the exponential distribution to $Y_n$ and $c_n$
probability statistics
asked May 5 '16 at 16:01
StevenSteven
2217
$endgroup$
I am trying to find the power function for a test.
I know that the power function is calculated by $beta(0) = P_0(x in R)$ where $R$ is the rejection region. What I know about this test is that $X sim Exp(beta) \H_0: beta = beta_0 quad H_1: beta > beta_0 \ Y_n = min(X_1,..,X_n) quad Y sim Exp(frac{beta}{n})$
What I have tried until now, is that I think that $P_0(x in R) = P_0(Y_n > c_n) $ since in this scenario the null hypothesis is rejected if all values for $X$ are higher than $C_n$. However I am unsure how to apply the properties of the exponential distribution to $Y_n$ and $c_n$
probability statistics
probability statistics
asked May 5 '16 at 16:01
StevenSteven
2217
asked May 5 '16 at 16:01
StevenSteven
2217
edited May 5 '16 at 17:28
Steven
asked May 5 '16 at 16:01
StevenSteven
2217
asked May 5 '16 at 16:01
StevenSteven
2217
asked May 5 '16 at 16:01
StevenSteven
2217
2217
$begingroup$
I'm not saying it is wrong, but how do you conclude that the power function is $mathbb{P}_0(Y_n < c_n)$? There is no description of what kind of test.
$endgroup$
– Siron
May 5 '16 at 16:26
$begingroup$
@Siron I think I have specified in my question that $X_1,..,X_n sim Exp(beta)$ I concluded this because for this scenario $Y_n = min(X_1,..,X_n)$ thus I thought that .. although I don't have proof for it yet
$endgroup$
– Steven
May 5 '16 at 16:35
$begingroup$
Yes, that is the reason that I asked my question. Why do you think $mathbb{P}(Y_n > c_n)$? If this is indeed true then it is not hard to compute the power function.
$endgroup$
– Siron
May 5 '16 at 21:09
$begingroup$
Ok. Yes I gained more insight. For now I know that this is the region for which $H_0$ will be rejected. Because $H_0$ whill be rejected when the value for the test statistic of the hypothesis's distribution is smaller than the alpha level. So thus I should calculate this region using the information I know about this is distributed? But will I need to calculate test statistics of the exponential distribution? I know what the region means but I don"t get how to get further..
$endgroup$
– Steven
May 5 '16 at 21:23
add a comment |
$begingroup$
I'm not saying it is wrong, but how do you conclude that the power function is $mathbb{P}_0(Y_n < c_n)$? There is no description of what kind of test.
$endgroup$
– Siron
May 5 '16 at 16:26
$begingroup$
@Siron I think I have specified in my question that $X_1,..,X_n sim Exp(beta)$ I concluded this because for this scenario $Y_n = min(X_1,..,X_n)$ thus I thought that .. although I don't have proof for it yet
$endgroup$
– Steven
May 5 '16 at 16:35
$begingroup$
Yes, that is the reason that I asked my question. Why do you think $mathbb{P}(Y_n > c_n)$? If this is indeed true then it is not hard to compute the power function.
$endgroup$
– Siron
May 5 '16 at 21:09
$begingroup$
Ok. Yes I gained more insight. For now I know that this is the region for which $H_0$ will be rejected. Because $H_0$ whill be rejected when the value for the test statistic of the hypothesis's distribution is smaller than the alpha level. So thus I should calculate this region using the information I know about this is distributed? But will I need to calculate test statistics of the exponential distribution? I know what the region means but I don"t get how to get further..
$endgroup$
– Steven
May 5 '16 at 21:23
$begingroup$
I'm not saying it is wrong, but how do you conclude that the power function is $mathbb{P}_0(Y_n < c_n)$? There is no description of what kind of test.
$endgroup$
– Siron
May 5 '16 at 16:26
$begingroup$
I'm not saying it is wrong, but how do you conclude that the power function is $mathbb{P}_0(Y_n < c_n)$? There is no description of what kind of test.
$endgroup$
– Siron
May 5 '16 at 16:26
$begingroup$
@Siron I think I have specified in my question that $X_1,..,X_n sim Exp(beta)$ I concluded this because for this scenario $Y_n = min(X_1,..,X_n)$ thus I thought that .. although I don't have proof for it yet
$endgroup$
– Steven
May 5 '16 at 16:35
$begingroup$
@Siron I think I have specified in my question that $X_1,..,X_n sim Exp(beta)$ I concluded this because for this scenario $Y_n = min(X_1,..,X_n)$ thus I thought that .. although I don't have proof for it yet
$endgroup$
– Steven
May 5 '16 at 16:35
$begingroup$
Yes, that is the reason that I asked my question. Why do you think $mathbb{P}(Y_n > c_n)$? If this is indeed true then it is not hard to compute the power function.
$endgroup$
– Siron
May 5 '16 at 21:09
$begingroup$
Yes, that is the reason that I asked my question. Why do you think $mathbb{P}(Y_n > c_n)$? If this is indeed true then it is not hard to compute the power function.
$endgroup$
– Siron
May 5 '16 at 21:09
$begingroup$
Ok. Yes I gained more insight. For now I know that this is the region for which $H_0$ will be rejected. Because $H_0$ whill be rejected when the value for the test statistic of the hypothesis's distribution is smaller than the alpha level. So thus I should calculate this region using the information I know about this is distributed? But will I need to calculate test statistics of the exponential distribution? I know what the region means but I don"t get how to get further..
$endgroup$
– Steven
May 5 '16 at 21:23
$begingroup$
Ok. Yes I gained more insight. For now I know that this is the region for which $H_0$ will be rejected. Because $H_0$ whill be rejected when the value for the test statistic of the hypothesis's distribution is smaller than the alpha level. So thus I should calculate this region using the information I know about this is distributed? But will I need to calculate test statistics of the exponential distribution? I know what the region means but I don"t get how to get further..
$endgroup$
– Steven
May 5 '16 at 21:23
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Based on your information I think the question has to be interpret as
Suppose $X_1,ldots,X_n$ are i.i.d exponentially distributed with an unknown parameter $beta$. Suppose we want to test $H_0: beta = beta_0$ versus $H_1: beta > beta_0$. Suppose the test is performed by rejecting $H_0$ when $Y = min{X_1,ldots,X_n}>c$. Compute the power function for this test.
Since $Y sim mbox{Exp}left(frac{beta}{n}right)$, we can easily compute the power function
$$mathbb{P}(Y_n > c) = 1 - mathbb{P}(Y_n < c) = 1- F_{Y_n}(c),$$
where $F_{Y_n}$ is the distribution function of an exponential distribution with parameter $beta/n$. Please correct me if my interpretation of the question is incorrect.
answered May 5 '16 at 22:11
SironSiron
1,205514
$endgroup$
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%2f1772904%2fpower-function-exponential-distribution%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
$begingroup$
Based on your information I think the question has to be interpret as
Suppose $X_1,ldots,X_n$ are i.i.d exponentially distributed with an unknown parameter $beta$. Suppose we want to test $H_0: beta = beta_0$ versus $H_1: beta > beta_0$. Suppose the test is performed by rejecting $H_0$ when $Y = min{X_1,ldots,X_n}>c$. Compute the power function for this test.
Since $Y sim mbox{Exp}left(frac{beta}{n}right)$, we can easily compute the power function
$$mathbb{P}(Y_n > c) = 1 - mathbb{P}(Y_n < c) = 1- F_{Y_n}(c),$$
where $F_{Y_n}$ is the distribution function of an exponential distribution with parameter $beta/n$. Please correct me if my interpretation of the question is incorrect.
answered May 5 '16 at 22:11
SironSiron
1,205514
$endgroup$
add a comment |
$begingroup$
Based on your information I think the question has to be interpret as
Suppose $X_1,ldots,X_n$ are i.i.d exponentially distributed with an unknown parameter $beta$. Suppose we want to test $H_0: beta = beta_0$ versus $H_1: beta > beta_0$. Suppose the test is performed by rejecting $H_0$ when $Y = min{X_1,ldots,X_n}>c$. Compute the power function for this test.
Since $Y sim mbox{Exp}left(frac{beta}{n}right)$, we can easily compute the power function
$$mathbb{P}(Y_n > c) = 1 - mathbb{P}(Y_n < c) = 1- F_{Y_n}(c),$$
where $F_{Y_n}$ is the distribution function of an exponential distribution with parameter $beta/n$. Please correct me if my interpretation of the question is incorrect.
answered May 5 '16 at 22:11
SironSiron
1,205514
$endgroup$
add a comment |
$begingroup$
Based on your information I think the question has to be interpret as
Suppose $X_1,ldots,X_n$ are i.i.d exponentially distributed with an unknown parameter $beta$. Suppose we want to test $H_0: beta = beta_0$ versus $H_1: beta > beta_0$. Suppose the test is performed by rejecting $H_0$ when $Y = min{X_1,ldots,X_n}>c$. Compute the power function for this test.
Since $Y sim mbox{Exp}left(frac{beta}{n}right)$, we can easily compute the power function
$$mathbb{P}(Y_n > c) = 1 - mathbb{P}(Y_n < c) = 1- F_{Y_n}(c),$$
where $F_{Y_n}$ is the distribution function of an exponential distribution with parameter $beta/n$. Please correct me if my interpretation of the question is incorrect.
answered May 5 '16 at 22:11
SironSiron
1,205514
$endgroup$
Based on your information I think the question has to be interpret as
Suppose $X_1,ldots,X_n$ are i.i.d exponentially distributed with an unknown parameter $beta$. Suppose we want to test $H_0: beta = beta_0$ versus $H_1: beta > beta_0$. Suppose the test is performed by rejecting $H_0$ when $Y = min{X_1,ldots,X_n}>c$. Compute the power function for this test.
Since $Y sim mbox{Exp}left(frac{beta}{n}right)$, we can easily compute the power function
$$mathbb{P}(Y_n > c) = 1 - mathbb{P}(Y_n < c) = 1- F_{Y_n}(c),$$
where $F_{Y_n}$ is the distribution function of an exponential distribution with parameter $beta/n$. Please correct me if my interpretation of the question is incorrect.
answered May 5 '16 at 22:11
SironSiron
1,205514
answered May 5 '16 at 22:11
SironSiron
1,205514
answered May 5 '16 at 22:11
SironSiron
1,205514
answered May 5 '16 at 22:11
SironSiron
1,205514
1,205514
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%2f1772904%2fpower-function-exponential-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
$begingroup$
I'm not saying it is wrong, but how do you conclude that the power function is $mathbb{P}_0(Y_n < c_n)$? There is no description of what kind of test.
$endgroup$
– Siron
May 5 '16 at 16:26
$begingroup$
@Siron I think I have specified in my question that $X_1,..,X_n sim Exp(beta)$ I concluded this because for this scenario $Y_n = min(X_1,..,X_n)$ thus I thought that .. although I don't have proof for it yet
$endgroup$
– Steven
May 5 '16 at 16:35
$begingroup$
Yes, that is the reason that I asked my question. Why do you think $mathbb{P}(Y_n > c_n)$? If this is indeed true then it is not hard to compute the power function.
$endgroup$
– Siron
May 5 '16 at 21:09
$begingroup$
Ok. Yes I gained more insight. For now I know that this is the region for which $H_0$ will be rejected. Because $H_0$ whill be rejected when the value for the test statistic of the hypothesis's distribution is smaller than the alpha level. So thus I should calculate this region using the information I know about this is distributed? But will I need to calculate test statistics of the exponential distribution? I know what the region means but I don"t get how to get further..
$endgroup$
– Steven
May 5 '16 at 21:23