Exact Confidence Interval for Uniform Parameter The 2019 Stack Overflow Developer Survey Results Are In$100(1-alpha)$% approximate confidence intervalConfidence interval for Poisson distribution coefficientConfidence interval of a uniform distributionConfidence interval of the parameter of $exp$ and normal distribution from MLE?Confidence interval for a density function parameterConfidence interval for parameterConstructing a symmetrical $100(1-a)%$ confidence interval for $theta$.confidence interval with MLE estimatorHow to compute asymptotic confidence interval for linear regression?
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Exact Confidence Interval for Uniform Parameter
The 2019 Stack Overflow Developer Survey Results Are In$100(1-alpha)$% approximate confidence intervalConfidence interval for Poisson distribution coefficientConfidence interval of a uniform distributionConfidence interval of the parameter of $exp$ and normal distribution from MLE?Confidence interval for a density function parameterConfidence interval for parameterConstructing a symmetrical $100(1-a)%$ confidence interval for $theta$.confidence interval with MLE estimatorHow to compute asymptotic confidence interval for linear regression?
$begingroup$
I am given the following.
$$X_1, X_2,...,X_n sim U(0,theta)$$
and I want to get an exact confidence interval of $theta$ without using normal approximation.
What I know so far is that $Y_n=max(x_1,x_2,...,x_n)$ is the MLE for $theta $ and $fracn+1nY_n$ is the unbiased estimator.
I have been instructed to find the confidence interval based on this estimator and this is what I have tried.
Let $W=fracn+1nY_n/theta$ then
$$Pr[l<W<u]=1-alpha$$
$$Pr[W<l]=alpha/2$$
$$Pr[W<u]=1-alpha/2$$
where $l$ and $u$ represent the lower and the upper bound of the confidence interval.
Now here is what confuses me.
When I solve for the confidence interval I get
$$fracY_n^nsqrt1-alpha/2 <theta<fracY_n^nsqrtalpha/2 $$
and algebraically the $fracn+1n$ became irrelevant.
even though I got $l=fracn+1n^nsqrtalpha/2$ and $u=fracn+1n^nsqrt1-alpha/2$
Trying to clarify my question as much as possible, what I probably want to say is that if the unbiased estimator $fracn+1nY_n$ is superior to just the MLE, $Y_n$, how is it that the confident interval is not centered around that unbiased estimator?
I know that this is a very odd question and I could not find something similar to my argument online, so I would really appreciate your help.
statistics confidence-interval
asked Mar 30 at 21:39
hyg17hyg17
2,00422044
$endgroup$
add a comment |
$begingroup$
I am given the following.
$$X_1, X_2,...,X_n sim U(0,theta)$$
and I want to get an exact confidence interval of $theta$ without using normal approximation.
What I know so far is that $Y_n=max(x_1,x_2,...,x_n)$ is the MLE for $theta $ and $fracn+1nY_n$ is the unbiased estimator.
I have been instructed to find the confidence interval based on this estimator and this is what I have tried.
Let $W=fracn+1nY_n/theta$ then
$$Pr[l<W<u]=1-alpha$$
$$Pr[W<l]=alpha/2$$
$$Pr[W<u]=1-alpha/2$$
where $l$ and $u$ represent the lower and the upper bound of the confidence interval.
Now here is what confuses me.
When I solve for the confidence interval I get
$$fracY_n^nsqrt1-alpha/2 <theta<fracY_n^nsqrtalpha/2 $$
and algebraically the $fracn+1n$ became irrelevant.
even though I got $l=fracn+1n^nsqrtalpha/2$ and $u=fracn+1n^nsqrt1-alpha/2$
Trying to clarify my question as much as possible, what I probably want to say is that if the unbiased estimator $fracn+1nY_n$ is superior to just the MLE, $Y_n$, how is it that the confident interval is not centered around that unbiased estimator?
I know that this is a very odd question and I could not find something similar to my argument online, so I would really appreciate your help.
statistics confidence-interval
asked Mar 30 at 21:39
hyg17hyg17
2,00422044
$endgroup$
$begingroup$
See Wikipedia on Uniform Dist'n, CI for maximum.
$endgroup$
– BruceET
Mar 31 at 6:30
$begingroup$
What is important here is that $Y_n/theta$ is a pivot. And we expect the CI based on this pivot to be 'good' as $Y_n$ is a sufficient statistic (and MLE also). Why is the unbiased estimator superior to the MLE? Unbiasedness is irrelevant here.
$endgroup$
– StubbornAtom
Mar 31 at 6:37
$begingroup$
I am thinking that the MLE will never get close enough to the actual parameter, so that is why it is not useful. Can you tell me why that has nothing to do with the confidence interval? What I want to say here is that, I tried to create a confidence interval using the unbiased estimator, but the result is no different from just using the MLE. That is counter intuitive to me.
$endgroup$
– hyg17
Apr 1 at 18:40
add a comment |
$begingroup$
I am given the following.
$$X_1, X_2,...,X_n sim U(0,theta)$$
and I want to get an exact confidence interval of $theta$ without using normal approximation.
What I know so far is that $Y_n=max(x_1,x_2,...,x_n)$ is the MLE for $theta $ and $fracn+1nY_n$ is the unbiased estimator.
I have been instructed to find the confidence interval based on this estimator and this is what I have tried.
Let $W=fracn+1nY_n/theta$ then
$$Pr[l<W<u]=1-alpha$$
$$Pr[W<l]=alpha/2$$
$$Pr[W<u]=1-alpha/2$$
where $l$ and $u$ represent the lower and the upper bound of the confidence interval.
Now here is what confuses me.
When I solve for the confidence interval I get
$$fracY_n^nsqrt1-alpha/2 <theta<fracY_n^nsqrtalpha/2 $$
and algebraically the $fracn+1n$ became irrelevant.
even though I got $l=fracn+1n^nsqrtalpha/2$ and $u=fracn+1n^nsqrt1-alpha/2$
Trying to clarify my question as much as possible, what I probably want to say is that if the unbiased estimator $fracn+1nY_n$ is superior to just the MLE, $Y_n$, how is it that the confident interval is not centered around that unbiased estimator?
I know that this is a very odd question and I could not find something similar to my argument online, so I would really appreciate your help.
statistics confidence-interval
asked Mar 30 at 21:39
hyg17hyg17
2,00422044
$endgroup$
I am given the following.
$$X_1, X_2,...,X_n sim U(0,theta)$$
and I want to get an exact confidence interval of $theta$ without using normal approximation.
What I know so far is that $Y_n=max(x_1,x_2,...,x_n)$ is the MLE for $theta $ and $fracn+1nY_n$ is the unbiased estimator.
I have been instructed to find the confidence interval based on this estimator and this is what I have tried.
Let $W=fracn+1nY_n/theta$ then
$$Pr[l<W<u]=1-alpha$$
$$Pr[W<l]=alpha/2$$
$$Pr[W<u]=1-alpha/2$$
where $l$ and $u$ represent the lower and the upper bound of the confidence interval.
Now here is what confuses me.
When I solve for the confidence interval I get
$$fracY_n^nsqrt1-alpha/2 <theta<fracY_n^nsqrtalpha/2 $$
and algebraically the $fracn+1n$ became irrelevant.
even though I got $l=fracn+1n^nsqrtalpha/2$ and $u=fracn+1n^nsqrt1-alpha/2$
Trying to clarify my question as much as possible, what I probably want to say is that if the unbiased estimator $fracn+1nY_n$ is superior to just the MLE, $Y_n$, how is it that the confident interval is not centered around that unbiased estimator?
I know that this is a very odd question and I could not find something similar to my argument online, so I would really appreciate your help.
statistics confidence-interval
statistics confidence-interval
asked Mar 30 at 21:39
hyg17hyg17
2,00422044
asked Mar 30 at 21:39
hyg17hyg17
2,00422044
asked Mar 30 at 21:39
hyg17hyg17
2,00422044
asked Mar 30 at 21:39
hyg17hyg17
2,00422044
asked Mar 30 at 21:39
hyg17hyg17
2,00422044
2,00422044
$begingroup$
See Wikipedia on Uniform Dist'n, CI for maximum.
$endgroup$
– BruceET
Mar 31 at 6:30
$begingroup$
What is important here is that $Y_n/theta$ is a pivot. And we expect the CI based on this pivot to be 'good' as $Y_n$ is a sufficient statistic (and MLE also). Why is the unbiased estimator superior to the MLE? Unbiasedness is irrelevant here.
$endgroup$
– StubbornAtom
Mar 31 at 6:37
$begingroup$
I am thinking that the MLE will never get close enough to the actual parameter, so that is why it is not useful. Can you tell me why that has nothing to do with the confidence interval? What I want to say here is that, I tried to create a confidence interval using the unbiased estimator, but the result is no different from just using the MLE. That is counter intuitive to me.
$endgroup$
– hyg17
Apr 1 at 18:40
add a comment |
$begingroup$
See Wikipedia on Uniform Dist'n, CI for maximum.
$endgroup$
– BruceET
Mar 31 at 6:30
$begingroup$
What is important here is that $Y_n/theta$ is a pivot. And we expect the CI based on this pivot to be 'good' as $Y_n$ is a sufficient statistic (and MLE also). Why is the unbiased estimator superior to the MLE? Unbiasedness is irrelevant here.
$endgroup$
– StubbornAtom
Mar 31 at 6:37
$begingroup$
I am thinking that the MLE will never get close enough to the actual parameter, so that is why it is not useful. Can you tell me why that has nothing to do with the confidence interval? What I want to say here is that, I tried to create a confidence interval using the unbiased estimator, but the result is no different from just using the MLE. That is counter intuitive to me.
$endgroup$
– hyg17
Apr 1 at 18:40
$begingroup$
See Wikipedia on Uniform Dist'n, CI for maximum.
$endgroup$
– BruceET
Mar 31 at 6:30
$begingroup$
See Wikipedia on Uniform Dist'n, CI for maximum.
$endgroup$
– BruceET
Mar 31 at 6:30
$begingroup$
What is important here is that $Y_n/theta$ is a pivot. And we expect the CI based on this pivot to be 'good' as $Y_n$ is a sufficient statistic (and MLE also). Why is the unbiased estimator superior to the MLE? Unbiasedness is irrelevant here.
$endgroup$
– StubbornAtom
Mar 31 at 6:37
$begingroup$
What is important here is that $Y_n/theta$ is a pivot. And we expect the CI based on this pivot to be 'good' as $Y_n$ is a sufficient statistic (and MLE also). Why is the unbiased estimator superior to the MLE? Unbiasedness is irrelevant here.
$endgroup$
– StubbornAtom
Mar 31 at 6:37
$begingroup$
I am thinking that the MLE will never get close enough to the actual parameter, so that is why it is not useful. Can you tell me why that has nothing to do with the confidence interval? What I want to say here is that, I tried to create a confidence interval using the unbiased estimator, but the result is no different from just using the MLE. That is counter intuitive to me.
$endgroup$
– hyg17
Apr 1 at 18:40
$begingroup$
I am thinking that the MLE will never get close enough to the actual parameter, so that is why it is not useful. Can you tell me why that has nothing to do with the confidence interval? What I want to say here is that, I tried to create a confidence interval using the unbiased estimator, but the result is no different from just using the MLE. That is counter intuitive to me.
$endgroup$
– hyg17
Apr 1 at 18:40
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Be "centered" you assume that $Y_n$ has some symmetric distribution (density), however $Y_n$ is clearly asymmetrically distributed, hence there is no reason why an exact CI will be symmetric (and thus there is no meaning to talk about "centering" it).
answered Mar 31 at 7:19
V. VancakV. Vancak
11.4k31026
$endgroup$
$begingroup$
So, what I want to know is this. As a point estimator, $fracn+1nX_(n)$ is unbiased. But when I find the confidence interval using this estimator and make it a pivot by saying $W=fracn+1nX_(n)/ theta$ the $fracn+1n$ is not reflected in the final solution compared to when I simply use $W=fracX_(n)theta$
$endgroup$
– hyg17
Apr 3 at 23:20
add a comment |
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1 Answer
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oldest
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1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Be "centered" you assume that $Y_n$ has some symmetric distribution (density), however $Y_n$ is clearly asymmetrically distributed, hence there is no reason why an exact CI will be symmetric (and thus there is no meaning to talk about "centering" it).
answered Mar 31 at 7:19
V. VancakV. Vancak
11.4k31026
$endgroup$
$begingroup$
So, what I want to know is this. As a point estimator, $fracn+1nX_(n)$ is unbiased. But when I find the confidence interval using this estimator and make it a pivot by saying $W=fracn+1nX_(n)/ theta$ the $fracn+1n$ is not reflected in the final solution compared to when I simply use $W=fracX_(n)theta$
$endgroup$
– hyg17
Apr 3 at 23:20
add a comment |
$begingroup$
Be "centered" you assume that $Y_n$ has some symmetric distribution (density), however $Y_n$ is clearly asymmetrically distributed, hence there is no reason why an exact CI will be symmetric (and thus there is no meaning to talk about "centering" it).
answered Mar 31 at 7:19
V. VancakV. Vancak
11.4k31026
$endgroup$
$begingroup$
So, what I want to know is this. As a point estimator, $fracn+1nX_(n)$ is unbiased. But when I find the confidence interval using this estimator and make it a pivot by saying $W=fracn+1nX_(n)/ theta$ the $fracn+1n$ is not reflected in the final solution compared to when I simply use $W=fracX_(n)theta$
$endgroup$
– hyg17
Apr 3 at 23:20
add a comment |
$begingroup$
Be "centered" you assume that $Y_n$ has some symmetric distribution (density), however $Y_n$ is clearly asymmetrically distributed, hence there is no reason why an exact CI will be symmetric (and thus there is no meaning to talk about "centering" it).
answered Mar 31 at 7:19
V. VancakV. Vancak
11.4k31026
$endgroup$
Be "centered" you assume that $Y_n$ has some symmetric distribution (density), however $Y_n$ is clearly asymmetrically distributed, hence there is no reason why an exact CI will be symmetric (and thus there is no meaning to talk about "centering" it).
answered Mar 31 at 7:19
V. VancakV. Vancak
11.4k31026
answered Mar 31 at 7:19
V. VancakV. Vancak
11.4k31026
answered Mar 31 at 7:19
V. VancakV. Vancak
11.4k31026
answered Mar 31 at 7:19
V. VancakV. Vancak
11.4k31026
11.4k31026
$begingroup$
So, what I want to know is this. As a point estimator, $fracn+1nX_(n)$ is unbiased. But when I find the confidence interval using this estimator and make it a pivot by saying $W=fracn+1nX_(n)/ theta$ the $fracn+1n$ is not reflected in the final solution compared to when I simply use $W=fracX_(n)theta$
$endgroup$
– hyg17
Apr 3 at 23:20
add a comment |
$begingroup$
So, what I want to know is this. As a point estimator, $fracn+1nX_(n)$ is unbiased. But when I find the confidence interval using this estimator and make it a pivot by saying $W=fracn+1nX_(n)/ theta$ the $fracn+1n$ is not reflected in the final solution compared to when I simply use $W=fracX_(n)theta$
$endgroup$
– hyg17
Apr 3 at 23:20
$begingroup$
So, what I want to know is this. As a point estimator, $fracn+1nX_(n)$ is unbiased. But when I find the confidence interval using this estimator and make it a pivot by saying $W=fracn+1nX_(n)/ theta$ the $fracn+1n$ is not reflected in the final solution compared to when I simply use $W=fracX_(n)theta$
$endgroup$
– hyg17
Apr 3 at 23:20
$begingroup$
So, what I want to know is this. As a point estimator, $fracn+1nX_(n)$ is unbiased. But when I find the confidence interval using this estimator and make it a pivot by saying $W=fracn+1nX_(n)/ theta$ the $fracn+1n$ is not reflected in the final solution compared to when I simply use $W=fracX_(n)theta$
$endgroup$
– hyg17
Apr 3 at 23:20
add a comment |
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$begingroup$
See Wikipedia on Uniform Dist'n, CI for maximum.
$endgroup$
– BruceET
Mar 31 at 6:30
$begingroup$
What is important here is that $Y_n/theta$ is a pivot. And we expect the CI based on this pivot to be 'good' as $Y_n$ is a sufficient statistic (and MLE also). Why is the unbiased estimator superior to the MLE? Unbiasedness is irrelevant here.
$endgroup$
– StubbornAtom
Mar 31 at 6:37
$begingroup$
I am thinking that the MLE will never get close enough to the actual parameter, so that is why it is not useful. Can you tell me why that has nothing to do with the confidence interval? What I want to say here is that, I tried to create a confidence interval using the unbiased estimator, but the result is no different from just using the MLE. That is counter intuitive to me.
$endgroup$
– hyg17
Apr 1 at 18:40