How to compute the confidence interval, given an unbiased estimator, for $theta$, for the distribution...
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Given an unbiased estimator, how can one compute confidence interval? Consider $Y$, with distribution given by the largest order statistic for a sample of size $2$ with $X_{i}$ sampled from the uniform distribution on $[0,theta]$, that is, pdf of $Y$ is $f_Y(t)=2t/theta^2$. How to compute the $.05$ confidence interval for $theta$, using the unbiased estimator $hattheta=3(Y_1+...+Y_n)/2n$? The example in Wikipedia is limited to normal distribution, so it's not very helpful.
EDIT: Ok, here's some thoughts I have on how to proceed, thanks to https://math.stackexchange.com/a/568579/627534. By central limit theorem, $text{pdf }(2sqrt{2n}(hattheta-theta)/theta)=mathcal{N}(0,1)$ for large enough $n$, so $$.95=P(-1.96leq 2sqrt{2n}(hattheta-theta)/thetaleq 1.96)=P(-1.96leq 2sqrt{2n}hattheta/theta-2sqrt{2n}leq 1.96)=P(-1.96+2sqrt{2n}leq 2sqrt{2n}hattheta/thetaleq 1.96+2sqrt{2n})=P(1/(-1.96+2sqrt{2n})geq theta/2sqrt{2n}hatthetageq 1/(1.96+2sqrt{2n}))=P(2sqrt{2n}hattheta/(-1.96+2sqrt{2n})geq thetageq 2sqrt{2n}hattheta/(1.96+2sqrt{2n})).$$ Is this right?
statistics
asked Dec 18 '18 at 13:25
mckmccmckmcc
11
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add a comment |
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Given an unbiased estimator, how can one compute confidence interval? Consider $Y$, with distribution given by the largest order statistic for a sample of size $2$ with $X_{i}$ sampled from the uniform distribution on $[0,theta]$, that is, pdf of $Y$ is $f_Y(t)=2t/theta^2$. How to compute the $.05$ confidence interval for $theta$, using the unbiased estimator $hattheta=3(Y_1+...+Y_n)/2n$? The example in Wikipedia is limited to normal distribution, so it's not very helpful.
EDIT: Ok, here's some thoughts I have on how to proceed, thanks to https://math.stackexchange.com/a/568579/627534. By central limit theorem, $text{pdf }(2sqrt{2n}(hattheta-theta)/theta)=mathcal{N}(0,1)$ for large enough $n$, so $$.95=P(-1.96leq 2sqrt{2n}(hattheta-theta)/thetaleq 1.96)=P(-1.96leq 2sqrt{2n}hattheta/theta-2sqrt{2n}leq 1.96)=P(-1.96+2sqrt{2n}leq 2sqrt{2n}hattheta/thetaleq 1.96+2sqrt{2n})=P(1/(-1.96+2sqrt{2n})geq theta/2sqrt{2n}hatthetageq 1/(1.96+2sqrt{2n}))=P(2sqrt{2n}hattheta/(-1.96+2sqrt{2n})geq thetageq 2sqrt{2n}hattheta/(1.96+2sqrt{2n})).$$ Is this right?
statistics
asked Dec 18 '18 at 13:25
mckmccmckmcc
11
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There's a big difference between "what is the definition of X" and "how do you compute X". You could give a much better title.
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– David K
Dec 18 '18 at 13:37
add a comment |
$begingroup$
Given an unbiased estimator, how can one compute confidence interval? Consider $Y$, with distribution given by the largest order statistic for a sample of size $2$ with $X_{i}$ sampled from the uniform distribution on $[0,theta]$, that is, pdf of $Y$ is $f_Y(t)=2t/theta^2$. How to compute the $.05$ confidence interval for $theta$, using the unbiased estimator $hattheta=3(Y_1+...+Y_n)/2n$? The example in Wikipedia is limited to normal distribution, so it's not very helpful.
EDIT: Ok, here's some thoughts I have on how to proceed, thanks to https://math.stackexchange.com/a/568579/627534. By central limit theorem, $text{pdf }(2sqrt{2n}(hattheta-theta)/theta)=mathcal{N}(0,1)$ for large enough $n$, so $$.95=P(-1.96leq 2sqrt{2n}(hattheta-theta)/thetaleq 1.96)=P(-1.96leq 2sqrt{2n}hattheta/theta-2sqrt{2n}leq 1.96)=P(-1.96+2sqrt{2n}leq 2sqrt{2n}hattheta/thetaleq 1.96+2sqrt{2n})=P(1/(-1.96+2sqrt{2n})geq theta/2sqrt{2n}hatthetageq 1/(1.96+2sqrt{2n}))=P(2sqrt{2n}hattheta/(-1.96+2sqrt{2n})geq thetageq 2sqrt{2n}hattheta/(1.96+2sqrt{2n})).$$ Is this right?
statistics
asked Dec 18 '18 at 13:25
mckmccmckmcc
11
$endgroup$
Given an unbiased estimator, how can one compute confidence interval? Consider $Y$, with distribution given by the largest order statistic for a sample of size $2$ with $X_{i}$ sampled from the uniform distribution on $[0,theta]$, that is, pdf of $Y$ is $f_Y(t)=2t/theta^2$. How to compute the $.05$ confidence interval for $theta$, using the unbiased estimator $hattheta=3(Y_1+...+Y_n)/2n$? The example in Wikipedia is limited to normal distribution, so it's not very helpful.
EDIT: Ok, here's some thoughts I have on how to proceed, thanks to https://math.stackexchange.com/a/568579/627534. By central limit theorem, $text{pdf }(2sqrt{2n}(hattheta-theta)/theta)=mathcal{N}(0,1)$ for large enough $n$, so $$.95=P(-1.96leq 2sqrt{2n}(hattheta-theta)/thetaleq 1.96)=P(-1.96leq 2sqrt{2n}hattheta/theta-2sqrt{2n}leq 1.96)=P(-1.96+2sqrt{2n}leq 2sqrt{2n}hattheta/thetaleq 1.96+2sqrt{2n})=P(1/(-1.96+2sqrt{2n})geq theta/2sqrt{2n}hatthetageq 1/(1.96+2sqrt{2n}))=P(2sqrt{2n}hattheta/(-1.96+2sqrt{2n})geq thetageq 2sqrt{2n}hattheta/(1.96+2sqrt{2n})).$$ Is this right?
statistics
statistics
asked Dec 18 '18 at 13:25
mckmccmckmcc
11
asked Dec 18 '18 at 13:25
mckmccmckmcc
11
edited Dec 18 '18 at 14:22
mckmcc
asked Dec 18 '18 at 13:25
mckmccmckmcc
11
asked Dec 18 '18 at 13:25
mckmccmckmcc
11
asked Dec 18 '18 at 13:25
mckmccmckmcc
11
11
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There's a big difference between "what is the definition of X" and "how do you compute X". You could give a much better title.
$endgroup$
– David K
Dec 18 '18 at 13:37
add a comment |
$begingroup$
There's a big difference between "what is the definition of X" and "how do you compute X". You could give a much better title.
$endgroup$
– David K
Dec 18 '18 at 13:37
$begingroup$
There's a big difference between "what is the definition of X" and "how do you compute X". You could give a much better title.
$endgroup$
– David K
Dec 18 '18 at 13:37
$begingroup$
There's a big difference between "what is the definition of X" and "how do you compute X". You could give a much better title.
$endgroup$
– David K
Dec 18 '18 at 13:37
add a comment |
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There's a big difference between "what is the definition of X" and "how do you compute X". You could give a much better title.
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– David K
Dec 18 '18 at 13:37