Find a lower bound of 95% confidence for θ, of $U(0,θ)$ distribution
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Consider a random sample of observations of a $U(0,θ)$ distribution. Find a lower bound of 95% confidence for θ
My idea was:
Let $X_ {1}, X_ {2}, ..., X_ {N}$ be a random sample of a uniform population in the interval $(0, θ)$ and $X_ {n}$ the maximum.
So
$(frac{1}{a}) ^ n - (frac{1}{b}) ^ n$ is the confidence coefficient of the parameter.
But I do not know how to continue to solve this problem, can anybody help me?
statistical-inference parameter-estimation
asked Nov 21 at 4:42
Jazmín Jones
548
add a comment |
up vote
0
down vote
favorite
Consider a random sample of observations of a $U(0,θ)$ distribution. Find a lower bound of 95% confidence for θ
My idea was:
Let $X_ {1}, X_ {2}, ..., X_ {N}$ be a random sample of a uniform population in the interval $(0, θ)$ and $X_ {n}$ the maximum.
So
$(frac{1}{a}) ^ n - (frac{1}{b}) ^ n$ is the confidence coefficient of the parameter.
But I do not know how to continue to solve this problem, can anybody help me?
statistical-inference parameter-estimation
asked Nov 21 at 4:42
Jazmín Jones
548
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
Consider a random sample of observations of a $U(0,θ)$ distribution. Find a lower bound of 95% confidence for θ
My idea was:
Let $X_ {1}, X_ {2}, ..., X_ {N}$ be a random sample of a uniform population in the interval $(0, θ)$ and $X_ {n}$ the maximum.
So
$(frac{1}{a}) ^ n - (frac{1}{b}) ^ n$ is the confidence coefficient of the parameter.
But I do not know how to continue to solve this problem, can anybody help me?
statistical-inference parameter-estimation
asked Nov 21 at 4:42
Jazmín Jones
548
Consider a random sample of observations of a $U(0,θ)$ distribution. Find a lower bound of 95% confidence for θ
My idea was:
Let $X_ {1}, X_ {2}, ..., X_ {N}$ be a random sample of a uniform population in the interval $(0, θ)$ and $X_ {n}$ the maximum.
So
$(frac{1}{a}) ^ n - (frac{1}{b}) ^ n$ is the confidence coefficient of the parameter.
But I do not know how to continue to solve this problem, can anybody help me?
statistical-inference parameter-estimation
statistical-inference parameter-estimation
asked Nov 21 at 4:42
Jazmín Jones
548
asked Nov 21 at 4:42
Jazmín Jones
548
asked Nov 21 at 4:42
Jazmín Jones
548
asked Nov 21 at 4:42
Jazmín Jones
548
asked Nov 21 at 4:42
Jazmín Jones
548
548
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
up vote
0
down vote
The maximum is a sufficient statistic here, so a good choice for a test statistic for your CI. (However I'm not sure what you mean with the $(1/a^n-1/b^n)$... it may just be terminology I'm not aware of).
To get a CI we need to first find the distribution of the test statistic: $$ P(M_nle m) = P(Xle m)^n = (m/theta)^n.$$ Intuitively, we know the $theta>M_n$ so it makes sense to set the lower bound of the CI to be $M_n$ and then let the upper end of the interval be $M_n(1+c^*)$ for some choice of $c^*.$ (Note as always there is no unique confidence interval and this is just a sensible and convenient choice). Then we can choose $c^*$ so we have $1-alpha$ coverage (where $alpha=0.05)$: $$ 1-alpha=P(theta < (1+c^*)M_n) = 1- Pleft(M_nle frac{theta}{1+c^*}right)= 1-(1+c^*)^{-n}$$ which gives $$ c^* = alpha^{-1/n}-1$$
EDIT
It appears I misinterpreted the problem here. If we want a lower bound that has $1-alpha$ confidence (i.e. a minimal statistic $x$ such that $P(theta>x) le 1-alpha$) then we can choose $x=(1+c^*)M_n$ and choose $c^*$ such that $$ 1-alpha = P((1+c^*)M_n< theta)= (1+c^*)^{-n},$$ so this gives $$ c^* = (1-alpha)^{-1/n}-1,$$ which is a good deal smaller than the $c^*$ derived above for an upper bound.
answered Nov 21 at 5:35
spaceisdarkgreen
31.6k21552
sorry, but what do you mean with CI?
– Jazmín Jones
Nov 21 at 5:41
@JazmínJones I mean confidence interval. Perhaps I misinterpreted the problem
– spaceisdarkgreen
Nov 21 at 5:43
@JazmínJones I edited in what I think is an answer to the problem. My interpretation is that you want to know how far above $M_n$ we can go while still having $ge 95%$ confidence that $theta$ is larger than this value. (My initial answer gives an upper bound instead.)
– spaceisdarkgreen
Nov 21 at 5:58
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
The maximum is a sufficient statistic here, so a good choice for a test statistic for your CI. (However I'm not sure what you mean with the $(1/a^n-1/b^n)$... it may just be terminology I'm not aware of).
To get a CI we need to first find the distribution of the test statistic: $$ P(M_nle m) = P(Xle m)^n = (m/theta)^n.$$ Intuitively, we know the $theta>M_n$ so it makes sense to set the lower bound of the CI to be $M_n$ and then let the upper end of the interval be $M_n(1+c^*)$ for some choice of $c^*.$ (Note as always there is no unique confidence interval and this is just a sensible and convenient choice). Then we can choose $c^*$ so we have $1-alpha$ coverage (where $alpha=0.05)$: $$ 1-alpha=P(theta < (1+c^*)M_n) = 1- Pleft(M_nle frac{theta}{1+c^*}right)= 1-(1+c^*)^{-n}$$ which gives $$ c^* = alpha^{-1/n}-1$$
EDIT
It appears I misinterpreted the problem here. If we want a lower bound that has $1-alpha$ confidence (i.e. a minimal statistic $x$ such that $P(theta>x) le 1-alpha$) then we can choose $x=(1+c^*)M_n$ and choose $c^*$ such that $$ 1-alpha = P((1+c^*)M_n< theta)= (1+c^*)^{-n},$$ so this gives $$ c^* = (1-alpha)^{-1/n}-1,$$ which is a good deal smaller than the $c^*$ derived above for an upper bound.
answered Nov 21 at 5:35
spaceisdarkgreen
31.6k21552
sorry, but what do you mean with CI?
– Jazmín Jones
Nov 21 at 5:41
@JazmínJones I mean confidence interval. Perhaps I misinterpreted the problem
– spaceisdarkgreen
Nov 21 at 5:43
@JazmínJones I edited in what I think is an answer to the problem. My interpretation is that you want to know how far above $M_n$ we can go while still having $ge 95%$ confidence that $theta$ is larger than this value. (My initial answer gives an upper bound instead.)
– spaceisdarkgreen
Nov 21 at 5:58
add a comment |
up vote
0
down vote
The maximum is a sufficient statistic here, so a good choice for a test statistic for your CI. (However I'm not sure what you mean with the $(1/a^n-1/b^n)$... it may just be terminology I'm not aware of).
To get a CI we need to first find the distribution of the test statistic: $$ P(M_nle m) = P(Xle m)^n = (m/theta)^n.$$ Intuitively, we know the $theta>M_n$ so it makes sense to set the lower bound of the CI to be $M_n$ and then let the upper end of the interval be $M_n(1+c^*)$ for some choice of $c^*.$ (Note as always there is no unique confidence interval and this is just a sensible and convenient choice). Then we can choose $c^*$ so we have $1-alpha$ coverage (where $alpha=0.05)$: $$ 1-alpha=P(theta < (1+c^*)M_n) = 1- Pleft(M_nle frac{theta}{1+c^*}right)= 1-(1+c^*)^{-n}$$ which gives $$ c^* = alpha^{-1/n}-1$$
EDIT
It appears I misinterpreted the problem here. If we want a lower bound that has $1-alpha$ confidence (i.e. a minimal statistic $x$ such that $P(theta>x) le 1-alpha$) then we can choose $x=(1+c^*)M_n$ and choose $c^*$ such that $$ 1-alpha = P((1+c^*)M_n< theta)= (1+c^*)^{-n},$$ so this gives $$ c^* = (1-alpha)^{-1/n}-1,$$ which is a good deal smaller than the $c^*$ derived above for an upper bound.
answered Nov 21 at 5:35
spaceisdarkgreen
31.6k21552
sorry, but what do you mean with CI?
– Jazmín Jones
Nov 21 at 5:41
@JazmínJones I mean confidence interval. Perhaps I misinterpreted the problem
– spaceisdarkgreen
Nov 21 at 5:43
@JazmínJones I edited in what I think is an answer to the problem. My interpretation is that you want to know how far above $M_n$ we can go while still having $ge 95%$ confidence that $theta$ is larger than this value. (My initial answer gives an upper bound instead.)
– spaceisdarkgreen
Nov 21 at 5:58
add a comment |
up vote
0
down vote
up vote
0
down vote
The maximum is a sufficient statistic here, so a good choice for a test statistic for your CI. (However I'm not sure what you mean with the $(1/a^n-1/b^n)$... it may just be terminology I'm not aware of).
To get a CI we need to first find the distribution of the test statistic: $$ P(M_nle m) = P(Xle m)^n = (m/theta)^n.$$ Intuitively, we know the $theta>M_n$ so it makes sense to set the lower bound of the CI to be $M_n$ and then let the upper end of the interval be $M_n(1+c^*)$ for some choice of $c^*.$ (Note as always there is no unique confidence interval and this is just a sensible and convenient choice). Then we can choose $c^*$ so we have $1-alpha$ coverage (where $alpha=0.05)$: $$ 1-alpha=P(theta < (1+c^*)M_n) = 1- Pleft(M_nle frac{theta}{1+c^*}right)= 1-(1+c^*)^{-n}$$ which gives $$ c^* = alpha^{-1/n}-1$$
EDIT
It appears I misinterpreted the problem here. If we want a lower bound that has $1-alpha$ confidence (i.e. a minimal statistic $x$ such that $P(theta>x) le 1-alpha$) then we can choose $x=(1+c^*)M_n$ and choose $c^*$ such that $$ 1-alpha = P((1+c^*)M_n< theta)= (1+c^*)^{-n},$$ so this gives $$ c^* = (1-alpha)^{-1/n}-1,$$ which is a good deal smaller than the $c^*$ derived above for an upper bound.
answered Nov 21 at 5:35
spaceisdarkgreen
31.6k21552
The maximum is a sufficient statistic here, so a good choice for a test statistic for your CI. (However I'm not sure what you mean with the $(1/a^n-1/b^n)$... it may just be terminology I'm not aware of).
To get a CI we need to first find the distribution of the test statistic: $$ P(M_nle m) = P(Xle m)^n = (m/theta)^n.$$ Intuitively, we know the $theta>M_n$ so it makes sense to set the lower bound of the CI to be $M_n$ and then let the upper end of the interval be $M_n(1+c^*)$ for some choice of $c^*.$ (Note as always there is no unique confidence interval and this is just a sensible and convenient choice). Then we can choose $c^*$ so we have $1-alpha$ coverage (where $alpha=0.05)$: $$ 1-alpha=P(theta < (1+c^*)M_n) = 1- Pleft(M_nle frac{theta}{1+c^*}right)= 1-(1+c^*)^{-n}$$ which gives $$ c^* = alpha^{-1/n}-1$$
EDIT
It appears I misinterpreted the problem here. If we want a lower bound that has $1-alpha$ confidence (i.e. a minimal statistic $x$ such that $P(theta>x) le 1-alpha$) then we can choose $x=(1+c^*)M_n$ and choose $c^*$ such that $$ 1-alpha = P((1+c^*)M_n< theta)= (1+c^*)^{-n},$$ so this gives $$ c^* = (1-alpha)^{-1/n}-1,$$ which is a good deal smaller than the $c^*$ derived above for an upper bound.
answered Nov 21 at 5:35
spaceisdarkgreen
31.6k21552
edited Nov 21 at 5:53
answered Nov 21 at 5:35
spaceisdarkgreen
31.6k21552
answered Nov 21 at 5:35
spaceisdarkgreen
31.6k21552
answered Nov 21 at 5:35
spaceisdarkgreen
31.6k21552
31.6k21552
sorry, but what do you mean with CI?
– Jazmín Jones
Nov 21 at 5:41
@JazmínJones I mean confidence interval. Perhaps I misinterpreted the problem
– spaceisdarkgreen
Nov 21 at 5:43
@JazmínJones I edited in what I think is an answer to the problem. My interpretation is that you want to know how far above $M_n$ we can go while still having $ge 95%$ confidence that $theta$ is larger than this value. (My initial answer gives an upper bound instead.)
– spaceisdarkgreen
Nov 21 at 5:58
add a comment |
sorry, but what do you mean with CI?
– Jazmín Jones
Nov 21 at 5:41
@JazmínJones I mean confidence interval. Perhaps I misinterpreted the problem
– spaceisdarkgreen
Nov 21 at 5:43
@JazmínJones I edited in what I think is an answer to the problem. My interpretation is that you want to know how far above $M_n$ we can go while still having $ge 95%$ confidence that $theta$ is larger than this value. (My initial answer gives an upper bound instead.)
– spaceisdarkgreen
Nov 21 at 5:58
sorry, but what do you mean with CI?
– Jazmín Jones
Nov 21 at 5:41
sorry, but what do you mean with CI?
– Jazmín Jones
Nov 21 at 5:41
@JazmínJones I mean confidence interval. Perhaps I misinterpreted the problem
– spaceisdarkgreen
Nov 21 at 5:43
@JazmínJones I mean confidence interval. Perhaps I misinterpreted the problem
– spaceisdarkgreen
Nov 21 at 5:43
@JazmínJones I edited in what I think is an answer to the problem. My interpretation is that you want to know how far above $M_n$ we can go while still having $ge 95%$ confidence that $theta$ is larger than this value. (My initial answer gives an upper bound instead.)
– spaceisdarkgreen
Nov 21 at 5:58
@JazmínJones I edited in what I think is an answer to the problem. My interpretation is that you want to know how far above $M_n$ we can go while still having $ge 95%$ confidence that $theta$ is larger than this value. (My initial answer gives an upper bound instead.)
– spaceisdarkgreen
Nov 21 at 5:58
add a comment |
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