Mixing
Minimum Sufficient Statistic for Uniform(θ,θ+1) distribution?
0
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I saw that (min(X), max(X)) are minimal sufficient statistic for this.
But I was wondering, why can't we just have min(X) OR max(X)?
If distribution is Uniform(θ1,θ2), then it makes sense to have two dimensional minimum sufficient statistic, but if not then why can't we recover θ from min(X) OR max(X)?
statistics data-sufficiency
asked Jan 28 at 16:42
user1953366user1953366
1011
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add a comment |
0
$begingroup$
I saw that (min(X), max(X)) are minimal sufficient statistic for this.
But I was wondering, why can't we just have min(X) OR max(X)?
If distribution is Uniform(θ1,θ2), then it makes sense to have two dimensional minimum sufficient statistic, but if not then why can't we recover θ from min(X) OR max(X)?
statistics data-sufficiency
asked Jan 28 at 16:42
user1953366user1953366
1011
$endgroup$
$begingroup$
Because of the constraint $theta<min x_i,,, max x_i<theta+1$, i.e. $max x_i-1<theta<min x_i$ in the joint density.
$endgroup$
– StubbornAtom
Jan 28 at 17:01
$begingroup$
@StubbornAtom why would θ < minx?? I guess u meant >. But again why can't we recover θ from min X OR max X?
$endgroup$
– user1953366
Jan 28 at 17:03
$begingroup$
It follows from $theta<x_1,x_2,ldots,x_n<theta+1$. If you leave out any one of $min x_i$ or $max x_i$ you lose information about $theta$ (because of the constraint), and $min x_i$ or $max x_i$ no longer remain minimal sufficient on their own.
$endgroup$
– StubbornAtom
Jan 28 at 18:01
$begingroup$
I get why θ<minxi. But given a data sample, and knowing that it is from Uniform(θ, θ+1), my approximation for θ will be as good from just one statistic, which is: (min(x) + max(x))/2, as from two different statistic. Thus just a single statistic of (min(x) + max(x))/2 should be sufficient to recover θ.
$endgroup$
– user1953366
Jan 28 at 22:02
add a comment |
0
0
0
$begingroup$
I saw that (min(X), max(X)) are minimal sufficient statistic for this.
But I was wondering, why can't we just have min(X) OR max(X)?
If distribution is Uniform(θ1,θ2), then it makes sense to have two dimensional minimum sufficient statistic, but if not then why can't we recover θ from min(X) OR max(X)?
statistics data-sufficiency
asked Jan 28 at 16:42
user1953366user1953366
1011
$endgroup$
I saw that (min(X), max(X)) are minimal sufficient statistic for this.
But I was wondering, why can't we just have min(X) OR max(X)?
If distribution is Uniform(θ1,θ2), then it makes sense to have two dimensional minimum sufficient statistic, but if not then why can't we recover θ from min(X) OR max(X)?
statistics data-sufficiency
statistics data-sufficiency
asked Jan 28 at 16:42
user1953366user1953366
1011
asked Jan 28 at 16:42
user1953366user1953366
1011
asked Jan 28 at 16:42
user1953366user1953366
1011
asked Jan 28 at 16:42
user1953366user1953366
1011
asked Jan 28 at 16:42
user1953366user1953366
1011
1011
$begingroup$
Because of the constraint $theta<min x_i,,, max x_i<theta+1$, i.e. $max x_i-1<theta<min x_i$ in the joint density.
$endgroup$
– StubbornAtom
Jan 28 at 17:01
$begingroup$
@StubbornAtom why would θ < minx?? I guess u meant >. But again why can't we recover θ from min X OR max X?
$endgroup$
– user1953366
Jan 28 at 17:03
$begingroup$
It follows from $theta<x_1,x_2,ldots,x_n<theta+1$. If you leave out any one of $min x_i$ or $max x_i$ you lose information about $theta$ (because of the constraint), and $min x_i$ or $max x_i$ no longer remain minimal sufficient on their own.
$endgroup$
– StubbornAtom
Jan 28 at 18:01
$begingroup$
I get why θ<minxi. But given a data sample, and knowing that it is from Uniform(θ, θ+1), my approximation for θ will be as good from just one statistic, which is: (min(x) + max(x))/2, as from two different statistic. Thus just a single statistic of (min(x) + max(x))/2 should be sufficient to recover θ.
$endgroup$
– user1953366
Jan 28 at 22:02
add a comment |
$begingroup$
Because of the constraint $theta<min x_i,,, max x_i<theta+1$, i.e. $max x_i-1<theta<min x_i$ in the joint density.
$endgroup$
– StubbornAtom
Jan 28 at 17:01
$begingroup$
@StubbornAtom why would θ < minx?? I guess u meant >. But again why can't we recover θ from min X OR max X?
$endgroup$
– user1953366
Jan 28 at 17:03
$begingroup$
It follows from $theta<x_1,x_2,ldots,x_n<theta+1$. If you leave out any one of $min x_i$ or $max x_i$ you lose information about $theta$ (because of the constraint), and $min x_i$ or $max x_i$ no longer remain minimal sufficient on their own.
$endgroup$
– StubbornAtom
Jan 28 at 18:01
$begingroup$
I get why θ<minxi. But given a data sample, and knowing that it is from Uniform(θ, θ+1), my approximation for θ will be as good from just one statistic, which is: (min(x) + max(x))/2, as from two different statistic. Thus just a single statistic of (min(x) + max(x))/2 should be sufficient to recover θ.
$endgroup$
– user1953366
Jan 28 at 22:02
$begingroup$
Because of the constraint $theta<min x_i,,, max x_i<theta+1$, i.e. $max x_i-1<theta<min x_i$ in the joint density.
$endgroup$
– StubbornAtom
Jan 28 at 17:01
$begingroup$
Because of the constraint $theta<min x_i,,, max x_i<theta+1$, i.e. $max x_i-1<theta<min x_i$ in the joint density.
$endgroup$
– StubbornAtom
Jan 28 at 17:01
$begingroup$
@StubbornAtom why would θ < minx?? I guess u meant >. But again why can't we recover θ from min X OR max X?
$endgroup$
– user1953366
Jan 28 at 17:03
$begingroup$
@StubbornAtom why would θ < minx?? I guess u meant >. But again why can't we recover θ from min X OR max X?
$endgroup$
– user1953366
Jan 28 at 17:03
$begingroup$
It follows from $theta<x_1,x_2,ldots,x_n<theta+1$. If you leave out any one of $min x_i$ or $max x_i$ you lose information about $theta$ (because of the constraint), and $min x_i$ or $max x_i$ no longer remain minimal sufficient on their own.
$endgroup$
– StubbornAtom
Jan 28 at 18:01
$begingroup$
It follows from $theta<x_1,x_2,ldots,x_n<theta+1$. If you leave out any one of $min x_i$ or $max x_i$ you lose information about $theta$ (because of the constraint), and $min x_i$ or $max x_i$ no longer remain minimal sufficient on their own.
$endgroup$
– StubbornAtom
Jan 28 at 18:01
$begingroup$
I get why θ<minxi. But given a data sample, and knowing that it is from Uniform(θ, θ+1), my approximation for θ will be as good from just one statistic, which is: (min(x) + max(x))/2, as from two different statistic. Thus just a single statistic of (min(x) + max(x))/2 should be sufficient to recover θ.
$endgroup$
– user1953366
Jan 28 at 22:02
$begingroup$
I get why θ<minxi. But given a data sample, and knowing that it is from Uniform(θ, θ+1), my approximation for θ will be as good from just one statistic, which is: (min(x) + max(x))/2, as from two different statistic. Thus just a single statistic of (min(x) + max(x))/2 should be sufficient to recover θ.
$endgroup$
– user1953366
Jan 28 at 22:02
add a comment |
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$begingroup$
Because of the constraint $theta<min x_i,,, max x_i<theta+1$, i.e. $max x_i-1<theta<min x_i$ in the joint density.
$endgroup$
– StubbornAtom
Jan 28 at 17:01
$begingroup$
@StubbornAtom why would θ < minx?? I guess u meant >. But again why can't we recover θ from min X OR max X?
$endgroup$
– user1953366
Jan 28 at 17:03
$begingroup$
It follows from $theta<x_1,x_2,ldots,x_n<theta+1$. If you leave out any one of $min x_i$ or $max x_i$ you lose information about $theta$ (because of the constraint), and $min x_i$ or $max x_i$ no longer remain minimal sufficient on their own.
$endgroup$
– StubbornAtom
Jan 28 at 18:01
$begingroup$
I get why θ<minxi. But given a data sample, and knowing that it is from Uniform(θ, θ+1), my approximation for θ will be as good from just one statistic, which is: (min(x) + max(x))/2, as from two different statistic. Thus just a single statistic of (min(x) + max(x))/2 should be sufficient to recover θ.
$endgroup$
– user1953366
Jan 28 at 22:02