Mixing
What is the minimum height among probability densities having mean 0 and variance 1?
1
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It seems to me that the U($-sqrt{3}, sqrt{3}$) density provides this minimum, with height $1/(2sqrt{3})$. Is this true? And if so, how does the proof look?
probability-distributions optimization
asked Jan 30 at 20:29
Peter WestfallPeter Westfall
1465
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add a comment |
1
$begingroup$
It seems to me that the U($-sqrt{3}, sqrt{3}$) density provides this minimum, with height $1/(2sqrt{3})$. Is this true? And if so, how does the proof look?
probability-distributions optimization
asked Jan 30 at 20:29
Peter WestfallPeter Westfall
1465
$endgroup$
$begingroup$
how do you define the height of a probability distribution?
$endgroup$
– LinAlg
Jan 30 at 20:33
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The proof would be in two steps. First show that uniform over an interval gives minimum height and then calculate what the interval would be.
$endgroup$
– herb steinberg
Jan 30 at 21:36
$begingroup$
I mean height to be $max_x p(x)$.
$endgroup$
– Peter Westfall
Jan 30 at 23:18
$begingroup$
Herb, it's the first step that I don't see how to do. The second step is trivial by the moment constraint. Clues?
$endgroup$
– Peter Westfall
Jan 31 at 23:59
add a comment |
1
1
1
$begingroup$
It seems to me that the U($-sqrt{3}, sqrt{3}$) density provides this minimum, with height $1/(2sqrt{3})$. Is this true? And if so, how does the proof look?
probability-distributions optimization
asked Jan 30 at 20:29
Peter WestfallPeter Westfall
1465
$endgroup$
It seems to me that the U($-sqrt{3}, sqrt{3}$) density provides this minimum, with height $1/(2sqrt{3})$. Is this true? And if so, how does the proof look?
probability-distributions optimization
probability-distributions optimization
asked Jan 30 at 20:29
Peter WestfallPeter Westfall
1465
asked Jan 30 at 20:29
Peter WestfallPeter Westfall
1465
asked Jan 30 at 20:29
Peter WestfallPeter Westfall
1465
asked Jan 30 at 20:29
Peter WestfallPeter Westfall
1465
asked Jan 30 at 20:29
Peter WestfallPeter Westfall
1465
1465
$begingroup$
how do you define the height of a probability distribution?
$endgroup$
– LinAlg
Jan 30 at 20:33
$begingroup$
The proof would be in two steps. First show that uniform over an interval gives minimum height and then calculate what the interval would be.
$endgroup$
– herb steinberg
Jan 30 at 21:36
$begingroup$
I mean height to be $max_x p(x)$.
$endgroup$
– Peter Westfall
Jan 30 at 23:18
$begingroup$
Herb, it's the first step that I don't see how to do. The second step is trivial by the moment constraint. Clues?
$endgroup$
– Peter Westfall
Jan 31 at 23:59
add a comment |
$begingroup$
how do you define the height of a probability distribution?
$endgroup$
– LinAlg
Jan 30 at 20:33
$begingroup$
The proof would be in two steps. First show that uniform over an interval gives minimum height and then calculate what the interval would be.
$endgroup$
– herb steinberg
Jan 30 at 21:36
$begingroup$
I mean height to be $max_x p(x)$.
$endgroup$
– Peter Westfall
Jan 30 at 23:18
$begingroup$
Herb, it's the first step that I don't see how to do. The second step is trivial by the moment constraint. Clues?
$endgroup$
– Peter Westfall
Jan 31 at 23:59
$begingroup$
how do you define the height of a probability distribution?
$endgroup$
– LinAlg
Jan 30 at 20:33
$begingroup$
how do you define the height of a probability distribution?
$endgroup$
– LinAlg
Jan 30 at 20:33
$begingroup$
The proof would be in two steps. First show that uniform over an interval gives minimum height and then calculate what the interval would be.
$endgroup$
– herb steinberg
Jan 30 at 21:36
$begingroup$
The proof would be in two steps. First show that uniform over an interval gives minimum height and then calculate what the interval would be.
$endgroup$
– herb steinberg
Jan 30 at 21:36
$begingroup$
I mean height to be $max_x p(x)$.
$endgroup$
– Peter Westfall
Jan 30 at 23:18
$begingroup$
I mean height to be $max_x p(x)$.
$endgroup$
– Peter Westfall
Jan 30 at 23:18
$begingroup$
Herb, it's the first step that I don't see how to do. The second step is trivial by the moment constraint. Clues?
$endgroup$
– Peter Westfall
Jan 31 at 23:59
$begingroup$
Herb, it's the first step that I don't see how to do. The second step is trivial by the moment constraint. Clues?
$endgroup$
– Peter Westfall
Jan 31 at 23:59
add a comment |
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$begingroup$
how do you define the height of a probability distribution?
$endgroup$
– LinAlg
Jan 30 at 20:33
$begingroup$
The proof would be in two steps. First show that uniform over an interval gives minimum height and then calculate what the interval would be.
$endgroup$
– herb steinberg
Jan 30 at 21:36
$begingroup$
I mean height to be $max_x p(x)$.
$endgroup$
– Peter Westfall
Jan 30 at 23:18
$begingroup$
Herb, it's the first step that I don't see how to do. The second step is trivial by the moment constraint. Clues?
$endgroup$
– Peter Westfall
Jan 31 at 23:59