Mixing
Is there any meaning of this “Median-mean”
$begingroup$
Given a data set ${a_1,cdots,a_n}$ with median M
Define the medimean to be the value of $x$ s.t.
$$left(frac{1}{n}sum_{n}{}{a_n}^x right)^{1/x}=M$$
Is this $x$ value useful / used at all in statistics?
Some examples
${frac{1}{5},frac{1}{4},frac{1}{3},frac{1}{2},frac{1}{1}} : x=-1$
${2,4,8,16,32} : x=0$
${sqrt{1},sqrt{2},sqrt{3},sqrt{4},sqrt{5}} : x=2$
${1,2,3,4,5} : x=1$
${1,4,9,16,25} : x=frac{1}{2}$
$frac{1}{x}$ seems to say something about how the numbers are distributed, apart from when $x=0$
Just an idea I was playing around with yesterday.
${1,1,1,2,2} : x=-infty$
${1,1,2,2,2} : x=infty$
statistics statistical-inference descriptive-statistics
asked Jan 18 at 10:06
Ben CrossleyBen Crossley
904418
$endgroup$
add a comment |
$begingroup$
Given a data set ${a_1,cdots,a_n}$ with median M
Define the medimean to be the value of $x$ s.t.
$$left(frac{1}{n}sum_{n}{}{a_n}^x right)^{1/x}=M$$
Is this $x$ value useful / used at all in statistics?
Some examples
${frac{1}{5},frac{1}{4},frac{1}{3},frac{1}{2},frac{1}{1}} : x=-1$
${2,4,8,16,32} : x=0$
${sqrt{1},sqrt{2},sqrt{3},sqrt{4},sqrt{5}} : x=2$
${1,2,3,4,5} : x=1$
${1,4,9,16,25} : x=frac{1}{2}$
$frac{1}{x}$ seems to say something about how the numbers are distributed, apart from when $x=0$
Just an idea I was playing around with yesterday.
${1,1,1,2,2} : x=-infty$
${1,1,2,2,2} : x=infty$
statistics statistical-inference descriptive-statistics
asked Jan 18 at 10:06
Ben CrossleyBen Crossley
904418
$endgroup$
$begingroup$
Does such an $x$ always exist? If so, is it always unique?
$endgroup$
– Rahul
Jan 18 at 10:18
3
$begingroup$
Yes and yes. $x=-infty$ produces the minimum, $x=infty$ produces the maximum this is the generalised mean which is a continuous non-descreasing function (when considered across the means) so the "medimean" will both exist and be unique.
$endgroup$
– Ben Crossley
Jan 18 at 10:52
$begingroup$
@Rahul (didnt tag you in previous comment)
$endgroup$
– Ben Crossley
Jan 18 at 11:59
add a comment |
$begingroup$
Given a data set ${a_1,cdots,a_n}$ with median M
Define the medimean to be the value of $x$ s.t.
$$left(frac{1}{n}sum_{n}{}{a_n}^x right)^{1/x}=M$$
Is this $x$ value useful / used at all in statistics?
Some examples
${frac{1}{5},frac{1}{4},frac{1}{3},frac{1}{2},frac{1}{1}} : x=-1$
${2,4,8,16,32} : x=0$
${sqrt{1},sqrt{2},sqrt{3},sqrt{4},sqrt{5}} : x=2$
${1,2,3,4,5} : x=1$
${1,4,9,16,25} : x=frac{1}{2}$
$frac{1}{x}$ seems to say something about how the numbers are distributed, apart from when $x=0$
Just an idea I was playing around with yesterday.
${1,1,1,2,2} : x=-infty$
${1,1,2,2,2} : x=infty$
statistics statistical-inference descriptive-statistics
asked Jan 18 at 10:06
Ben CrossleyBen Crossley
904418
$endgroup$
Given a data set ${a_1,cdots,a_n}$ with median M
Define the medimean to be the value of $x$ s.t.
$$left(frac{1}{n}sum_{n}{}{a_n}^x right)^{1/x}=M$$
Is this $x$ value useful / used at all in statistics?
Some examples
${frac{1}{5},frac{1}{4},frac{1}{3},frac{1}{2},frac{1}{1}} : x=-1$
${2,4,8,16,32} : x=0$
${sqrt{1},sqrt{2},sqrt{3},sqrt{4},sqrt{5}} : x=2$
${1,2,3,4,5} : x=1$
${1,4,9,16,25} : x=frac{1}{2}$
$frac{1}{x}$ seems to say something about how the numbers are distributed, apart from when $x=0$
Just an idea I was playing around with yesterday.
${1,1,1,2,2} : x=-infty$
${1,1,2,2,2} : x=infty$
statistics statistical-inference descriptive-statistics
statistics statistical-inference descriptive-statistics
asked Jan 18 at 10:06
Ben CrossleyBen Crossley
904418
asked Jan 18 at 10:06
Ben CrossleyBen Crossley
904418
asked Jan 18 at 10:06
Ben CrossleyBen Crossley
904418
asked Jan 18 at 10:06
Ben CrossleyBen Crossley
904418
asked Jan 18 at 10:06
Ben CrossleyBen Crossley
904418
904418
$begingroup$
Does such an $x$ always exist? If so, is it always unique?
$endgroup$
– Rahul
Jan 18 at 10:18
3
$begingroup$
Yes and yes. $x=-infty$ produces the minimum, $x=infty$ produces the maximum this is the generalised mean which is a continuous non-descreasing function (when considered across the means) so the "medimean" will both exist and be unique.
$endgroup$
– Ben Crossley
Jan 18 at 10:52
$begingroup$
@Rahul (didnt tag you in previous comment)
$endgroup$
– Ben Crossley
Jan 18 at 11:59
add a comment |
$begingroup$
Does such an $x$ always exist? If so, is it always unique?
$endgroup$
– Rahul
Jan 18 at 10:18
3
$begingroup$
Yes and yes. $x=-infty$ produces the minimum, $x=infty$ produces the maximum this is the generalised mean which is a continuous non-descreasing function (when considered across the means) so the "medimean" will both exist and be unique.
$endgroup$
– Ben Crossley
Jan 18 at 10:52
$begingroup$
@Rahul (didnt tag you in previous comment)
$endgroup$
– Ben Crossley
Jan 18 at 11:59
$begingroup$
Does such an $x$ always exist? If so, is it always unique?
$endgroup$
– Rahul
Jan 18 at 10:18
$begingroup$
Does such an $x$ always exist? If so, is it always unique?
$endgroup$
– Rahul
Jan 18 at 10:18
3
3
$begingroup$
Yes and yes. $x=-infty$ produces the minimum, $x=infty$ produces the maximum this is the generalised mean which is a continuous non-descreasing function (when considered across the means) so the "medimean" will both exist and be unique.
$endgroup$
– Ben Crossley
Jan 18 at 10:52
$begingroup$
Yes and yes. $x=-infty$ produces the minimum, $x=infty$ produces the maximum this is the generalised mean which is a continuous non-descreasing function (when considered across the means) so the "medimean" will both exist and be unique.
$endgroup$
– Ben Crossley
Jan 18 at 10:52
$begingroup$
@Rahul (didnt tag you in previous comment)
$endgroup$
– Ben Crossley
Jan 18 at 11:59
$begingroup$
@Rahul (didnt tag you in previous comment)
$endgroup$
– Ben Crossley
Jan 18 at 11:59
add a comment |
1 Answer
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oldest
votes
$begingroup$
This is an extended comment with thoughts on the matter rather than a complete answer (though honestly, I don't think it's possible the fully answer a question like "is that useful").
I haven't seen a medimean used for anything, and honestly, I would be surprised if it proved useful for something.
In what it describes, the medimean seems quite similar to skewness: both can be used to describe where is the median in relation to the mean. Skewness is negative, and medimean - positive, when the mean is less than the median; and vice versa. However, skewness has one enormous advantage over medimean: it's much easier to calculate. Thus, I would expect skewness to be a better choice whenever medimean would be useful.
That said, there are cases where skewness simply couldn't be used, so maybe medimean could work in some of those?
Suppose we have a sample of positive numbers, and we want to do inference based on medimean.
First thing first, medimean seems well defined when we work with positive values. Real powers of negative numbers, however, are complex and infinitely many, and it's not at all obvious to me that medimean exists at all if we allow negative numbers.
We probably don't want the sample medimean, but rather the population medimean. That means, we would like to estimate population medimeans from a sample. Is the sample medimean a good estimator to population medimean?
If we have a model for the population, can we compute the medimean of the model distribution and use the sample medimean to fir the model? Unfortunately, the most popular model - the normal - admits negative values.
I think the most promising case for using medimean could be for some fat tailed positive models, like Pareto distribution, that don't have skewness, but maybe have medimeans? Still, I would be very surprised if someone could derive those medimeans as function of parameters.
answered Jan 22 at 15:40
Todor MarkovTodor Markov
2,410412
$endgroup$
add a comment |
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1 Answer
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1 Answer
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active
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$begingroup$
This is an extended comment with thoughts on the matter rather than a complete answer (though honestly, I don't think it's possible the fully answer a question like "is that useful").
I haven't seen a medimean used for anything, and honestly, I would be surprised if it proved useful for something.
In what it describes, the medimean seems quite similar to skewness: both can be used to describe where is the median in relation to the mean. Skewness is negative, and medimean - positive, when the mean is less than the median; and vice versa. However, skewness has one enormous advantage over medimean: it's much easier to calculate. Thus, I would expect skewness to be a better choice whenever medimean would be useful.
That said, there are cases where skewness simply couldn't be used, so maybe medimean could work in some of those?
Suppose we have a sample of positive numbers, and we want to do inference based on medimean.
First thing first, medimean seems well defined when we work with positive values. Real powers of negative numbers, however, are complex and infinitely many, and it's not at all obvious to me that medimean exists at all if we allow negative numbers.
We probably don't want the sample medimean, but rather the population medimean. That means, we would like to estimate population medimeans from a sample. Is the sample medimean a good estimator to population medimean?
If we have a model for the population, can we compute the medimean of the model distribution and use the sample medimean to fir the model? Unfortunately, the most popular model - the normal - admits negative values.
I think the most promising case for using medimean could be for some fat tailed positive models, like Pareto distribution, that don't have skewness, but maybe have medimeans? Still, I would be very surprised if someone could derive those medimeans as function of parameters.
answered Jan 22 at 15:40
Todor MarkovTodor Markov
2,410412
$endgroup$
add a comment |
$begingroup$
This is an extended comment with thoughts on the matter rather than a complete answer (though honestly, I don't think it's possible the fully answer a question like "is that useful").
I haven't seen a medimean used for anything, and honestly, I would be surprised if it proved useful for something.
In what it describes, the medimean seems quite similar to skewness: both can be used to describe where is the median in relation to the mean. Skewness is negative, and medimean - positive, when the mean is less than the median; and vice versa. However, skewness has one enormous advantage over medimean: it's much easier to calculate. Thus, I would expect skewness to be a better choice whenever medimean would be useful.
That said, there are cases where skewness simply couldn't be used, so maybe medimean could work in some of those?
Suppose we have a sample of positive numbers, and we want to do inference based on medimean.
First thing first, medimean seems well defined when we work with positive values. Real powers of negative numbers, however, are complex and infinitely many, and it's not at all obvious to me that medimean exists at all if we allow negative numbers.
We probably don't want the sample medimean, but rather the population medimean. That means, we would like to estimate population medimeans from a sample. Is the sample medimean a good estimator to population medimean?
If we have a model for the population, can we compute the medimean of the model distribution and use the sample medimean to fir the model? Unfortunately, the most popular model - the normal - admits negative values.
I think the most promising case for using medimean could be for some fat tailed positive models, like Pareto distribution, that don't have skewness, but maybe have medimeans? Still, I would be very surprised if someone could derive those medimeans as function of parameters.
answered Jan 22 at 15:40
Todor MarkovTodor Markov
2,410412
$endgroup$
add a comment |
$begingroup$
This is an extended comment with thoughts on the matter rather than a complete answer (though honestly, I don't think it's possible the fully answer a question like "is that useful").
I haven't seen a medimean used for anything, and honestly, I would be surprised if it proved useful for something.
In what it describes, the medimean seems quite similar to skewness: both can be used to describe where is the median in relation to the mean. Skewness is negative, and medimean - positive, when the mean is less than the median; and vice versa. However, skewness has one enormous advantage over medimean: it's much easier to calculate. Thus, I would expect skewness to be a better choice whenever medimean would be useful.
That said, there are cases where skewness simply couldn't be used, so maybe medimean could work in some of those?
Suppose we have a sample of positive numbers, and we want to do inference based on medimean.
First thing first, medimean seems well defined when we work with positive values. Real powers of negative numbers, however, are complex and infinitely many, and it's not at all obvious to me that medimean exists at all if we allow negative numbers.
We probably don't want the sample medimean, but rather the population medimean. That means, we would like to estimate population medimeans from a sample. Is the sample medimean a good estimator to population medimean?
If we have a model for the population, can we compute the medimean of the model distribution and use the sample medimean to fir the model? Unfortunately, the most popular model - the normal - admits negative values.
I think the most promising case for using medimean could be for some fat tailed positive models, like Pareto distribution, that don't have skewness, but maybe have medimeans? Still, I would be very surprised if someone could derive those medimeans as function of parameters.
answered Jan 22 at 15:40
Todor MarkovTodor Markov
2,410412
$endgroup$
This is an extended comment with thoughts on the matter rather than a complete answer (though honestly, I don't think it's possible the fully answer a question like "is that useful").
I haven't seen a medimean used for anything, and honestly, I would be surprised if it proved useful for something.
In what it describes, the medimean seems quite similar to skewness: both can be used to describe where is the median in relation to the mean. Skewness is negative, and medimean - positive, when the mean is less than the median; and vice versa. However, skewness has one enormous advantage over medimean: it's much easier to calculate. Thus, I would expect skewness to be a better choice whenever medimean would be useful.
That said, there are cases where skewness simply couldn't be used, so maybe medimean could work in some of those?
Suppose we have a sample of positive numbers, and we want to do inference based on medimean.
First thing first, medimean seems well defined when we work with positive values. Real powers of negative numbers, however, are complex and infinitely many, and it's not at all obvious to me that medimean exists at all if we allow negative numbers.
We probably don't want the sample medimean, but rather the population medimean. That means, we would like to estimate population medimeans from a sample. Is the sample medimean a good estimator to population medimean?
If we have a model for the population, can we compute the medimean of the model distribution and use the sample medimean to fir the model? Unfortunately, the most popular model - the normal - admits negative values.
I think the most promising case for using medimean could be for some fat tailed positive models, like Pareto distribution, that don't have skewness, but maybe have medimeans? Still, I would be very surprised if someone could derive those medimeans as function of parameters.
answered Jan 22 at 15:40
Todor MarkovTodor Markov
2,410412
answered Jan 22 at 15:40
Todor MarkovTodor Markov
2,410412
answered Jan 22 at 15:40
Todor MarkovTodor Markov
2,410412
answered Jan 22 at 15:40
Todor MarkovTodor Markov
2,410412
2,410412
add a comment |
add a comment |
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$begingroup$
Does such an $x$ always exist? If so, is it always unique?
$endgroup$
– Rahul
Jan 18 at 10:18
3
$begingroup$
Yes and yes. $x=-infty$ produces the minimum, $x=infty$ produces the maximum this is the generalised mean which is a continuous non-descreasing function (when considered across the means) so the "medimean" will both exist and be unique.
$endgroup$
– Ben Crossley
Jan 18 at 10:52
$begingroup$
@Rahul (didnt tag you in previous comment)
$endgroup$
– Ben Crossley
Jan 18 at 11:59