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Can two distinct datasets of the same size have the same median and the same deviation from a real number?
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Can two distinct datasets of the same size have the same median and the same deviation from any real number?
For example Let $a_1 < a_2 < ··· < a_n$ and $b_1 < b_2 < ··· < b_n$ be real numbers such that $sum_i=1^n |a_i-x|=sum_i=1^n|b_i-x|$,where $x$ is any real.
Now can it be proved that $a_i=b_i$?
statistics standard-deviation
asked Jul 24 at 3:00
Legend Killer
1,500523
add a comment |Â
up vote
0
down vote
favorite
Can two distinct datasets of the same size have the same median and the same deviation from any real number?
For example Let $a_1 < a_2 < ··· < a_n$ and $b_1 < b_2 < ··· < b_n$ be real numbers such that $sum_i=1^n |a_i-x|=sum_i=1^n|b_i-x|$,where $x$ is any real.
Now can it be proved that $a_i=b_i$?
statistics standard-deviation
asked Jul 24 at 3:00
Legend Killer
1,500523
Is the deviation to be from one real number, as the first sentence suggests, or all real numbers, as the next to last suggests?
– Ross Millikan
Jul 24 at 3:06
Any real number $x$
– Legend Killer
Jul 24 at 3:07
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
Can two distinct datasets of the same size have the same median and the same deviation from any real number?
For example Let $a_1 < a_2 < ··· < a_n$ and $b_1 < b_2 < ··· < b_n$ be real numbers such that $sum_i=1^n |a_i-x|=sum_i=1^n|b_i-x|$,where $x$ is any real.
Now can it be proved that $a_i=b_i$?
statistics standard-deviation
asked Jul 24 at 3:00
Legend Killer
1,500523
Can two distinct datasets of the same size have the same median and the same deviation from any real number?
For example Let $a_1 < a_2 < ··· < a_n$ and $b_1 < b_2 < ··· < b_n$ be real numbers such that $sum_i=1^n |a_i-x|=sum_i=1^n|b_i-x|$,where $x$ is any real.
Now can it be proved that $a_i=b_i$?
statistics standard-deviation
asked Jul 24 at 3:00
Legend Killer
1,500523
edited Jul 24 at 3:07
asked Jul 24 at 3:00
Legend Killer
1,500523
asked Jul 24 at 3:00
Legend Killer
1,500523
asked Jul 24 at 3:00
Legend Killer
1,500523
1,500523
Is the deviation to be from one real number, as the first sentence suggests, or all real numbers, as the next to last suggests?
– Ross Millikan
Jul 24 at 3:06
Any real number $x$
– Legend Killer
Jul 24 at 3:07
add a comment |Â
Is the deviation to be from one real number, as the first sentence suggests, or all real numbers, as the next to last suggests?
– Ross Millikan
Jul 24 at 3:06
Any real number $x$
– Legend Killer
Jul 24 at 3:07
Is the deviation to be from one real number, as the first sentence suggests, or all real numbers, as the next to last suggests?
– Ross Millikan
Jul 24 at 3:06
Is the deviation to be from one real number, as the first sentence suggests, or all real numbers, as the next to last suggests?
– Ross Millikan
Jul 24 at 3:06
Any real number $x$
– Legend Killer
Jul 24 at 3:07
Any real number $x$
– Legend Killer
Jul 24 at 3:07
add a comment |Â
1 Answer
1
active
oldest
votes
up vote
1
down vote
accepted
The total deviation constraint is enough without the requirement on the median and we can prove that the size of the two sets is the same. Let $clt a_1,b_1$ and $d=c-1$. The total deviation of the $a$s from $d$ is greater than the total deviation of the $a$s from $c$ by the number of the $a$s because $c$ is $1$ closer to each one. If the total deviation of the $a$s from $d$ equals the total deviation of the $b$s from $d$ and similarly for $c$ there must be the same number of $a$s and $b$s. As you have done, call that number $n.$
Now assume $a_1 neq b_1$. WOLOG we can assume $a_1 lt b_1$ and define $e=min(frac 12 (b_1-a_1), frac 12(a_2-a_1))gt 0$. We are given that the total deviation of both the $a$s and $b$s from $a_1$ is some number $f$. The total deviation of the $a$s from $a_1+e$ is $f+e-(n-1)e=f-(n-2)e$ because we are going away from $a_1$ and towards all the other $a$s. The total deviation of the $b$s from $a_1+e$ is $f-ne$ because we are going towards all of them. This is a contradiction, so $a_1=b_1$. We can repeat the argument now for $a_2$ and $b_2$ and so on up the line.
answered Jul 24 at 3:05
Ross Millikan
275k21186351
How can you calculate the sum of deviations of $a$ from $a_1+e$?
– Legend Killer
Jul 24 at 3:58
Once I have the sum of deviations from $a_1$, which I defined as $f$, I look at how each term in the sum of deviations changes. When you go to $a_1+e$ you get $e$ further from $a_1$ and $e$ closer to all the rest. I chose $e$ small enough to ensure you don't move past any of the others to make this work. You also get $e$ closer to all the $b$s. That is the heart of the argument. Take the $a$s as $1,2,3$ and $e=0.1$ and do the calculation by hand to see how it works.
– Ross Millikan
Jul 24 at 4:02
It is a brilliant proof , sir.I actually proved the two datasets are equal in size by taking x to be the median of the set a
– Legend Killer
Jul 24 at 4:11
I don't see how the median can prove the datasets are equal in size. Take a dataset and add one value above the median and one below. You have a new dataset with the same median and more values. Maybe you have more information than that. It is true that the median has the property that it is a minimum in the absolute deviation and the slope of the absolute deviation reflects the size of the dataset. If the slopes are equal the size of the datasets are equal.
– Ross Millikan
Jul 24 at 4:19
My approach was $sum |a_i-x| $ is least when $x=a_m$ ,that is median of set $a$.As the two sums are equal $sum |b_i-x|$ is also least when $x=a_m$. But we know $sum|b_i - x|$ is least when $x=b_m$.So, in fact the two medians are same and so they have equal number of observations below them
– Legend Killer
Jul 24 at 4:25
 |Â
show 1 more comment
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
The total deviation constraint is enough without the requirement on the median and we can prove that the size of the two sets is the same. Let $clt a_1,b_1$ and $d=c-1$. The total deviation of the $a$s from $d$ is greater than the total deviation of the $a$s from $c$ by the number of the $a$s because $c$ is $1$ closer to each one. If the total deviation of the $a$s from $d$ equals the total deviation of the $b$s from $d$ and similarly for $c$ there must be the same number of $a$s and $b$s. As you have done, call that number $n.$
Now assume $a_1 neq b_1$. WOLOG we can assume $a_1 lt b_1$ and define $e=min(frac 12 (b_1-a_1), frac 12(a_2-a_1))gt 0$. We are given that the total deviation of both the $a$s and $b$s from $a_1$ is some number $f$. The total deviation of the $a$s from $a_1+e$ is $f+e-(n-1)e=f-(n-2)e$ because we are going away from $a_1$ and towards all the other $a$s. The total deviation of the $b$s from $a_1+e$ is $f-ne$ because we are going towards all of them. This is a contradiction, so $a_1=b_1$. We can repeat the argument now for $a_2$ and $b_2$ and so on up the line.
answered Jul 24 at 3:05
Ross Millikan
275k21186351
How can you calculate the sum of deviations of $a$ from $a_1+e$?
– Legend Killer
Jul 24 at 3:58
Once I have the sum of deviations from $a_1$, which I defined as $f$, I look at how each term in the sum of deviations changes. When you go to $a_1+e$ you get $e$ further from $a_1$ and $e$ closer to all the rest. I chose $e$ small enough to ensure you don't move past any of the others to make this work. You also get $e$ closer to all the $b$s. That is the heart of the argument. Take the $a$s as $1,2,3$ and $e=0.1$ and do the calculation by hand to see how it works.
– Ross Millikan
Jul 24 at 4:02
It is a brilliant proof , sir.I actually proved the two datasets are equal in size by taking x to be the median of the set a
– Legend Killer
Jul 24 at 4:11
I don't see how the median can prove the datasets are equal in size. Take a dataset and add one value above the median and one below. You have a new dataset with the same median and more values. Maybe you have more information than that. It is true that the median has the property that it is a minimum in the absolute deviation and the slope of the absolute deviation reflects the size of the dataset. If the slopes are equal the size of the datasets are equal.
– Ross Millikan
Jul 24 at 4:19
My approach was $sum |a_i-x| $ is least when $x=a_m$ ,that is median of set $a$.As the two sums are equal $sum |b_i-x|$ is also least when $x=a_m$. But we know $sum|b_i - x|$ is least when $x=b_m$.So, in fact the two medians are same and so they have equal number of observations below them
– Legend Killer
Jul 24 at 4:25
 |Â
show 1 more comment
up vote
1
down vote
accepted
The total deviation constraint is enough without the requirement on the median and we can prove that the size of the two sets is the same. Let $clt a_1,b_1$ and $d=c-1$. The total deviation of the $a$s from $d$ is greater than the total deviation of the $a$s from $c$ by the number of the $a$s because $c$ is $1$ closer to each one. If the total deviation of the $a$s from $d$ equals the total deviation of the $b$s from $d$ and similarly for $c$ there must be the same number of $a$s and $b$s. As you have done, call that number $n.$
Now assume $a_1 neq b_1$. WOLOG we can assume $a_1 lt b_1$ and define $e=min(frac 12 (b_1-a_1), frac 12(a_2-a_1))gt 0$. We are given that the total deviation of both the $a$s and $b$s from $a_1$ is some number $f$. The total deviation of the $a$s from $a_1+e$ is $f+e-(n-1)e=f-(n-2)e$ because we are going away from $a_1$ and towards all the other $a$s. The total deviation of the $b$s from $a_1+e$ is $f-ne$ because we are going towards all of them. This is a contradiction, so $a_1=b_1$. We can repeat the argument now for $a_2$ and $b_2$ and so on up the line.
answered Jul 24 at 3:05
Ross Millikan
275k21186351
How can you calculate the sum of deviations of $a$ from $a_1+e$?
– Legend Killer
Jul 24 at 3:58
Once I have the sum of deviations from $a_1$, which I defined as $f$, I look at how each term in the sum of deviations changes. When you go to $a_1+e$ you get $e$ further from $a_1$ and $e$ closer to all the rest. I chose $e$ small enough to ensure you don't move past any of the others to make this work. You also get $e$ closer to all the $b$s. That is the heart of the argument. Take the $a$s as $1,2,3$ and $e=0.1$ and do the calculation by hand to see how it works.
– Ross Millikan
Jul 24 at 4:02
It is a brilliant proof , sir.I actually proved the two datasets are equal in size by taking x to be the median of the set a
– Legend Killer
Jul 24 at 4:11
I don't see how the median can prove the datasets are equal in size. Take a dataset and add one value above the median and one below. You have a new dataset with the same median and more values. Maybe you have more information than that. It is true that the median has the property that it is a minimum in the absolute deviation and the slope of the absolute deviation reflects the size of the dataset. If the slopes are equal the size of the datasets are equal.
– Ross Millikan
Jul 24 at 4:19
My approach was $sum |a_i-x| $ is least when $x=a_m$ ,that is median of set $a$.As the two sums are equal $sum |b_i-x|$ is also least when $x=a_m$. But we know $sum|b_i - x|$ is least when $x=b_m$.So, in fact the two medians are same and so they have equal number of observations below them
– Legend Killer
Jul 24 at 4:25
 |Â
show 1 more comment
up vote
1
down vote
accepted
up vote
1
down vote
accepted
The total deviation constraint is enough without the requirement on the median and we can prove that the size of the two sets is the same. Let $clt a_1,b_1$ and $d=c-1$. The total deviation of the $a$s from $d$ is greater than the total deviation of the $a$s from $c$ by the number of the $a$s because $c$ is $1$ closer to each one. If the total deviation of the $a$s from $d$ equals the total deviation of the $b$s from $d$ and similarly for $c$ there must be the same number of $a$s and $b$s. As you have done, call that number $n.$
Now assume $a_1 neq b_1$. WOLOG we can assume $a_1 lt b_1$ and define $e=min(frac 12 (b_1-a_1), frac 12(a_2-a_1))gt 0$. We are given that the total deviation of both the $a$s and $b$s from $a_1$ is some number $f$. The total deviation of the $a$s from $a_1+e$ is $f+e-(n-1)e=f-(n-2)e$ because we are going away from $a_1$ and towards all the other $a$s. The total deviation of the $b$s from $a_1+e$ is $f-ne$ because we are going towards all of them. This is a contradiction, so $a_1=b_1$. We can repeat the argument now for $a_2$ and $b_2$ and so on up the line.
answered Jul 24 at 3:05
Ross Millikan
275k21186351
The total deviation constraint is enough without the requirement on the median and we can prove that the size of the two sets is the same. Let $clt a_1,b_1$ and $d=c-1$. The total deviation of the $a$s from $d$ is greater than the total deviation of the $a$s from $c$ by the number of the $a$s because $c$ is $1$ closer to each one. If the total deviation of the $a$s from $d$ equals the total deviation of the $b$s from $d$ and similarly for $c$ there must be the same number of $a$s and $b$s. As you have done, call that number $n.$
Now assume $a_1 neq b_1$. WOLOG we can assume $a_1 lt b_1$ and define $e=min(frac 12 (b_1-a_1), frac 12(a_2-a_1))gt 0$. We are given that the total deviation of both the $a$s and $b$s from $a_1$ is some number $f$. The total deviation of the $a$s from $a_1+e$ is $f+e-(n-1)e=f-(n-2)e$ because we are going away from $a_1$ and towards all the other $a$s. The total deviation of the $b$s from $a_1+e$ is $f-ne$ because we are going towards all of them. This is a contradiction, so $a_1=b_1$. We can repeat the argument now for $a_2$ and $b_2$ and so on up the line.
answered Jul 24 at 3:05
Ross Millikan
275k21186351
edited Jul 24 at 3:19
answered Jul 24 at 3:05
Ross Millikan
275k21186351
answered Jul 24 at 3:05
Ross Millikan
275k21186351
answered Jul 24 at 3:05
Ross Millikan
275k21186351
275k21186351
How can you calculate the sum of deviations of $a$ from $a_1+e$?
– Legend Killer
Jul 24 at 3:58
Once I have the sum of deviations from $a_1$, which I defined as $f$, I look at how each term in the sum of deviations changes. When you go to $a_1+e$ you get $e$ further from $a_1$ and $e$ closer to all the rest. I chose $e$ small enough to ensure you don't move past any of the others to make this work. You also get $e$ closer to all the $b$s. That is the heart of the argument. Take the $a$s as $1,2,3$ and $e=0.1$ and do the calculation by hand to see how it works.
– Ross Millikan
Jul 24 at 4:02
It is a brilliant proof , sir.I actually proved the two datasets are equal in size by taking x to be the median of the set a
– Legend Killer
Jul 24 at 4:11
I don't see how the median can prove the datasets are equal in size. Take a dataset and add one value above the median and one below. You have a new dataset with the same median and more values. Maybe you have more information than that. It is true that the median has the property that it is a minimum in the absolute deviation and the slope of the absolute deviation reflects the size of the dataset. If the slopes are equal the size of the datasets are equal.
– Ross Millikan
Jul 24 at 4:19
My approach was $sum |a_i-x| $ is least when $x=a_m$ ,that is median of set $a$.As the two sums are equal $sum |b_i-x|$ is also least when $x=a_m$. But we know $sum|b_i - x|$ is least when $x=b_m$.So, in fact the two medians are same and so they have equal number of observations below them
– Legend Killer
Jul 24 at 4:25
 |Â
show 1 more comment
How can you calculate the sum of deviations of $a$ from $a_1+e$?
– Legend Killer
Jul 24 at 3:58
Once I have the sum of deviations from $a_1$, which I defined as $f$, I look at how each term in the sum of deviations changes. When you go to $a_1+e$ you get $e$ further from $a_1$ and $e$ closer to all the rest. I chose $e$ small enough to ensure you don't move past any of the others to make this work. You also get $e$ closer to all the $b$s. That is the heart of the argument. Take the $a$s as $1,2,3$ and $e=0.1$ and do the calculation by hand to see how it works.
– Ross Millikan
Jul 24 at 4:02
It is a brilliant proof , sir.I actually proved the two datasets are equal in size by taking x to be the median of the set a
– Legend Killer
Jul 24 at 4:11
I don't see how the median can prove the datasets are equal in size. Take a dataset and add one value above the median and one below. You have a new dataset with the same median and more values. Maybe you have more information than that. It is true that the median has the property that it is a minimum in the absolute deviation and the slope of the absolute deviation reflects the size of the dataset. If the slopes are equal the size of the datasets are equal.
– Ross Millikan
Jul 24 at 4:19
My approach was $sum |a_i-x| $ is least when $x=a_m$ ,that is median of set $a$.As the two sums are equal $sum |b_i-x|$ is also least when $x=a_m$. But we know $sum|b_i - x|$ is least when $x=b_m$.So, in fact the two medians are same and so they have equal number of observations below them
– Legend Killer
Jul 24 at 4:25
How can you calculate the sum of deviations of $a$ from $a_1+e$?
– Legend Killer
Jul 24 at 3:58
How can you calculate the sum of deviations of $a$ from $a_1+e$?
– Legend Killer
Jul 24 at 3:58
Once I have the sum of deviations from $a_1$, which I defined as $f$, I look at how each term in the sum of deviations changes. When you go to $a_1+e$ you get $e$ further from $a_1$ and $e$ closer to all the rest. I chose $e$ small enough to ensure you don't move past any of the others to make this work. You also get $e$ closer to all the $b$s. That is the heart of the argument. Take the $a$s as $1,2,3$ and $e=0.1$ and do the calculation by hand to see how it works.
– Ross Millikan
Jul 24 at 4:02
Once I have the sum of deviations from $a_1$, which I defined as $f$, I look at how each term in the sum of deviations changes. When you go to $a_1+e$ you get $e$ further from $a_1$ and $e$ closer to all the rest. I chose $e$ small enough to ensure you don't move past any of the others to make this work. You also get $e$ closer to all the $b$s. That is the heart of the argument. Take the $a$s as $1,2,3$ and $e=0.1$ and do the calculation by hand to see how it works.
– Ross Millikan
Jul 24 at 4:02
It is a brilliant proof , sir.I actually proved the two datasets are equal in size by taking x to be the median of the set a
– Legend Killer
Jul 24 at 4:11
It is a brilliant proof , sir.I actually proved the two datasets are equal in size by taking x to be the median of the set a
– Legend Killer
Jul 24 at 4:11
I don't see how the median can prove the datasets are equal in size. Take a dataset and add one value above the median and one below. You have a new dataset with the same median and more values. Maybe you have more information than that. It is true that the median has the property that it is a minimum in the absolute deviation and the slope of the absolute deviation reflects the size of the dataset. If the slopes are equal the size of the datasets are equal.
– Ross Millikan
Jul 24 at 4:19
I don't see how the median can prove the datasets are equal in size. Take a dataset and add one value above the median and one below. You have a new dataset with the same median and more values. Maybe you have more information than that. It is true that the median has the property that it is a minimum in the absolute deviation and the slope of the absolute deviation reflects the size of the dataset. If the slopes are equal the size of the datasets are equal.
– Ross Millikan
Jul 24 at 4:19
My approach was $sum |a_i-x| $ is least when $x=a_m$ ,that is median of set $a$.As the two sums are equal $sum |b_i-x|$ is also least when $x=a_m$. But we know $sum|b_i - x|$ is least when $x=b_m$.So, in fact the two medians are same and so they have equal number of observations below them
– Legend Killer
Jul 24 at 4:25
My approach was $sum |a_i-x| $ is least when $x=a_m$ ,that is median of set $a$.As the two sums are equal $sum |b_i-x|$ is also least when $x=a_m$. But we know $sum|b_i - x|$ is least when $x=b_m$.So, in fact the two medians are same and so they have equal number of observations below them
– Legend Killer
Jul 24 at 4:25
 |Â
show 1 more comment
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Is the deviation to be from one real number, as the first sentence suggests, or all real numbers, as the next to last suggests?
– Ross Millikan
Jul 24 at 3:06
Any real number $x$
– Legend Killer
Jul 24 at 3:07