Inequality of difference of medians of two datasets.
$begingroup$
For ${x_k}_{k=1}^m$ and ${y_k}_{k=1}^minmathbb{R}$ with $x_k-y_k=d_k$
define $x=lim_{deltarightarrow 0}left(argmin frac{1}{m}sum_{k=1}^m|x_k-x|+delta x^2right)$
and $y=lim_{deltarightarrow 0}left(argmin frac{1}{m}sum_{k=1}^m|y_k-y|+delta y^2right).$ Is there any way to get the inequality of this form $|x-y|leq C|frac1msum_{k=1}^md_k|?$ I have tried in several ways, but had problem with $|x_k-x|$ and $|y_k-y|$ terms. All I could come up with $|x-y|leq frac C msum_{k=1}^m|d_k|.$ Any help would be appreciated. Thanks in advance!
inequality median
asked Dec 21 '18 at 15:55
John_WickJohn_Wick
1,616111
$endgroup$
add a comment |
$begingroup$
For ${x_k}_{k=1}^m$ and ${y_k}_{k=1}^minmathbb{R}$ with $x_k-y_k=d_k$
define $x=lim_{deltarightarrow 0}left(argmin frac{1}{m}sum_{k=1}^m|x_k-x|+delta x^2right)$
and $y=lim_{deltarightarrow 0}left(argmin frac{1}{m}sum_{k=1}^m|y_k-y|+delta y^2right).$ Is there any way to get the inequality of this form $|x-y|leq C|frac1msum_{k=1}^md_k|?$ I have tried in several ways, but had problem with $|x_k-x|$ and $|y_k-y|$ terms. All I could come up with $|x-y|leq frac C msum_{k=1}^m|d_k|.$ Any help would be appreciated. Thanks in advance!
inequality median
asked Dec 21 '18 at 15:55
John_WickJohn_Wick
1,616111
$endgroup$
add a comment |
$begingroup$
For ${x_k}_{k=1}^m$ and ${y_k}_{k=1}^minmathbb{R}$ with $x_k-y_k=d_k$
define $x=lim_{deltarightarrow 0}left(argmin frac{1}{m}sum_{k=1}^m|x_k-x|+delta x^2right)$
and $y=lim_{deltarightarrow 0}left(argmin frac{1}{m}sum_{k=1}^m|y_k-y|+delta y^2right).$ Is there any way to get the inequality of this form $|x-y|leq C|frac1msum_{k=1}^md_k|?$ I have tried in several ways, but had problem with $|x_k-x|$ and $|y_k-y|$ terms. All I could come up with $|x-y|leq frac C msum_{k=1}^m|d_k|.$ Any help would be appreciated. Thanks in advance!
inequality median
asked Dec 21 '18 at 15:55
John_WickJohn_Wick
1,616111
$endgroup$
For ${x_k}_{k=1}^m$ and ${y_k}_{k=1}^minmathbb{R}$ with $x_k-y_k=d_k$
define $x=lim_{deltarightarrow 0}left(argmin frac{1}{m}sum_{k=1}^m|x_k-x|+delta x^2right)$
and $y=lim_{deltarightarrow 0}left(argmin frac{1}{m}sum_{k=1}^m|y_k-y|+delta y^2right).$ Is there any way to get the inequality of this form $|x-y|leq C|frac1msum_{k=1}^md_k|?$ I have tried in several ways, but had problem with $|x_k-x|$ and $|y_k-y|$ terms. All I could come up with $|x-y|leq frac C msum_{k=1}^m|d_k|.$ Any help would be appreciated. Thanks in advance!
inequality median
inequality median
asked Dec 21 '18 at 15:55
John_WickJohn_Wick
1,616111
asked Dec 21 '18 at 15:55
John_WickJohn_Wick
1,616111
edited Dec 21 '18 at 16:00
John_Wick
asked Dec 21 '18 at 15:55
John_WickJohn_Wick
1,616111
asked Dec 21 '18 at 15:55
John_WickJohn_Wick
1,616111
asked Dec 21 '18 at 15:55
John_WickJohn_Wick
1,616111
1,616111
add a comment |
add a comment |
1 Answer
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Unfortunately, there is no such way because otherwise if $sum d_k=0$ then we should have $x=y$. This obviously fails, even in case $sum x_k=sum y_k=0$. For instance, put $k=4$, $x_1=x_2=1$, $x_3=x_4=-1$, $y_1=y_2=y_3=1$, $y_4=-3$. Then $x=0$, but $y=1$.
answered Dec 22 '18 at 14:09
Alex RavskyAlex Ravsky
42.4k32383
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1 Answer
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1 Answer
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$begingroup$
Unfortunately, there is no such way because otherwise if $sum d_k=0$ then we should have $x=y$. This obviously fails, even in case $sum x_k=sum y_k=0$. For instance, put $k=4$, $x_1=x_2=1$, $x_3=x_4=-1$, $y_1=y_2=y_3=1$, $y_4=-3$. Then $x=0$, but $y=1$.
answered Dec 22 '18 at 14:09
Alex RavskyAlex Ravsky
42.4k32383
$endgroup$
add a comment |
$begingroup$
Unfortunately, there is no such way because otherwise if $sum d_k=0$ then we should have $x=y$. This obviously fails, even in case $sum x_k=sum y_k=0$. For instance, put $k=4$, $x_1=x_2=1$, $x_3=x_4=-1$, $y_1=y_2=y_3=1$, $y_4=-3$. Then $x=0$, but $y=1$.
answered Dec 22 '18 at 14:09
Alex RavskyAlex Ravsky
42.4k32383
$endgroup$
add a comment |
$begingroup$
Unfortunately, there is no such way because otherwise if $sum d_k=0$ then we should have $x=y$. This obviously fails, even in case $sum x_k=sum y_k=0$. For instance, put $k=4$, $x_1=x_2=1$, $x_3=x_4=-1$, $y_1=y_2=y_3=1$, $y_4=-3$. Then $x=0$, but $y=1$.
answered Dec 22 '18 at 14:09
Alex RavskyAlex Ravsky
42.4k32383
$endgroup$
Unfortunately, there is no such way because otherwise if $sum d_k=0$ then we should have $x=y$. This obviously fails, even in case $sum x_k=sum y_k=0$. For instance, put $k=4$, $x_1=x_2=1$, $x_3=x_4=-1$, $y_1=y_2=y_3=1$, $y_4=-3$. Then $x=0$, but $y=1$.
answered Dec 22 '18 at 14:09
Alex RavskyAlex Ravsky
42.4k32383
answered Dec 22 '18 at 14:09
Alex RavskyAlex Ravsky
42.4k32383
answered Dec 22 '18 at 14:09
Alex RavskyAlex Ravsky
42.4k32383
answered Dec 22 '18 at 14:09
Alex RavskyAlex Ravsky
42.4k32383
42.4k32383
add a comment |
add a comment |
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