Dgdrxrt

Inexact numbers as keys in Association?

How many extra stops do monopods offer for tele photographs?

IC has pull-down resistors on SMBus lines?

If Nick Fury and Coulson already knew about aliens (Kree and Skrull) why did they wait until Thor's appearance to start making weapons?

Can I calculate next year's exemptions based on this year's refund/amount owed?

How to set page number in right side in chapter title page?

My ex-girlfriend uses my Apple ID to login to her iPad, do I have to give her my Apple ID password to reset it?

Is there a reasonable and studied concept of reduction between regular languages?

Is it okay to majorly distort historical facts while writing a fiction story?

Strange use of "whether ... than ..." in official text

Why is information "lost" when it got into a black hole?

(How) Could a medieval fantasy world survive a magic-induced "nuclear winter"?

Help/tips for a first time writer?

In the "Harry Potter and the Order of the Phoenix" videogame, what potion is used to sabotage Umbridge's speakers?

Audio Conversion With ADS1243

Sulfuric acid symmetry point group

What is the difference between Statistical Mechanics and Quantum Mechanics

How did Beeri the Hittite come up with naming his daughter Yehudit?

Calculate the Mean mean of two numbers

Does destroying a Lich's phylactery destroy the soul within it?

Easy to read palindrome checker

Reference request: Grassmannian and Plucker coordinates in type B, C, D

When "be it" is at the beginning of a sentence, what kind of structure do you call it?

Why don't programming languages automatically manage the synchronous/asynchronous problem?

Writing median rather than mean

0

$begingroup$

I have some code to calculate the mean bias between two arrays.

$$mean_bias=fracsum_i=1^na_i-b_in$$

and was hoping to represent the median instead:

$$median_bias=Median(a-b)$$

Does something like this hold true? But it's not sorted so index will not be the middle:

$$median_bias=sum_i=n/2^n/2+1a_i-b_i$$

How can I represent this?

Thanks

average means median

share|cite|improve this question

edited Mar 28 at 1:40

Ocean Scientist

asked Mar 28 at 0:52

Ocean ScientistOcean Scientist

32

New contributor

Ocean Scientist is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  This question does not concern mathematics and would be better suited for TeX - LaTeX Stackexchange.
  $endgroup$
  – Brian
  Mar 28 at 1:08
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I tried there first and was suggested to ask here instead.
  $endgroup$
  – Ocean Scientist
  Mar 28 at 1:14
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  I don't understand the question
  $endgroup$
  – Jorge Fernández Hidalgo
  Mar 28 at 1:14
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  You can write median bias = median a - median b
  $endgroup$
  – Jorge Fernández Hidalgo
  Mar 28 at 1:15
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  The formula only, I can convert it into Tex myself.
  $endgroup$
  – Ocean Scientist
  Mar 28 at 1:18
  

  
  
  
  

 | 
show 4 more comments

0

$begingroup$

I have some code to calculate the mean bias between two arrays.

$$mean_bias=fracsum_i=1^na_i-b_in$$

and was hoping to represent the median instead:

$$median_bias=Median(a-b)$$

Does something like this hold true? But it's not sorted so index will not be the middle:

$$median_bias=sum_i=n/2^n/2+1a_i-b_i$$

How can I represent this?

Thanks

average means median

share|cite|improve this question

edited Mar 28 at 1:40

Ocean Scientist

asked Mar 28 at 0:52

Ocean ScientistOcean Scientist

32

New contributor

Ocean Scientist is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  This question does not concern mathematics and would be better suited for TeX - LaTeX Stackexchange.
  $endgroup$
  – Brian
  Mar 28 at 1:08
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I tried there first and was suggested to ask here instead.
  $endgroup$
  – Ocean Scientist
  Mar 28 at 1:14
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  I don't understand the question
  $endgroup$
  – Jorge Fernández Hidalgo
  Mar 28 at 1:14
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  You can write median bias = median a - median b
  $endgroup$
  – Jorge Fernández Hidalgo
  Mar 28 at 1:15
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  The formula only, I can convert it into Tex myself.
  $endgroup$
  – Ocean Scientist
  Mar 28 at 1:18
  

  
  
  
  

 | 
show 4 more comments

$begingroup$

I have some code to calculate the mean bias between two arrays.

$$mean_bias=fracsum_i=1^na_i-b_in$$

and was hoping to represent the median instead:

$$median_bias=Median(a-b)$$

Does something like this hold true? But it's not sorted so index will not be the middle:

$$median_bias=sum_i=n/2^n/2+1a_i-b_i$$

How can I represent this?

Thanks

average means median

share|cite|improve this question

edited Mar 28 at 1:40

Ocean Scientist

asked Mar 28 at 0:52

Ocean ScientistOcean Scientist

32

New contributor

Ocean Scientist is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

$endgroup$

share|cite|improve this question

edited Mar 28 at 1:40

Ocean Scientist

asked Mar 28 at 0:52

Ocean ScientistOcean Scientist

32

New contributor

Ocean Scientist is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

share|cite|improve this question

edited Mar 28 at 1:40

Ocean Scientist

asked Mar 28 at 0:52

Ocean ScientistOcean Scientist

32

New contributor

Ocean Scientist is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

asked Mar 28 at 0:52

Ocean ScientistOcean Scientist

32

New contributor

Ocean Scientist is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

asked Mar 28 at 0:52

Ocean ScientistOcean Scientist

32

1 Answer
1

active

oldest

votes

0

$begingroup$

You might be able to write something like this:

begingather
bias_textmedian
= frac12left(max_ilefta_i-b_i mid
leftlvertleftjmid a_i-b_i > a_j-b_jrightrightrvert < tfrac n2right \
qquadqquadqquadqquadqquad
+ min_ilefta_i-b_i mid
leftlvertleftjmid a_i-b_i < a_j-b_jrightrightrvert < tfrac n2right
right)
endgather

This is a way of saying that if $n$ is odd, you take the value $a_i - b_i$ that partitions the list of values $a_j - b_j$
(excluding $a_i - b_i$ itself) into two equal sets, one of values less than
$a_i - b_i$ and one of values greater; and if $n$ is even you take the mean
of the two values $a_i - b_i$ and $a_i' - b_i'$ that partition the list
of values $a_j - b_j$ (excluding those two values) into two equal sets.

But I think this is a rather horrible expression.

A better way is probably to define some notation of your own, for example,

The notation $textmedian(S)$ represents the median of the set $S$,

followed at some later time by an expression such as

$$ textmedian a_i - b_i mid 1 leq i leq n. $$

share|cite|improve this answer

answered Mar 28 at 3:47

David KDavid K

55.4k345120

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  So there isn't a nice way of doing it. Interesting. I agree that the first one does the job, but it's a bit much! I think I will just define that median means median like your suggestion 2. Thanks very much!
  $endgroup$
  – Ocean Scientist
  Mar 28 at 4:12
  

  
  
  
  

add a comment | 

StackExchange.ifUsing("editor", function ()
return StackExchange.using("mathjaxEditing", function ()
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix)
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
);
);
, "mathjax-editing");

StackExchange.ready(function()
var channelOptions =
tags: "".split(" "),
id: "69"
;
initTagRenderer("".split(" "), "".split(" "), channelOptions);

StackExchange.using("externalEditor", function()
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled)
StackExchange.using("snippets", function()
createEditor();
);

else
createEditor();

);

function createEditor()
StackExchange.prepareEditor(
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader:
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
,
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
);



);

Ocean Scientist is a new contributor. Be nice, and check out our Code of Conduct.

draft saved

draft discarded

Sign up or log in

StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);

Sign up using Google

Sign up using Facebook

Sign up using Email and Password

Post as a guest

Name

Email

Required, but never shown

StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3165333%2fwriting-median-rather-than-mean%23new-answer', 'question_page');

);

Post as a guest

Name

Email

Required, but never shown

0

$begingroup$

You might be able to write something like this:

begingather
bias_textmedian
= frac12left(max_ilefta_i-b_i mid
leftlvertleftjmid a_i-b_i > a_j-b_jrightrightrvert < tfrac n2right \
qquadqquadqquadqquadqquad
+ min_ilefta_i-b_i mid
leftlvertleftjmid a_i-b_i < a_j-b_jrightrightrvert < tfrac n2right
right)
endgather

This is a way of saying that if $n$ is odd, you take the value $a_i - b_i$ that partitions the list of values $a_j - b_j$
(excluding $a_i - b_i$ itself) into two equal sets, one of values less than
$a_i - b_i$ and one of values greater; and if $n$ is even you take the mean
of the two values $a_i - b_i$ and $a_i' - b_i'$ that partition the list
of values $a_j - b_j$ (excluding those two values) into two equal sets.

But I think this is a rather horrible expression.

A better way is probably to define some notation of your own, for example,

The notation $textmedian(S)$ represents the median of the set $S$,

followed at some later time by an expression such as

$$ textmedian a_i - b_i mid 1 leq i leq n. $$

share|cite|improve this answer

answered Mar 28 at 3:47

David KDavid K

55.4k345120

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  So there isn't a nice way of doing it. Interesting. I agree that the first one does the job, but it's a bit much! I think I will just define that median means median like your suggestion 2. Thanks very much!
  $endgroup$
  – Ocean Scientist
  Mar 28 at 4:12
  

  
  
  
  

add a comment | 

0

$begingroup$

You might be able to write something like this:

begingather
bias_textmedian
= frac12left(max_ilefta_i-b_i mid
leftlvertleftjmid a_i-b_i > a_j-b_jrightrightrvert < tfrac n2right \
qquadqquadqquadqquadqquad
+ min_ilefta_i-b_i mid
leftlvertleftjmid a_i-b_i < a_j-b_jrightrightrvert < tfrac n2right
right)
endgather

This is a way of saying that if $n$ is odd, you take the value $a_i - b_i$ that partitions the list of values $a_j - b_j$
(excluding $a_i - b_i$ itself) into two equal sets, one of values less than
$a_i - b_i$ and one of values greater; and if $n$ is even you take the mean
of the two values $a_i - b_i$ and $a_i' - b_i'$ that partition the list
of values $a_j - b_j$ (excluding those two values) into two equal sets.

But I think this is a rather horrible expression.

A better way is probably to define some notation of your own, for example,

The notation $textmedian(S)$ represents the median of the set $S$,

followed at some later time by an expression such as

$$ textmedian a_i - b_i mid 1 leq i leq n. $$

share|cite|improve this answer

answered Mar 28 at 3:47

David KDavid K

55.4k345120

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  So there isn't a nice way of doing it. Interesting. I agree that the first one does the job, but it's a bit much! I think I will just define that median means median like your suggestion 2. Thanks very much!
  $endgroup$
  – Ocean Scientist
  Mar 28 at 4:12
  

  
  
  
  

add a comment | 

$begingroup$

You might be able to write something like this:

begingather
bias_textmedian
= frac12left(max_ilefta_i-b_i mid
leftlvertleftjmid a_i-b_i > a_j-b_jrightrightrvert < tfrac n2right \
qquadqquadqquadqquadqquad
+ min_ilefta_i-b_i mid
leftlvertleftjmid a_i-b_i < a_j-b_jrightrightrvert < tfrac n2right
right)
endgather

This is a way of saying that if $n$ is odd, you take the value $a_i - b_i$ that partitions the list of values $a_j - b_j$
(excluding $a_i - b_i$ itself) into two equal sets, one of values less than
$a_i - b_i$ and one of values greater; and if $n$ is even you take the mean
of the two values $a_i - b_i$ and $a_i' - b_i'$ that partition the list
of values $a_j - b_j$ (excluding those two values) into two equal sets.

But I think this is a rather horrible expression.

A better way is probably to define some notation of your own, for example,

The notation $textmedian(S)$ represents the median of the set $S$,

followed at some later time by an expression such as

$$ textmedian a_i - b_i mid 1 leq i leq n. $$

share|cite|improve this answer

answered Mar 28 at 3:47

David KDavid K

55.4k345120

$endgroup$

share|cite|improve this answer

answered Mar 28 at 3:47

David KDavid K

55.4k345120

answered Mar 28 at 3:47

David KDavid K

55.4k345120

answered Mar 28 at 3:47

David KDavid K

55.4k345120