A nice variance inequality

Clash Royale CLAN TAG#URR8PPP

up vote
0
down vote

favorite

Let $X$, $Y$ be random variables. Is there a name for the inequality
$$
mathrmVar(X)leq2mathrmVar(Y)+2E[(X-Y)^2]?
$$

probability

share|cite|improve this question

edited Jul 30 at 15:34

asked Jul 30 at 13:07

julian

312110

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Where did you find this inequality? Have you tried to prove it? Playing with it, it seems like it could be a result of the Cauchy-Schwartz inequality (in either of the forms $(E | XY |)^2 leq E(X^2)E(Y^2)$ or $mathrmCov(X,Y)^2 leq mathrmVar(X) mathrmVar(Y)$ ).
  – Malkin
  Jul 30 at 13:32
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  If we first assume $E[X]=E[Y]=0$, then the asserted inequality is equivalent to $0leq E[X^2]+4E[Y^2]-4E[XY]$. By Cauchy-Schwarz, $E[X^2]+4E[Y^2]-4E[XY]geq E[X^2]+4E[Y^2]-4E[X^2]^1/2E[Y^2]^1/2=(E[X^2]^1/2-2E[Y^2]^1/2)^2geq 0$.
  – julian
  Jul 30 at 13:59
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  The general case is similar. Here, one uses the special case for $X-E[X]$ and $Y-E[Y]$ and exploits $mathrmCov(X,Y)leq mathrmVar(X)^1/2mathrmVar(Y)^1/2$. Probably, there is no specific name for the inequality.
  – julian
  Jul 30 at 14:09
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  A lovely trick. Thanks for sharing. Yes, I expect similar. "A corollary to Cauchy-Schwartz", perhaps.
  – Malkin
  Jul 30 at 14:14
  

  
  
  
  

add a comment | 

up vote
0
down vote

favorite

Let $X$, $Y$ be random variables. Is there a name for the inequality
$$
mathrmVar(X)leq2mathrmVar(Y)+2E[(X-Y)^2]?
$$

probability

share|cite|improve this question

edited Jul 30 at 15:34

asked Jul 30 at 13:07

julian

312110

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Where did you find this inequality? Have you tried to prove it? Playing with it, it seems like it could be a result of the Cauchy-Schwartz inequality (in either of the forms $(E | XY |)^2 leq E(X^2)E(Y^2)$ or $mathrmCov(X,Y)^2 leq mathrmVar(X) mathrmVar(Y)$ ).
  – Malkin
  Jul 30 at 13:32
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  If we first assume $E[X]=E[Y]=0$, then the asserted inequality is equivalent to $0leq E[X^2]+4E[Y^2]-4E[XY]$. By Cauchy-Schwarz, $E[X^2]+4E[Y^2]-4E[XY]geq E[X^2]+4E[Y^2]-4E[X^2]^1/2E[Y^2]^1/2=(E[X^2]^1/2-2E[Y^2]^1/2)^2geq 0$.
  – julian
  Jul 30 at 13:59
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  The general case is similar. Here, one uses the special case for $X-E[X]$ and $Y-E[Y]$ and exploits $mathrmCov(X,Y)leq mathrmVar(X)^1/2mathrmVar(Y)^1/2$. Probably, there is no specific name for the inequality.
  – julian
  Jul 30 at 14:09
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  A lovely trick. Thanks for sharing. Yes, I expect similar. "A corollary to Cauchy-Schwartz", perhaps.
  – Malkin
  Jul 30 at 14:14
  

  
  
  
  

add a comment | 

up vote
0
down vote

favorite

up vote
0
down vote

favorite

Let $X$, $Y$ be random variables. Is there a name for the inequality
$$
mathrmVar(X)leq2mathrmVar(Y)+2E[(X-Y)^2]?
$$

probability

share|cite|improve this question

edited Jul 30 at 15:34

asked Jul 30 at 13:07

julian

312110

Let $X$, $Y$ be random variables. Is there a name for the inequality
$$
mathrmVar(X)leq2mathrmVar(Y)+2E[(X-Y)^2]?
$$

probability

share|cite|improve this question

edited Jul 30 at 15:34

asked Jul 30 at 13:07

julian

312110

share|cite|improve this question

share|cite|improve this question

edited Jul 30 at 15:34

edited Jul 30 at 15:34

edited Jul 30 at 15:34

asked Jul 30 at 13:07

julian

312110

asked Jul 30 at 13:07

julian

312110

asked Jul 30 at 13:07

julian

312110

312110

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Where did you find this inequality? Have you tried to prove it? Playing with it, it seems like it could be a result of the Cauchy-Schwartz inequality (in either of the forms $(E | XY |)^2 leq E(X^2)E(Y^2)$ or $mathrmCov(X,Y)^2 leq mathrmVar(X) mathrmVar(Y)$ ).
  – Malkin
  Jul 30 at 13:32
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  If we first assume $E[X]=E[Y]=0$, then the asserted inequality is equivalent to $0leq E[X^2]+4E[Y^2]-4E[XY]$. By Cauchy-Schwarz, $E[X^2]+4E[Y^2]-4E[XY]geq E[X^2]+4E[Y^2]-4E[X^2]^1/2E[Y^2]^1/2=(E[X^2]^1/2-2E[Y^2]^1/2)^2geq 0$.
  – julian
  Jul 30 at 13:59
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  The general case is similar. Here, one uses the special case for $X-E[X]$ and $Y-E[Y]$ and exploits $mathrmCov(X,Y)leq mathrmVar(X)^1/2mathrmVar(Y)^1/2$. Probably, there is no specific name for the inequality.
  – julian
  Jul 30 at 14:09
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  A lovely trick. Thanks for sharing. Yes, I expect similar. "A corollary to Cauchy-Schwartz", perhaps.
  – Malkin
  Jul 30 at 14:14
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Where did you find this inequality? Have you tried to prove it? Playing with it, it seems like it could be a result of the Cauchy-Schwartz inequality (in either of the forms $(E | XY |)^2 leq E(X^2)E(Y^2)$ or $mathrmCov(X,Y)^2 leq mathrmVar(X) mathrmVar(Y)$ ).
  – Malkin
  Jul 30 at 13:32
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  If we first assume $E[X]=E[Y]=0$, then the asserted inequality is equivalent to $0leq E[X^2]+4E[Y^2]-4E[XY]$. By Cauchy-Schwarz, $E[X^2]+4E[Y^2]-4E[XY]geq E[X^2]+4E[Y^2]-4E[X^2]^1/2E[Y^2]^1/2=(E[X^2]^1/2-2E[Y^2]^1/2)^2geq 0$.
  – julian
  Jul 30 at 13:59
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  The general case is similar. Here, one uses the special case for $X-E[X]$ and $Y-E[Y]$ and exploits $mathrmCov(X,Y)leq mathrmVar(X)^1/2mathrmVar(Y)^1/2$. Probably, there is no specific name for the inequality.
  – julian
  Jul 30 at 14:09
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  A lovely trick. Thanks for sharing. Yes, I expect similar. "A corollary to Cauchy-Schwartz", perhaps.
  – Malkin
  Jul 30 at 14:14
  

  
  
  
  

Where did you find this inequality? Have you tried to prove it? Playing with it, it seems like it could be a result of the Cauchy-Schwartz inequality (in either of the forms $(E | XY |)^2 leq E(X^2)E(Y^2)$ or $mathrmCov(X,Y)^2 leq mathrmVar(X) mathrmVar(Y)$ ).
– Malkin
Jul 30 at 13:32

Where did you find this inequality? Have you tried to prove it? Playing with it, it seems like it could be a result of the Cauchy-Schwartz inequality (in either of the forms $(E | XY |)^2 leq E(X^2)E(Y^2)$ or $mathrmCov(X,Y)^2 leq mathrmVar(X) mathrmVar(Y)$ ).
– Malkin
Jul 30 at 13:32

1

1

If we first assume $E[X]=E[Y]=0$, then the asserted inequality is equivalent to $0leq E[X^2]+4E[Y^2]-4E[XY]$. By Cauchy-Schwarz, $E[X^2]+4E[Y^2]-4E[XY]geq E[X^2]+4E[Y^2]-4E[X^2]^1/2E[Y^2]^1/2=(E[X^2]^1/2-2E[Y^2]^1/2)^2geq 0$.
– julian
Jul 30 at 13:59

If we first assume $E[X]=E[Y]=0$, then the asserted inequality is equivalent to $0leq E[X^2]+4E[Y^2]-4E[XY]$. By Cauchy-Schwarz, $E[X^2]+4E[Y^2]-4E[XY]geq E[X^2]+4E[Y^2]-4E[X^2]^1/2E[Y^2]^1/2=(E[X^2]^1/2-2E[Y^2]^1/2)^2geq 0$.
– julian
Jul 30 at 13:59

The general case is similar. Here, one uses the special case for $X-E[X]$ and $Y-E[Y]$ and exploits $mathrmCov(X,Y)leq mathrmVar(X)^1/2mathrmVar(Y)^1/2$. Probably, there is no specific name for the inequality.
– julian
Jul 30 at 14:09

The general case is similar. Here, one uses the special case for $X-E[X]$ and $Y-E[Y]$ and exploits $mathrmCov(X,Y)leq mathrmVar(X)^1/2mathrmVar(Y)^1/2$. Probably, there is no specific name for the inequality.
– julian
Jul 30 at 14:09

A lovely trick. Thanks for sharing. Yes, I expect similar. "A corollary to Cauchy-Schwartz", perhaps.
– Malkin
Jul 30 at 14:14

A lovely trick. Thanks for sharing. Yes, I expect similar. "A corollary to Cauchy-Schwartz", perhaps.
– Malkin
Jul 30 at 14:14

add a comment | 

1 Answer
1

active

oldest

votes

up vote
0
down vote



accepted

Let me elaborate the comment in the hope that someone might find this inequality useful.

The inequality is equivalent to
beginalign*
0&leq 2mathrmVar(Y)+E[X^2]+2E[Y^2]+E[X]^2-4E[XY]\
&=4mathrmVar(Y)+mathrmVar(X)+2(E[X]^2+E[Y]^2)-4E[XY].
endalign*
Now by the elementary inequality $2ableq a^2+b^2$ and Cauchy-Schwarz, we get
beginalign*
4mathrmVar(Y)+mathrmVar(X)+2(E[X]^2+E[Y]^2)-4E[XY]&geq 4mathrmVar(Y)+mathrmVar(X)-4mathrmCov(X,Y)\
&geq (2sqrtmathrmVar(Y)-sqrtmathrmVar(X))^2\
&geq 0.
endalign*

share|cite|improve this answer

answered Jul 30 at 15:33

julian

312110

add a comment | 

Your Answer

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',
convertImagesToLinks: true,
noModals: false,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
);



);

 

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

StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2867005%2fa-nice-variance-inequality%23new-answer', 'question_page');

);

Post as a guest

Name Email

1 Answer
1

active

oldest

votes

1 Answer
1

active

oldest

votes

active

oldest

votes

active

oldest

votes

up vote
0
down vote



accepted

Let me elaborate the comment in the hope that someone might find this inequality useful.

The inequality is equivalent to
beginalign*
0&leq 2mathrmVar(Y)+E[X^2]+2E[Y^2]+E[X]^2-4E[XY]\
&=4mathrmVar(Y)+mathrmVar(X)+2(E[X]^2+E[Y]^2)-4E[XY].
endalign*
Now by the elementary inequality $2ableq a^2+b^2$ and Cauchy-Schwarz, we get
beginalign*
4mathrmVar(Y)+mathrmVar(X)+2(E[X]^2+E[Y]^2)-4E[XY]&geq 4mathrmVar(Y)+mathrmVar(X)-4mathrmCov(X,Y)\
&geq (2sqrtmathrmVar(Y)-sqrtmathrmVar(X))^2\
&geq 0.
endalign*

share|cite|improve this answer

answered Jul 30 at 15:33

julian

312110

add a comment | 

up vote
0
down vote



accepted

Let me elaborate the comment in the hope that someone might find this inequality useful.

The inequality is equivalent to
beginalign*
0&leq 2mathrmVar(Y)+E[X^2]+2E[Y^2]+E[X]^2-4E[XY]\
&=4mathrmVar(Y)+mathrmVar(X)+2(E[X]^2+E[Y]^2)-4E[XY].
endalign*
Now by the elementary inequality $2ableq a^2+b^2$ and Cauchy-Schwarz, we get
beginalign*
4mathrmVar(Y)+mathrmVar(X)+2(E[X]^2+E[Y]^2)-4E[XY]&geq 4mathrmVar(Y)+mathrmVar(X)-4mathrmCov(X,Y)\
&geq (2sqrtmathrmVar(Y)-sqrtmathrmVar(X))^2\
&geq 0.
endalign*

share|cite|improve this answer

answered Jul 30 at 15:33

julian

312110

add a comment | 

up vote
0
down vote



accepted

up vote
0
down vote



accepted

Let me elaborate the comment in the hope that someone might find this inequality useful.

The inequality is equivalent to
beginalign*
0&leq 2mathrmVar(Y)+E[X^2]+2E[Y^2]+E[X]^2-4E[XY]\
&=4mathrmVar(Y)+mathrmVar(X)+2(E[X]^2+E[Y]^2)-4E[XY].
endalign*
Now by the elementary inequality $2ableq a^2+b^2$ and Cauchy-Schwarz, we get
beginalign*
4mathrmVar(Y)+mathrmVar(X)+2(E[X]^2+E[Y]^2)-4E[XY]&geq 4mathrmVar(Y)+mathrmVar(X)-4mathrmCov(X,Y)\
&geq (2sqrtmathrmVar(Y)-sqrtmathrmVar(X))^2\
&geq 0.
endalign*

share|cite|improve this answer

answered Jul 30 at 15:33

julian

312110

Let me elaborate the comment in the hope that someone might find this inequality useful.

The inequality is equivalent to
beginalign*
0&leq 2mathrmVar(Y)+E[X^2]+2E[Y^2]+E[X]^2-4E[XY]\
&=4mathrmVar(Y)+mathrmVar(X)+2(E[X]^2+E[Y]^2)-4E[XY].
endalign*
Now by the elementary inequality $2ableq a^2+b^2$ and Cauchy-Schwarz, we get
beginalign*
4mathrmVar(Y)+mathrmVar(X)+2(E[X]^2+E[Y]^2)-4E[XY]&geq 4mathrmVar(Y)+mathrmVar(X)-4mathrmCov(X,Y)\
&geq (2sqrtmathrmVar(Y)-sqrtmathrmVar(X))^2\
&geq 0.
endalign*

share|cite|improve this answer

answered Jul 30 at 15:33

julian

312110

share|cite|improve this answer

share|cite|improve this answer

answered Jul 30 at 15:33

julian

312110

answered Jul 30 at 15:33

julian

312110

answered Jul 30 at 15:33

julian

312110

312110

add a comment | 

add a comment | 

 

draft saved

draft discarded

 

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

StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2867005%2fa-nice-variance-inequality%23new-answer', 'question_page');

);

Post as a guest

Name Email

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

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

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

Sign up using Google

Sign up using Facebook

Sign up using Email and Password

Post as a guest

Name Email

Name Email

Name

Email

Name Email

Name

Email