Correlation between two linear combinations of random variables

Clash Royale CLAN TAG#URR8PPP

up vote
1
down vote

favorite

Let $X_1, X_2, X_3, X_4$ be independent random variables with $operatornamevar(X_i)=1$, and

$$U = 2X_1+X_2+X_3$$
$$ V = X_2+X_3 + 2X_4$$

Find $operatornamecorr(U, V)$

In general, how can I calculate the correlation between two linear combinations of independent $X_i$ such as $U$ and $V$ knowing only $operatornamevar(X_i)$?

Or what if they weren't independent, but I had their covariance or correlation matrix?

probability statistics random-variables correlation

share|cite|improve this question

edited Jul 30 at 1:21

Michael Hardy

204k23185461

asked Jul 28 at 22:27

Duars

155

• 
  
  
  

  
  

  
  
  
  
  
  

  
  You can just go to this question. The answer are great.
  – M. Cris
  Jul 28 at 22:30
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  This question has no answer unless you know how $X_i$'s depend on each other.
  – Kavi Rama Murthy
  Jul 28 at 23:11
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @KaviRamaMurthy I'm sorry! I forgot to mention the X's are independent
  – Duars
  Jul 28 at 23:14
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  I think Henry's answer below is too complicated. See my answer.
  – Michael Hardy
  Jul 30 at 1:35
  

  
  
  
  

add a comment | 

up vote
1
down vote

favorite

Let $X_1, X_2, X_3, X_4$ be independent random variables with $operatornamevar(X_i)=1$, and

$$U = 2X_1+X_2+X_3$$
$$ V = X_2+X_3 + 2X_4$$

Find $operatornamecorr(U, V)$

In general, how can I calculate the correlation between two linear combinations of independent $X_i$ such as $U$ and $V$ knowing only $operatornamevar(X_i)$?

Or what if they weren't independent, but I had their covariance or correlation matrix?

probability statistics random-variables correlation

share|cite|improve this question

edited Jul 30 at 1:21

Michael Hardy

204k23185461

asked Jul 28 at 22:27

Duars

155

• 
  
  
  

  
  

  
  
  
  
  
  

  
  You can just go to this question. The answer are great.
  – M. Cris
  Jul 28 at 22:30
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  This question has no answer unless you know how $X_i$'s depend on each other.
  – Kavi Rama Murthy
  Jul 28 at 23:11
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @KaviRamaMurthy I'm sorry! I forgot to mention the X's are independent
  – Duars
  Jul 28 at 23:14
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  I think Henry's answer below is too complicated. See my answer.
  – Michael Hardy
  Jul 30 at 1:35
  

  
  
  
  

add a comment | 

up vote
1
down vote

favorite

up vote
1
down vote

favorite

Let $X_1, X_2, X_3, X_4$ be independent random variables with $operatornamevar(X_i)=1$, and

$$U = 2X_1+X_2+X_3$$
$$ V = X_2+X_3 + 2X_4$$

Find $operatornamecorr(U, V)$

In general, how can I calculate the correlation between two linear combinations of independent $X_i$ such as $U$ and $V$ knowing only $operatornamevar(X_i)$?

Or what if they weren't independent, but I had their covariance or correlation matrix?

probability statistics random-variables correlation

share|cite|improve this question

edited Jul 30 at 1:21

Michael Hardy

204k23185461

asked Jul 28 at 22:27

Duars

155

Let $X_1, X_2, X_3, X_4$ be independent random variables with $operatornamevar(X_i)=1$, and

$$U = 2X_1+X_2+X_3$$
$$ V = X_2+X_3 + 2X_4$$

Find $operatornamecorr(U, V)$

In general, how can I calculate the correlation between two linear combinations of independent $X_i$ such as $U$ and $V$ knowing only $operatornamevar(X_i)$?

Or what if they weren't independent, but I had their covariance or correlation matrix?

probability statistics random-variables correlation

share|cite|improve this question

edited Jul 30 at 1:21

Michael Hardy

204k23185461

asked Jul 28 at 22:27

Duars

155

share|cite|improve this question

share|cite|improve this question

edited Jul 30 at 1:21

Michael Hardy

204k23185461

edited Jul 30 at 1:21

Michael Hardy

204k23185461

edited Jul 30 at 1:21

Michael Hardy

204k23185461

204k23185461

asked Jul 28 at 22:27

Duars

155

asked Jul 28 at 22:27

Duars

155

asked Jul 28 at 22:27

Duars

155

155

• 
  
  
  

  
  

  
  
  
  
  
  

  
  You can just go to this question. The answer are great.
  – M. Cris
  Jul 28 at 22:30
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  This question has no answer unless you know how $X_i$'s depend on each other.
  – Kavi Rama Murthy
  Jul 28 at 23:11
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @KaviRamaMurthy I'm sorry! I forgot to mention the X's are independent
  – Duars
  Jul 28 at 23:14
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  I think Henry's answer below is too complicated. See my answer.
  – Michael Hardy
  Jul 30 at 1:35
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  

  
  
  
  
  
  

  
  You can just go to this question. The answer are great.
  – M. Cris
  Jul 28 at 22:30
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  This question has no answer unless you know how $X_i$'s depend on each other.
  – Kavi Rama Murthy
  Jul 28 at 23:11
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @KaviRamaMurthy I'm sorry! I forgot to mention the X's are independent
  – Duars
  Jul 28 at 23:14
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  I think Henry's answer below is too complicated. See my answer.
  – Michael Hardy
  Jul 30 at 1:35
  

  
  
  
  

You can just go to this question. The answer are great.
– M. Cris
Jul 28 at 22:30

You can just go to this question. The answer are great.
– M. Cris
Jul 28 at 22:30

This question has no answer unless you know how $X_i$'s depend on each other.
– Kavi Rama Murthy
Jul 28 at 23:11

This question has no answer unless you know how $X_i$'s depend on each other.
– Kavi Rama Murthy
Jul 28 at 23:11

@KaviRamaMurthy I'm sorry! I forgot to mention the X's are independent
– Duars
Jul 28 at 23:14

@KaviRamaMurthy I'm sorry! I forgot to mention the X's are independent
– Duars
Jul 28 at 23:14

I think Henry's answer below is too complicated. See my answer.
– Michael Hardy
Jul 30 at 1:35

I think Henry's answer below is too complicated. See my answer.
– Michael Hardy
Jul 30 at 1:35

add a comment | 

3 Answers
3

active

oldest

votes

up vote
1
down vote

Hints:

• You do not know the means of the $X_i$, but life would be simpler of you assumed they were $0$; if they are not, then consider $X_i-E[X_i]$ instead, with the same variances and covariances




• If the means are $0$ then $operatornamevar(A)= E[A^2]$ and $operatornamecov[A,B]=E[AB]$ and $operatornamecorr(A,B) = fracoperatornamecov[A,B]sqrtoperatornamevar(A)sqrtoperatornamevar(B)$




• If $A$ and $B$ are independent with positive finite variances then $operatornamecov[A,B]=0$ and $operatornamecorr(A,B) = 0$




• $E[nC+mD]=nE[C]+mE[D]$

So finding $operatornamecorr(U,V)$ is just a matter of substitution, multiplying and tidying up

share|cite|improve this answer

edited Jul 30 at 1:24

Michael Hardy

204k23185461

answered Jul 28 at 23:52

Henry

92.8k469147

• 
  
  
  

  
  

  
  
  
  
  
  

  
  I cleaned up your MathJax usage. See my edits for proper usage.
  – Michael Hardy
  Jul 30 at 1:25
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Certainly someone doing problems like this should rely on the bilinearity of covariance.
  – Michael Hardy
  Jul 30 at 1:34
  

  
  
  
  

add a comment | 

up vote
0
down vote

beginalign
& operatornamecov(U,V) \[10pt]
= & operatornamecov(2X_1+X_2+X_3,X_2+X_3 + 2X_4) \[10pt]
= & 2operatornamecov(X_1,, X_2+X_3 + 2X_4) + operatornamecov(X_2,,X_2+X_3 + 2X_4) + operatornamecov(X_3,,X_2+X_3+2X_4)
endalign
I.e. covariance is linear in the first argument. Then for something like
$operatornamecov(X_1,,X_2+X_3+2X_4),$ write
beginalign
& operatornamecov(X_1,,X_2+X_3+2X_4) \[10pt]
= & operatornamecov(X_1,,X_2) + operatornamecov(X_1,,X_3) + 2operatornamecov(X_1,,X_4)
endalign
and so on.

For $operatornamevar (U),$ you have
$$
operatornamevar(2X_1+X_2+X_3) = 2^2operatornamevar(X_1) +operatornamevar(X_2) + operatornamevar(X_3).
$$

share|cite|improve this answer

answered Jul 30 at 1:32

Michael Hardy

204k23185461

add a comment | 

up vote
0
down vote

When $A,B,C$ are pairwise independent, the bilinearity of covariance states:

$$mathsfCov(A+B,B+C)~=mathsf Cov(A,B)+mathsf Cov(A,C)+mathsf Cov(B,B)+mathsf Cov(B,C)\=0+0+mathsfVar(B)+0$$

Just apply this principle to to find the covariance for your $U,V$.

...and of course, as $mathsfVar(A+B)=mathsf Cov(A+B,A+B)$, then the variances of $U,V$ may be found in the same manner.

---

PS: Of course, if the $X_i$ variables are not uncorrelated, then those covariances will not be zero.   However, the bilinearity rule is still usable.

---

share|cite|improve this answer

answered Jul 30 at 1:43

Graham Kemp

80k43275

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%2f2865621%2fcorrelation-between-two-linear-combinations-of-random-variables%23new-answer', 'question_page');

);

Post as a guest

Name Email

3 Answers
3

active

oldest

votes

3 Answers
3

active

oldest

votes

active

oldest

votes

active

oldest

votes

up vote
1
down vote

Hints:

• You do not know the means of the $X_i$, but life would be simpler of you assumed they were $0$; if they are not, then consider $X_i-E[X_i]$ instead, with the same variances and covariances




• If the means are $0$ then $operatornamevar(A)= E[A^2]$ and $operatornamecov[A,B]=E[AB]$ and $operatornamecorr(A,B) = fracoperatornamecov[A,B]sqrtoperatornamevar(A)sqrtoperatornamevar(B)$




• If $A$ and $B$ are independent with positive finite variances then $operatornamecov[A,B]=0$ and $operatornamecorr(A,B) = 0$




• $E[nC+mD]=nE[C]+mE[D]$

So finding $operatornamecorr(U,V)$ is just a matter of substitution, multiplying and tidying up

share|cite|improve this answer

edited Jul 30 at 1:24

Michael Hardy

204k23185461

answered Jul 28 at 23:52

Henry

92.8k469147

• 
  
  
  

  
  

  
  
  
  
  
  

  
  I cleaned up your MathJax usage. See my edits for proper usage.
  – Michael Hardy
  Jul 30 at 1:25
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Certainly someone doing problems like this should rely on the bilinearity of covariance.
  – Michael Hardy
  Jul 30 at 1:34
  

  
  
  
  

add a comment | 

up vote
1
down vote

Hints:

• You do not know the means of the $X_i$, but life would be simpler of you assumed they were $0$; if they are not, then consider $X_i-E[X_i]$ instead, with the same variances and covariances




• If the means are $0$ then $operatornamevar(A)= E[A^2]$ and $operatornamecov[A,B]=E[AB]$ and $operatornamecorr(A,B) = fracoperatornamecov[A,B]sqrtoperatornamevar(A)sqrtoperatornamevar(B)$




• If $A$ and $B$ are independent with positive finite variances then $operatornamecov[A,B]=0$ and $operatornamecorr(A,B) = 0$




• $E[nC+mD]=nE[C]+mE[D]$

So finding $operatornamecorr(U,V)$ is just a matter of substitution, multiplying and tidying up

share|cite|improve this answer

edited Jul 30 at 1:24

Michael Hardy

204k23185461

answered Jul 28 at 23:52

Henry

92.8k469147

• 
  
  
  

  
  

  
  
  
  
  
  

  
  I cleaned up your MathJax usage. See my edits for proper usage.
  – Michael Hardy
  Jul 30 at 1:25
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Certainly someone doing problems like this should rely on the bilinearity of covariance.
  – Michael Hardy
  Jul 30 at 1:34
  

  
  
  
  

add a comment | 

up vote
1
down vote

up vote
1
down vote

Hints:

• You do not know the means of the $X_i$, but life would be simpler of you assumed they were $0$; if they are not, then consider $X_i-E[X_i]$ instead, with the same variances and covariances




• If the means are $0$ then $operatornamevar(A)= E[A^2]$ and $operatornamecov[A,B]=E[AB]$ and $operatornamecorr(A,B) = fracoperatornamecov[A,B]sqrtoperatornamevar(A)sqrtoperatornamevar(B)$




• If $A$ and $B$ are independent with positive finite variances then $operatornamecov[A,B]=0$ and $operatornamecorr(A,B) = 0$




• $E[nC+mD]=nE[C]+mE[D]$

So finding $operatornamecorr(U,V)$ is just a matter of substitution, multiplying and tidying up

share|cite|improve this answer

edited Jul 30 at 1:24

Michael Hardy

204k23185461

answered Jul 28 at 23:52

Henry

92.8k469147

Hints:

• You do not know the means of the $X_i$, but life would be simpler of you assumed they were $0$; if they are not, then consider $X_i-E[X_i]$ instead, with the same variances and covariances




• If the means are $0$ then $operatornamevar(A)= E[A^2]$ and $operatornamecov[A,B]=E[AB]$ and $operatornamecorr(A,B) = fracoperatornamecov[A,B]sqrtoperatornamevar(A)sqrtoperatornamevar(B)$




• If $A$ and $B$ are independent with positive finite variances then $operatornamecov[A,B]=0$ and $operatornamecorr(A,B) = 0$




• $E[nC+mD]=nE[C]+mE[D]$

So finding $operatornamecorr(U,V)$ is just a matter of substitution, multiplying and tidying up

share|cite|improve this answer

edited Jul 30 at 1:24

Michael Hardy

204k23185461

answered Jul 28 at 23:52

Henry

92.8k469147

share|cite|improve this answer

share|cite|improve this answer

edited Jul 30 at 1:24

Michael Hardy

204k23185461

edited Jul 30 at 1:24

Michael Hardy

204k23185461

edited Jul 30 at 1:24

Michael Hardy

204k23185461

204k23185461

answered Jul 28 at 23:52

Henry

92.8k469147

answered Jul 28 at 23:52

Henry

92.8k469147

answered Jul 28 at 23:52

Henry

92.8k469147

92.8k469147

• 
  
  
  

  
  

  
  
  
  
  
  

  
  I cleaned up your MathJax usage. See my edits for proper usage.
  – Michael Hardy
  Jul 30 at 1:25
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Certainly someone doing problems like this should rely on the bilinearity of covariance.
  – Michael Hardy
  Jul 30 at 1:34
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  

  
  
  
  
  
  

  
  I cleaned up your MathJax usage. See my edits for proper usage.
  – Michael Hardy
  Jul 30 at 1:25
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Certainly someone doing problems like this should rely on the bilinearity of covariance.
  – Michael Hardy
  Jul 30 at 1:34
  

  
  
  
  

I cleaned up your MathJax usage. See my edits for proper usage.
– Michael Hardy
Jul 30 at 1:25

I cleaned up your MathJax usage. See my edits for proper usage.
– Michael Hardy
Jul 30 at 1:25

Certainly someone doing problems like this should rely on the bilinearity of covariance.
– Michael Hardy
Jul 30 at 1:34

Certainly someone doing problems like this should rely on the bilinearity of covariance.
– Michael Hardy
Jul 30 at 1:34

add a comment | 

up vote
0
down vote

beginalign
& operatornamecov(U,V) \[10pt]
= & operatornamecov(2X_1+X_2+X_3,X_2+X_3 + 2X_4) \[10pt]
= & 2operatornamecov(X_1,, X_2+X_3 + 2X_4) + operatornamecov(X_2,,X_2+X_3 + 2X_4) + operatornamecov(X_3,,X_2+X_3+2X_4)
endalign
I.e. covariance is linear in the first argument. Then for something like
$operatornamecov(X_1,,X_2+X_3+2X_4),$ write
beginalign
& operatornamecov(X_1,,X_2+X_3+2X_4) \[10pt]
= & operatornamecov(X_1,,X_2) + operatornamecov(X_1,,X_3) + 2operatornamecov(X_1,,X_4)
endalign
and so on.

For $operatornamevar (U),$ you have
$$
operatornamevar(2X_1+X_2+X_3) = 2^2operatornamevar(X_1) +operatornamevar(X_2) + operatornamevar(X_3).
$$

share|cite|improve this answer

answered Jul 30 at 1:32

Michael Hardy

204k23185461

add a comment | 

up vote
0
down vote

beginalign
& operatornamecov(U,V) \[10pt]
= & operatornamecov(2X_1+X_2+X_3,X_2+X_3 + 2X_4) \[10pt]
= & 2operatornamecov(X_1,, X_2+X_3 + 2X_4) + operatornamecov(X_2,,X_2+X_3 + 2X_4) + operatornamecov(X_3,,X_2+X_3+2X_4)
endalign
I.e. covariance is linear in the first argument. Then for something like
$operatornamecov(X_1,,X_2+X_3+2X_4),$ write
beginalign
& operatornamecov(X_1,,X_2+X_3+2X_4) \[10pt]
= & operatornamecov(X_1,,X_2) + operatornamecov(X_1,,X_3) + 2operatornamecov(X_1,,X_4)
endalign
and so on.

For $operatornamevar (U),$ you have
$$
operatornamevar(2X_1+X_2+X_3) = 2^2operatornamevar(X_1) +operatornamevar(X_2) + operatornamevar(X_3).
$$

share|cite|improve this answer

answered Jul 30 at 1:32

Michael Hardy

204k23185461

add a comment | 

up vote
0
down vote

up vote
0
down vote

beginalign
& operatornamecov(U,V) \[10pt]
= & operatornamecov(2X_1+X_2+X_3,X_2+X_3 + 2X_4) \[10pt]
= & 2operatornamecov(X_1,, X_2+X_3 + 2X_4) + operatornamecov(X_2,,X_2+X_3 + 2X_4) + operatornamecov(X_3,,X_2+X_3+2X_4)
endalign
I.e. covariance is linear in the first argument. Then for something like
$operatornamecov(X_1,,X_2+X_3+2X_4),$ write
beginalign
& operatornamecov(X_1,,X_2+X_3+2X_4) \[10pt]
= & operatornamecov(X_1,,X_2) + operatornamecov(X_1,,X_3) + 2operatornamecov(X_1,,X_4)
endalign
and so on.

For $operatornamevar (U),$ you have
$$
operatornamevar(2X_1+X_2+X_3) = 2^2operatornamevar(X_1) +operatornamevar(X_2) + operatornamevar(X_3).
$$

share|cite|improve this answer

answered Jul 30 at 1:32

Michael Hardy

204k23185461

beginalign
& operatornamecov(U,V) \[10pt]
= & operatornamecov(2X_1+X_2+X_3,X_2+X_3 + 2X_4) \[10pt]
= & 2operatornamecov(X_1,, X_2+X_3 + 2X_4) + operatornamecov(X_2,,X_2+X_3 + 2X_4) + operatornamecov(X_3,,X_2+X_3+2X_4)
endalign
I.e. covariance is linear in the first argument. Then for something like
$operatornamecov(X_1,,X_2+X_3+2X_4),$ write
beginalign
& operatornamecov(X_1,,X_2+X_3+2X_4) \[10pt]
= & operatornamecov(X_1,,X_2) + operatornamecov(X_1,,X_3) + 2operatornamecov(X_1,,X_4)
endalign
and so on.

For $operatornamevar (U),$ you have
$$
operatornamevar(2X_1+X_2+X_3) = 2^2operatornamevar(X_1) +operatornamevar(X_2) + operatornamevar(X_3).
$$

share|cite|improve this answer

answered Jul 30 at 1:32

Michael Hardy

204k23185461

share|cite|improve this answer

share|cite|improve this answer

answered Jul 30 at 1:32

Michael Hardy

204k23185461

answered Jul 30 at 1:32

Michael Hardy

204k23185461

answered Jul 30 at 1:32

Michael Hardy

204k23185461

204k23185461

add a comment | 

add a comment | 

up vote
0
down vote

When $A,B,C$ are pairwise independent, the bilinearity of covariance states:

$$mathsfCov(A+B,B+C)~=mathsf Cov(A,B)+mathsf Cov(A,C)+mathsf Cov(B,B)+mathsf Cov(B,C)\=0+0+mathsfVar(B)+0$$

Just apply this principle to to find the covariance for your $U,V$.

...and of course, as $mathsfVar(A+B)=mathsf Cov(A+B,A+B)$, then the variances of $U,V$ may be found in the same manner.

---

PS: Of course, if the $X_i$ variables are not uncorrelated, then those covariances will not be zero.   However, the bilinearity rule is still usable.

---

share|cite|improve this answer

answered Jul 30 at 1:43

Graham Kemp

80k43275

add a comment | 

up vote
0
down vote

When $A,B,C$ are pairwise independent, the bilinearity of covariance states:

$$mathsfCov(A+B,B+C)~=mathsf Cov(A,B)+mathsf Cov(A,C)+mathsf Cov(B,B)+mathsf Cov(B,C)\=0+0+mathsfVar(B)+0$$

Just apply this principle to to find the covariance for your $U,V$.

...and of course, as $mathsfVar(A+B)=mathsf Cov(A+B,A+B)$, then the variances of $U,V$ may be found in the same manner.

---

PS: Of course, if the $X_i$ variables are not uncorrelated, then those covariances will not be zero.   However, the bilinearity rule is still usable.

---

share|cite|improve this answer

answered Jul 30 at 1:43

Graham Kemp

80k43275

add a comment | 

up vote
0
down vote

up vote
0
down vote

When $A,B,C$ are pairwise independent, the bilinearity of covariance states:

$$mathsfCov(A+B,B+C)~=mathsf Cov(A,B)+mathsf Cov(A,C)+mathsf Cov(B,B)+mathsf Cov(B,C)\=0+0+mathsfVar(B)+0$$

Just apply this principle to to find the covariance for your $U,V$.

...and of course, as $mathsfVar(A+B)=mathsf Cov(A+B,A+B)$, then the variances of $U,V$ may be found in the same manner.

---

PS: Of course, if the $X_i$ variables are not uncorrelated, then those covariances will not be zero.   However, the bilinearity rule is still usable.

---

share|cite|improve this answer

answered Jul 30 at 1:43

Graham Kemp

80k43275

When $A,B,C$ are pairwise independent, the bilinearity of covariance states:

$$mathsfCov(A+B,B+C)~=mathsf Cov(A,B)+mathsf Cov(A,C)+mathsf Cov(B,B)+mathsf Cov(B,C)\=0+0+mathsfVar(B)+0$$

Just apply this principle to to find the covariance for your $U,V$.

...and of course, as $mathsfVar(A+B)=mathsf Cov(A+B,A+B)$, then the variances of $U,V$ may be found in the same manner.

---

PS: Of course, if the $X_i$ variables are not uncorrelated, then those covariances will not be zero.   However, the bilinearity rule is still usable.

---

share|cite|improve this answer

answered Jul 30 at 1:43

Graham Kemp

80k43275

share|cite|improve this answer

share|cite|improve this answer

answered Jul 30 at 1:43

Graham Kemp

80k43275

answered Jul 30 at 1:43

Graham Kemp

80k43275

answered Jul 30 at 1:43

Graham Kemp

80k43275

80k43275

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%2f2865621%2fcorrelation-between-two-linear-combinations-of-random-variables%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