Finding variance using the matrix method [closed]

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If the random variables $X,Y$ and $Z$ have means: $$barx=2 ;;; bary=-3 ;;; barz=4$$ and variances $$operatornameVar(x)=1 ;;; operatornameVar(Y)=5 ;;; operatornameVar(Z)=2$$ and
co-variance $$operatornameCov(X,Y)= -2;;; operatornameCov(X,Z)=-1;;;operatornameCov(Y,Z)=1$$ Find

1. Mean of $W=3X-Y+2Z$


2. The variance of $W=3X-Y+2Z$ using the matrix method
   

statistics covariance variance

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edited Jul 30 at 16:42

Davide Morgante

1,693220

asked Jul 30 at 16:17

Eric kioko

62

closed as off-topic by StubbornAtom, Mostafa Ayaz, Strants, amWhy, Key Flex Jul 31 at 0:37

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – StubbornAtom, Mostafa Ayaz, Strants, amWhy, Key Flex

If this question can be reworded to fit the rules in the help center, please edit the question.

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Welcome to MSE. For a proficient interaction, please take a few minutes to learn MathJax: math.meta.stackexchange.com/questions/5020/…. Also providing the attempt that you've made would be much appreciated.
  – xbh
  Jul 30 at 16:25
  

  
  
  
  

add a comment | 

up vote
-1
down vote

favorite

1

If the random variables $X,Y$ and $Z$ have means: $$barx=2 ;;; bary=-3 ;;; barz=4$$ and variances $$operatornameVar(x)=1 ;;; operatornameVar(Y)=5 ;;; operatornameVar(Z)=2$$ and
co-variance $$operatornameCov(X,Y)= -2;;; operatornameCov(X,Z)=-1;;;operatornameCov(Y,Z)=1$$ Find

1. Mean of $W=3X-Y+2Z$


2. The variance of $W=3X-Y+2Z$ using the matrix method
   

statistics covariance variance

share|cite|improve this question

edited Jul 30 at 16:42

Davide Morgante

1,693220

asked Jul 30 at 16:17

Eric kioko

62

closed as off-topic by StubbornAtom, Mostafa Ayaz, Strants, amWhy, Key Flex Jul 31 at 0:37

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – StubbornAtom, Mostafa Ayaz, Strants, amWhy, Key Flex

If this question can be reworded to fit the rules in the help center, please edit the question.

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Welcome to MSE. For a proficient interaction, please take a few minutes to learn MathJax: math.meta.stackexchange.com/questions/5020/…. Also providing the attempt that you've made would be much appreciated.
  – xbh
  Jul 30 at 16:25
  

  
  
  
  

add a comment | 

up vote
-1
down vote

favorite

1

up vote
-1
down vote

favorite

1

1

If the random variables $X,Y$ and $Z$ have means: $$barx=2 ;;; bary=-3 ;;; barz=4$$ and variances $$operatornameVar(x)=1 ;;; operatornameVar(Y)=5 ;;; operatornameVar(Z)=2$$ and
co-variance $$operatornameCov(X,Y)= -2;;; operatornameCov(X,Z)=-1;;;operatornameCov(Y,Z)=1$$ Find

1. Mean of $W=3X-Y+2Z$


2. The variance of $W=3X-Y+2Z$ using the matrix method
   

statistics covariance variance

share|cite|improve this question

edited Jul 30 at 16:42

Davide Morgante

1,693220

asked Jul 30 at 16:17

Eric kioko

62

If the random variables $X,Y$ and $Z$ have means: $$barx=2 ;;; bary=-3 ;;; barz=4$$ and variances $$operatornameVar(x)=1 ;;; operatornameVar(Y)=5 ;;; operatornameVar(Z)=2$$ and
co-variance $$operatornameCov(X,Y)= -2;;; operatornameCov(X,Z)=-1;;;operatornameCov(Y,Z)=1$$ Find

1. Mean of $W=3X-Y+2Z$


2. The variance of $W=3X-Y+2Z$ using the matrix method
   

statistics covariance variance

share|cite|improve this question

edited Jul 30 at 16:42

Davide Morgante

1,693220

asked Jul 30 at 16:17

Eric kioko

62

share|cite|improve this question

share|cite|improve this question

edited Jul 30 at 16:42

Davide Morgante

1,693220

edited Jul 30 at 16:42

Davide Morgante

1,693220

edited Jul 30 at 16:42

Davide Morgante

1,693220

1,693220

asked Jul 30 at 16:17

Eric kioko

62

asked Jul 30 at 16:17

Eric kioko

62

asked Jul 30 at 16:17

Eric kioko

62

62

closed as off-topic by StubbornAtom, Mostafa Ayaz, Strants, amWhy, Key Flex Jul 31 at 0:37

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – StubbornAtom, Mostafa Ayaz, Strants, amWhy, Key Flex

If this question can be reworded to fit the rules in the help center, please edit the question.

closed as off-topic by StubbornAtom, Mostafa Ayaz, Strants, amWhy, Key Flex Jul 31 at 0:37

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – StubbornAtom, Mostafa Ayaz, Strants, amWhy, Key Flex

If this question can be reworded to fit the rules in the help center, please edit the question.

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Welcome to MSE. For a proficient interaction, please take a few minutes to learn MathJax: math.meta.stackexchange.com/questions/5020/…. Also providing the attempt that you've made would be much appreciated.
  – xbh
  Jul 30 at 16:25
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Welcome to MSE. For a proficient interaction, please take a few minutes to learn MathJax: math.meta.stackexchange.com/questions/5020/…. Also providing the attempt that you've made would be much appreciated.
  – xbh
  Jul 30 at 16:25
  

  
  
  
  

Welcome to MSE. For a proficient interaction, please take a few minutes to learn MathJax: math.meta.stackexchange.com/questions/5020/…. Also providing the attempt that you've made would be much appreciated.
– xbh
Jul 30 at 16:25

Welcome to MSE. For a proficient interaction, please take a few minutes to learn MathJax: math.meta.stackexchange.com/questions/5020/…. Also providing the attempt that you've made would be much appreciated.
– xbh
Jul 30 at 16:25

add a comment | 

1 Answer
1

active

oldest

votes

up vote
1
down vote



accepted

Hint

$$E[aX+b] = aE[X]+b$$
$$operatornameVar[aX+bY] = mathbfv^TmathbfMmathbfv$$ where $mathbfv = left(beginmatrixa\bendmatrixright)$ and $mathbfM$ is the matrix of covariance defined (in the case two variables) as $$left(beginmatrixoperatornameVar[X]&operatornameCov[X,Y]\operatornameCov[Y,X]&operatornameVar[Y]endmatrixright)$$

Solution

The fist question is the easiest, from the linearity of the mean you can simply add together the means with their respective coefficientent given by the formula for $W$, mainly $$E[W] = E[3X-Y+2Z] = 3E[X]-E[Y]+2E[Z] = 3(2)-(-3)+2(4) = 17$$ The second, given the formula above is to be solved by simple matrix multiplication. In our case $$mathbfv = left(beginmatrix3\-1\2endmatrixright);;;mathbfM=left(beginmatrixoperatornameVar[X]&operatornameCov[X,Y]&operatornameCov[X,Z]\operatornameCov[Y,X]&operatornameVar[Y]&operatornameCov[Y,Z]\operatornameCov[Z,X]&operatornameCov[Z,Y]&operatornameVar[Z]endmatrixright) = left(beginmatrix1&-2&-1\-2&5&1\-1&1&2endmatrixright)$$ Then we calculate the variance of $W$ as $$operatornameVar[W] = operatornameVar[3X-Y+2Z] = underbraceleft(beginmatrix3&-1&2endmatrixright)left(beginmatrix1&-2&-1\-2&5&1\-1&1&2endmatrixright)_textfirst multiplicationleft(beginmatrix3\-1\2endmatrixright)$$ let's do first the underbraced multiplication $$left(beginmatrix3&-1&2endmatrixright)left(beginmatrix1&-2&-1\-2&5&1\-1&1&2endmatrixright) = left(beginmatrix3&-9&6endmatrixright)$$ and now let's do the second multiplication $$left(beginmatrix3&-9&6endmatrixright)left(beginmatrix3\-1\2endmatrixright)=30$$ So in the end we have that $$operatornameVar[W]=30$$

share|cite|improve this answer

edited Jul 30 at 19:29

answered Jul 30 at 16:41

Davide Morgante

1,693220

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Please note that $mathbb E(aX+b)=acdot mathbb E(X)+colorredb$
  – callculus
  Jul 30 at 17:08
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @callculus Ops, my bad! Thank you
  – Davide Morgante
  Jul 30 at 17:09
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Is there any difference between Cov(Y,X) and Cov (X,Y) ?@DavideMorgante
  – Eric kioko
  Jul 30 at 18:30
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  No, covariance is symmetric! $$Cov[X,Y]=Cov[Y,X]$$
  – Davide Morgante
  Jul 30 at 18:31
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @DavideMorgante Thanks. So from the hint above I'm supposed to get the determinant of the matrix or? Kindly shed some light I'm new to the matrix method.
  – Eric kioko
  Jul 30 at 18:47
  

  
  
  
  

 | 
show 4 more comments

1 Answer
1

active

oldest

votes

1 Answer
1

active

oldest

votes

active

oldest

votes

active

oldest

votes

up vote
1
down vote



accepted

Hint

$$E[aX+b] = aE[X]+b$$
$$operatornameVar[aX+bY] = mathbfv^TmathbfMmathbfv$$ where $mathbfv = left(beginmatrixa\bendmatrixright)$ and $mathbfM$ is the matrix of covariance defined (in the case two variables) as $$left(beginmatrixoperatornameVar[X]&operatornameCov[X,Y]\operatornameCov[Y,X]&operatornameVar[Y]endmatrixright)$$

Solution

The fist question is the easiest, from the linearity of the mean you can simply add together the means with their respective coefficientent given by the formula for $W$, mainly $$E[W] = E[3X-Y+2Z] = 3E[X]-E[Y]+2E[Z] = 3(2)-(-3)+2(4) = 17$$ The second, given the formula above is to be solved by simple matrix multiplication. In our case $$mathbfv = left(beginmatrix3\-1\2endmatrixright);;;mathbfM=left(beginmatrixoperatornameVar[X]&operatornameCov[X,Y]&operatornameCov[X,Z]\operatornameCov[Y,X]&operatornameVar[Y]&operatornameCov[Y,Z]\operatornameCov[Z,X]&operatornameCov[Z,Y]&operatornameVar[Z]endmatrixright) = left(beginmatrix1&-2&-1\-2&5&1\-1&1&2endmatrixright)$$ Then we calculate the variance of $W$ as $$operatornameVar[W] = operatornameVar[3X-Y+2Z] = underbraceleft(beginmatrix3&-1&2endmatrixright)left(beginmatrix1&-2&-1\-2&5&1\-1&1&2endmatrixright)_textfirst multiplicationleft(beginmatrix3\-1\2endmatrixright)$$ let's do first the underbraced multiplication $$left(beginmatrix3&-1&2endmatrixright)left(beginmatrix1&-2&-1\-2&5&1\-1&1&2endmatrixright) = left(beginmatrix3&-9&6endmatrixright)$$ and now let's do the second multiplication $$left(beginmatrix3&-9&6endmatrixright)left(beginmatrix3\-1\2endmatrixright)=30$$ So in the end we have that $$operatornameVar[W]=30$$

share|cite|improve this answer

edited Jul 30 at 19:29

answered Jul 30 at 16:41

Davide Morgante

1,693220

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Please note that $mathbb E(aX+b)=acdot mathbb E(X)+colorredb$
  – callculus
  Jul 30 at 17:08
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @callculus Ops, my bad! Thank you
  – Davide Morgante
  Jul 30 at 17:09
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Is there any difference between Cov(Y,X) and Cov (X,Y) ?@DavideMorgante
  – Eric kioko
  Jul 30 at 18:30
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  No, covariance is symmetric! $$Cov[X,Y]=Cov[Y,X]$$
  – Davide Morgante
  Jul 30 at 18:31
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @DavideMorgante Thanks. So from the hint above I'm supposed to get the determinant of the matrix or? Kindly shed some light I'm new to the matrix method.
  – Eric kioko
  Jul 30 at 18:47
  

  
  
  
  

 | 
show 4 more comments

up vote
1
down vote



accepted

Hint

$$E[aX+b] = aE[X]+b$$
$$operatornameVar[aX+bY] = mathbfv^TmathbfMmathbfv$$ where $mathbfv = left(beginmatrixa\bendmatrixright)$ and $mathbfM$ is the matrix of covariance defined (in the case two variables) as $$left(beginmatrixoperatornameVar[X]&operatornameCov[X,Y]\operatornameCov[Y,X]&operatornameVar[Y]endmatrixright)$$

Solution

The fist question is the easiest, from the linearity of the mean you can simply add together the means with their respective coefficientent given by the formula for $W$, mainly $$E[W] = E[3X-Y+2Z] = 3E[X]-E[Y]+2E[Z] = 3(2)-(-3)+2(4) = 17$$ The second, given the formula above is to be solved by simple matrix multiplication. In our case $$mathbfv = left(beginmatrix3\-1\2endmatrixright);;;mathbfM=left(beginmatrixoperatornameVar[X]&operatornameCov[X,Y]&operatornameCov[X,Z]\operatornameCov[Y,X]&operatornameVar[Y]&operatornameCov[Y,Z]\operatornameCov[Z,X]&operatornameCov[Z,Y]&operatornameVar[Z]endmatrixright) = left(beginmatrix1&-2&-1\-2&5&1\-1&1&2endmatrixright)$$ Then we calculate the variance of $W$ as $$operatornameVar[W] = operatornameVar[3X-Y+2Z] = underbraceleft(beginmatrix3&-1&2endmatrixright)left(beginmatrix1&-2&-1\-2&5&1\-1&1&2endmatrixright)_textfirst multiplicationleft(beginmatrix3\-1\2endmatrixright)$$ let's do first the underbraced multiplication $$left(beginmatrix3&-1&2endmatrixright)left(beginmatrix1&-2&-1\-2&5&1\-1&1&2endmatrixright) = left(beginmatrix3&-9&6endmatrixright)$$ and now let's do the second multiplication $$left(beginmatrix3&-9&6endmatrixright)left(beginmatrix3\-1\2endmatrixright)=30$$ So in the end we have that $$operatornameVar[W]=30$$

share|cite|improve this answer

edited Jul 30 at 19:29

answered Jul 30 at 16:41

Davide Morgante

1,693220

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Please note that $mathbb E(aX+b)=acdot mathbb E(X)+colorredb$
  – callculus
  Jul 30 at 17:08
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @callculus Ops, my bad! Thank you
  – Davide Morgante
  Jul 30 at 17:09
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Is there any difference between Cov(Y,X) and Cov (X,Y) ?@DavideMorgante
  – Eric kioko
  Jul 30 at 18:30
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  No, covariance is symmetric! $$Cov[X,Y]=Cov[Y,X]$$
  – Davide Morgante
  Jul 30 at 18:31
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @DavideMorgante Thanks. So from the hint above I'm supposed to get the determinant of the matrix or? Kindly shed some light I'm new to the matrix method.
  – Eric kioko
  Jul 30 at 18:47
  

  
  
  
  

 | 
show 4 more comments

up vote
1
down vote



accepted

up vote
1
down vote



accepted

Hint

$$E[aX+b] = aE[X]+b$$
$$operatornameVar[aX+bY] = mathbfv^TmathbfMmathbfv$$ where $mathbfv = left(beginmatrixa\bendmatrixright)$ and $mathbfM$ is the matrix of covariance defined (in the case two variables) as $$left(beginmatrixoperatornameVar[X]&operatornameCov[X,Y]\operatornameCov[Y,X]&operatornameVar[Y]endmatrixright)$$

Solution

The fist question is the easiest, from the linearity of the mean you can simply add together the means with their respective coefficientent given by the formula for $W$, mainly $$E[W] = E[3X-Y+2Z] = 3E[X]-E[Y]+2E[Z] = 3(2)-(-3)+2(4) = 17$$ The second, given the formula above is to be solved by simple matrix multiplication. In our case $$mathbfv = left(beginmatrix3\-1\2endmatrixright);;;mathbfM=left(beginmatrixoperatornameVar[X]&operatornameCov[X,Y]&operatornameCov[X,Z]\operatornameCov[Y,X]&operatornameVar[Y]&operatornameCov[Y,Z]\operatornameCov[Z,X]&operatornameCov[Z,Y]&operatornameVar[Z]endmatrixright) = left(beginmatrix1&-2&-1\-2&5&1\-1&1&2endmatrixright)$$ Then we calculate the variance of $W$ as $$operatornameVar[W] = operatornameVar[3X-Y+2Z] = underbraceleft(beginmatrix3&-1&2endmatrixright)left(beginmatrix1&-2&-1\-2&5&1\-1&1&2endmatrixright)_textfirst multiplicationleft(beginmatrix3\-1\2endmatrixright)$$ let's do first the underbraced multiplication $$left(beginmatrix3&-1&2endmatrixright)left(beginmatrix1&-2&-1\-2&5&1\-1&1&2endmatrixright) = left(beginmatrix3&-9&6endmatrixright)$$ and now let's do the second multiplication $$left(beginmatrix3&-9&6endmatrixright)left(beginmatrix3\-1\2endmatrixright)=30$$ So in the end we have that $$operatornameVar[W]=30$$

share|cite|improve this answer

edited Jul 30 at 19:29

answered Jul 30 at 16:41

Davide Morgante

1,693220

Hint

$$E[aX+b] = aE[X]+b$$
$$operatornameVar[aX+bY] = mathbfv^TmathbfMmathbfv$$ where $mathbfv = left(beginmatrixa\bendmatrixright)$ and $mathbfM$ is the matrix of covariance defined (in the case two variables) as $$left(beginmatrixoperatornameVar[X]&operatornameCov[X,Y]\operatornameCov[Y,X]&operatornameVar[Y]endmatrixright)$$

Solution

The fist question is the easiest, from the linearity of the mean you can simply add together the means with their respective coefficientent given by the formula for $W$, mainly $$E[W] = E[3X-Y+2Z] = 3E[X]-E[Y]+2E[Z] = 3(2)-(-3)+2(4) = 17$$ The second, given the formula above is to be solved by simple matrix multiplication. In our case $$mathbfv = left(beginmatrix3\-1\2endmatrixright);;;mathbfM=left(beginmatrixoperatornameVar[X]&operatornameCov[X,Y]&operatornameCov[X,Z]\operatornameCov[Y,X]&operatornameVar[Y]&operatornameCov[Y,Z]\operatornameCov[Z,X]&operatornameCov[Z,Y]&operatornameVar[Z]endmatrixright) = left(beginmatrix1&-2&-1\-2&5&1\-1&1&2endmatrixright)$$ Then we calculate the variance of $W$ as $$operatornameVar[W] = operatornameVar[3X-Y+2Z] = underbraceleft(beginmatrix3&-1&2endmatrixright)left(beginmatrix1&-2&-1\-2&5&1\-1&1&2endmatrixright)_textfirst multiplicationleft(beginmatrix3\-1\2endmatrixright)$$ let's do first the underbraced multiplication $$left(beginmatrix3&-1&2endmatrixright)left(beginmatrix1&-2&-1\-2&5&1\-1&1&2endmatrixright) = left(beginmatrix3&-9&6endmatrixright)$$ and now let's do the second multiplication $$left(beginmatrix3&-9&6endmatrixright)left(beginmatrix3\-1\2endmatrixright)=30$$ So in the end we have that $$operatornameVar[W]=30$$

share|cite|improve this answer

edited Jul 30 at 19:29

answered Jul 30 at 16:41

Davide Morgante

1,693220

share|cite|improve this answer

share|cite|improve this answer

edited Jul 30 at 19:29

edited Jul 30 at 19:29

edited Jul 30 at 19:29

answered Jul 30 at 16:41

Davide Morgante

1,693220

answered Jul 30 at 16:41

Davide Morgante

1,693220

answered Jul 30 at 16:41

Davide Morgante

1,693220

1,693220

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Please note that $mathbb E(aX+b)=acdot mathbb E(X)+colorredb$
  – callculus
  Jul 30 at 17:08
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @callculus Ops, my bad! Thank you
  – Davide Morgante
  Jul 30 at 17:09
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Is there any difference between Cov(Y,X) and Cov (X,Y) ?@DavideMorgante
  – Eric kioko
  Jul 30 at 18:30
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  No, covariance is symmetric! $$Cov[X,Y]=Cov[Y,X]$$
  – Davide Morgante
  Jul 30 at 18:31
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @DavideMorgante Thanks. So from the hint above I'm supposed to get the determinant of the matrix or? Kindly shed some light I'm new to the matrix method.
  – Eric kioko
  Jul 30 at 18:47
  

  
  
  
  

 | 
show 4 more comments

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Please note that $mathbb E(aX+b)=acdot mathbb E(X)+colorredb$
  – callculus
  Jul 30 at 17:08
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @callculus Ops, my bad! Thank you
  – Davide Morgante
  Jul 30 at 17:09
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Is there any difference between Cov(Y,X) and Cov (X,Y) ?@DavideMorgante
  – Eric kioko
  Jul 30 at 18:30
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  No, covariance is symmetric! $$Cov[X,Y]=Cov[Y,X]$$
  – Davide Morgante
  Jul 30 at 18:31
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @DavideMorgante Thanks. So from the hint above I'm supposed to get the determinant of the matrix or? Kindly shed some light I'm new to the matrix method.
  – Eric kioko
  Jul 30 at 18:47
  

  
  
  
  

Please note that $mathbb E(aX+b)=acdot mathbb E(X)+colorredb$
– callculus
Jul 30 at 17:08

Please note that $mathbb E(aX+b)=acdot mathbb E(X)+colorredb$
– callculus
Jul 30 at 17:08

@callculus Ops, my bad! Thank you
– Davide Morgante
Jul 30 at 17:09

@callculus Ops, my bad! Thank you
– Davide Morgante
Jul 30 at 17:09

Is there any difference between Cov(Y,X) and Cov (X,Y) ?@DavideMorgante
– Eric kioko
Jul 30 at 18:30

Is there any difference between Cov(Y,X) and Cov (X,Y) ?@DavideMorgante
– Eric kioko
Jul 30 at 18:30

No, covariance is symmetric! $$Cov[X,Y]=Cov[Y,X]$$
– Davide Morgante
Jul 30 at 18:31

No, covariance is symmetric! $$Cov[X,Y]=Cov[Y,X]$$
– Davide Morgante
Jul 30 at 18:31

@DavideMorgante Thanks. So from the hint above I'm supposed to get the determinant of the matrix or? Kindly shed some light I'm new to the matrix method.
– Eric kioko
Jul 30 at 18:47

@DavideMorgante Thanks. So from the hint above I'm supposed to get the determinant of the matrix or? Kindly shed some light I'm new to the matrix method.
– Eric kioko
Jul 30 at 18:47

 | 
show 4 more comments