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calculus with sum sign - Pollak habit formation

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0

$begingroup$

i have no clue what is happening regarding the sum sign in the following:

$$ m- fracp_ia_i(x_i-b_i) sum_j=1^n a_j+p_jb_j=0 $$

$$ with sum_j=1^n a_j=1:$$

$$ fracp_ia_i(x_i-b_i) +sum_j=1^n p_jb_j=m$$

Could someone make me a bit smarter, please?
How gets the sum sign in front of $pb$ and why does it change the sign? what are the steps in between or how is this rule called? And why is there still the sum sign when the sum of aj eqauls one?

Thanks in advance.

Momo

calculus

share|cite|improve this question

edited Mar 11 at 12:01

YuiTo Cheng

2,0592637

asked Mar 11 at 11:52

PetePete

1

New contributor

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

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  You can separate the sum $sum (a_j + p_j b_j) = sum a_j + sum p_j b_j$. Now use the fact that $sum a_j = 1$. The change of sign is just due to the fact that a member of the equations has been moved to the other side of the equality.
  $endgroup$
  – TheSilverDoe
  Mar 11 at 12:04
  

  
  
  
  

add a comment | 

0

$begingroup$

i have no clue what is happening regarding the sum sign in the following:

$$ m- fracp_ia_i(x_i-b_i) sum_j=1^n a_j+p_jb_j=0 $$

$$ with sum_j=1^n a_j=1:$$

$$ fracp_ia_i(x_i-b_i) +sum_j=1^n p_jb_j=m$$

Could someone make me a bit smarter, please?
How gets the sum sign in front of $pb$ and why does it change the sign? what are the steps in between or how is this rule called? And why is there still the sum sign when the sum of aj eqauls one?

Thanks in advance.

Momo

calculus

share|cite|improve this question

edited Mar 11 at 12:01

YuiTo Cheng

2,0592637

asked Mar 11 at 11:52

PetePete

1

New contributor

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

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  You can separate the sum $sum (a_j + p_j b_j) = sum a_j + sum p_j b_j$. Now use the fact that $sum a_j = 1$. The change of sign is just due to the fact that a member of the equations has been moved to the other side of the equality.
  $endgroup$
  – TheSilverDoe
  Mar 11 at 12:04
  

  
  
  
  

add a comment | 

$begingroup$

i have no clue what is happening regarding the sum sign in the following:

$$ m- fracp_ia_i(x_i-b_i) sum_j=1^n a_j+p_jb_j=0 $$

$$ with sum_j=1^n a_j=1:$$

$$ fracp_ia_i(x_i-b_i) +sum_j=1^n p_jb_j=m$$

Could someone make me a bit smarter, please?
How gets the sum sign in front of $pb$ and why does it change the sign? what are the steps in between or how is this rule called? And why is there still the sum sign when the sum of aj eqauls one?

Thanks in advance.

Momo

calculus

share|cite|improve this question

edited Mar 11 at 12:01

YuiTo Cheng

2,0592637

asked Mar 11 at 11:52

PetePete

1

New contributor

Pete 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 11 at 12:01

YuiTo Cheng

2,0592637

asked Mar 11 at 11:52

PetePete

1

New contributor

Pete 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 11 at 12:01

YuiTo Cheng

2,0592637

asked Mar 11 at 11:52

PetePete

1

New contributor

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

edited Mar 11 at 12:01

YuiTo Cheng

2,0592637

edited Mar 11 at 12:01

YuiTo Cheng

2,0592637

asked Mar 11 at 11:52

PetePete

1

New contributor

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

asked Mar 11 at 11:52

PetePete

1