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Ridiculous manipulation question

  • #1
ECONOMETRICS CALCULUS.


1. The PRECEDING question and solution

Differentiate
351tuzq.jpg
with respect to b1

My solution:
df(b0,b1)/db1 = 2Σ[(yi-b0-b1xi)*(-xi)]

2. the ACTUAL question

By setting d,f(b0,b1)/d,b1 = 0, show that you obtain
qqadf4.jpg
.
(Hint: you will have to substitute b0 = yBAR - b1(xBAR), and use the results from the previous question.)

3. Some properties that we can assume while answering the question.

ouxr1x.jpg

eapesp.jpg

35d48wl.jpg


4. The attempt at a solution (I can't do it)

Letting 2Σ[(yi-b0-b1xi)*(-xi)] = 0
0= Σ[(yi-b0-b1xi)*(-xi)]
0= Σ(-yi xi)+Σb0xi+ Σb1xi ^2
0= -Σ (yi xi)+b0 Σxi+b1Σxi^2
b1Σxi^2=Σyi Σxi – (yBAR-b1xBAR)Σxi ............... (Re-arranging and substituting for b0)
b1Σxi=Σyi -yBAR+b1xBAR ............................ (dividing both sides by Σxi)
b1Σxi – b1xb = Σyi – yBAR
b1(Σxi – xb)=Σyi – yBAR
b1= (Σyi – yb)/(Σxi – xBAR)
= (Σyi – Σyi /n)(Σxi – Σxi/n)
 
Last edited:

Answers and Replies

  • #2
33,632
5,287
ECONOMETRICS CALCULUS.


1. The PRECEDING question and solution

Differentiate
351tuzq.jpg
with respect to b1

My solution:
df(b0,b1)/db1 = 2Σ[(yi-b0-b1xi)*(-xi)]

2. the ACTUAL question

By setting d,f(b0,b1)/d,b1 = 0, show that you obtain
qqadf4.jpg
.
(Hint: you will have to substitute b0 = yBAR - b1(xBAR), and use the results from the previous question.)

3. Some properties that we can assume while answering the question.

ouxr1x.jpg

eapesp.jpg

35d48wl.jpg


4. The attempt at a solution (I can't do it)

Letting 2Σ[(yi-b0-b1xi)*(-xi)] = 0
= Σ[(yi-b0-b1xi)*(-xi)]
The line above should be an equation. It should NOT start with =. The equation you're starting with says that
[tex]\frac{\partial f}{\partial b_1} = 0[/tex]
Each line after the first one should be an equation. Ultimately you should get an equation that starts with b1.
= Σ(-yi xi)+Σb0xi+ Σb1xi ^2
= -Σ (yi xi)+b0 Σxi+b1Σxi^2
b1Σxi^2=Σyi Σxi – (yBAR-b1xBAR)Σxi ............... (Re-arranging and substituting for b0)
b1Σxi=Σyi -yBAR+b1xBAR ............................ (dividing both sides by Σxi)
b1Σxi – b1xb = Σyi – yBAR
b1(Σxi – xb)=Σyi – yBAR
b1= (Σyi – yb)/(Σxi – xBAR)
= (Σyi – Σyi /n)(Σxi – Σxi/n)
 
  • #3
The line above should be an equation. It should NOT start with =. The equation you're starting with says that
[tex]\frac{\partial f}{\partial b_1} = 0[/tex]
Each line after the first one should be an equation. Ultimately you should get an equation that starts with b1.
That's just me being unformal though.

2=x
=y

is just 2=x=y.

I will formalise it when I figure out the answer.

Do you know where my actually working went wrong?
 
  • #4
33,632
5,287
That's just me being unformal though.

2=x
=y

is just 2=x=y.
If 2 = x, and x = y, then you can say 2 = y or 2 = x = y. This is different from what I'm talking about, and it's not just a matter of being informal vs. being formal. It's the difference between solving an equation and simplifying an expression, which are two entirely different operations.
I will formalise it when I figure out the answer.

Do you know where my actually working went wrong?
Start with 2Σ[(yi - b0 - b1xi)*(-xi)] = 0

and solve for b1, using the properties of summations.

Each step should be an equation!
 
  • #5
gabbagabbahey
Homework Helper
Gold Member
5,002
6
Hi operationres, it may seem that Mark is being nit-picky, but his comments are good advice. Even though you aren't handing in your forum post for marks, it is still in your best interests to make your workings as clear as possible. Doing so makes it easier for us to see where you are going wrong, and also gives you practice in writing things in a logical, mathematically correct manner (practice at this helps make things go quicker on your exams :wink: ).


Letting 2Σ[(yi-b0-b1xi)*(-xi)] = 0
0= Σ[(yi-b0-b1xi)*(-xi)]
0= Σ(-yi xi)+Σb0xi+ Σb1xi ^2
0= -Σ (yi xi)+b0 Σxi+b1Σxi^2
b1Σxi^2=Σyi Σxi – (yBAR-b1xBAR)Σxi ............... (Re-arranging and substituting for b0)
Here it looks like you are claiming that

[tex]\sum_{i=1}^nx_iy_i=\left(\sum_{i=1}^nx_i\right)\left(\sum_{i=1}^ny_i\right)[/tex]

Try plugging in some actual numbers to convince yourself that this is not correct.

b1Σxi=Σyi -yBAR+b1xBAR ............................ (dividing both sides by Σxi)
b1Σxi – b1xb = Σyi – yBAR
You also seem to think that

[tex]\frac{\sum_{i=1}^n(x_i)^2}{\sum_{i=1}^nx_i}=\sum_{i=1}^nx_i[/tex]

Which again isn't true (plug in some actual numbers to convince yourself of this).
 

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