Ridiculous manipulation question

[operationsres]

ECONOMETRICS CALCULUS.


1. The PRECEDING question and solution

Differentiate

with respect to b₁

My solution:
df(b₀,b₁)/db₁ = 2Σ[(y_i-b₀-b₁x_i)*(-x_i)]

2. the ACTUAL question

By setting d,f(b₀,b₁)/d,b₁ = 0, show that you obtain

.
(Hint: you will have to substitute b₀ = y^BAR - b₁(x^BAR), and use the results from the previous question.)

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

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

Letting 2Σ[(y_i-b₀-b₁x_i)*(-x_i)] = 0
0= Σ[(y_i-b₀-b₁x_i)*(-x_i)]
0= Σ(-y_i x_i)+Σb₀x_i+ Σb₁x_i ^2
0= -Σ (y_i x_i)+b₀ Σx_i+b₁Σx_i^2
b₁Σx_i^2=Σy_i Σx_i – (y^BAR-b₁x^BAR)Σx_i ... (Re-arranging and substituting for b₀)
b₁Σx_i=Σy_i -y^BAR+b₁x^BAR ...... (dividing both sides by Σx_i)
b₁Σx_i – b₁xb = Σy_i – y^BAR
b₁(Σx_i – xb)=Σy_i – y^BAR
b₁= (Σy_i – yb)/(Σx_i – x^BAR)
= (Σy_i – Σy_i /n)(Σx_i – Σx_i/n)

 

[Mark44]

ECONOMETRICS CALCULUS.


1. The PRECEDING question and solution

Differentiate

with respect to b₁

My solution:
df(b₀,b₁)/db₁ = 2Σ[(y_i-b₀-b₁x_i)*(-x_i)]

2. the ACTUAL question

By setting d,f(b₀,b₁)/d,b₁ = 0, show that you obtain

.
(Hint: you will have to substitute b₀ = y^BAR - b₁(x^BAR), and use the results from the previous question.)

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

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

Letting 2Σ[(y_i-b₀-b₁x_i)*(-x_i)] = 0
= Σ[(y_i-b₀-b₁x_i)*(-x_i)]

The line above should be an equation. It should NOT start with =. The equation you're starting with says that
\frac{\partial f}{\partial b_1} = 0
Each line after the first one should be an equation. Ultimately you should get an equation that starts with b₁.

= Σ(-y_i x_i)+Σb₀x_i+ Σb₁x_i ^2
= -Σ (y_i x_i)+b₀ Σx_i+b₁Σx_i^2
b₁Σx_i^2=Σy_i Σx_i – (y^BAR-b₁x^BAR)Σx_i ... (Re-arranging and substituting for b₀)
b₁Σx_i=Σy_i -y^BAR+b₁x^BAR ...... (dividing both sides by Σx_i)
b₁Σx_i – b₁xb = Σy_i – y^BAR
b₁(Σx_i – xb)=Σy_i – y^BAR
b₁= (Σy_i – yb)/(Σx_i – x^BAR)
= (Σy_i – Σy_i /n)(Σx_i – Σx_i/n)

 

[operationsres]

The line above should be an equation. It should NOT start with =. The equation you're starting with says that
\frac{\partial f}{\partial b_1} = 0
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?

 

[Mark44]

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[(y_i - b₀ - b₁x_i)*(-x_i)] = 0

and solve for b₁, using the properties of summations.

Each step should be an equation!

 

[gabbagabbahey]

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 ).

Letting 2Σ[(y_i-b₀-b₁x_i)*(-x_i)] = 0
0= Σ[(y_i-b₀-b₁x_i)*(-x_i)]
0= Σ(-y_i x_i)+Σb₀x_i+ Σb₁x_i ^2
0= -Σ (y_i x_i)+b₀ Σx_i+b₁Σx_i^2
b₁Σx_i^2=Σy_i Σx_i – (y^BAR-b₁x^BAR)Σx_i ... (Re-arranging and substituting for b₀)

Here it looks like you are claiming that

\sum_{i=1}^nx_iy_i=\left(\sum_{i=1}^nx_i\right)\left(\sum_{i=1}^ny_i\right)

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

b₁Σx_i=Σy_i -y^BAR+b₁x^BAR ...... (dividing both sides by Σx_i)
b₁Σx_i – b₁xb = Σy_i – y^BAR

You also seem to think that

\frac{\sum_{i=1}^n(x_i)^2}{\sum_{i=1}^nx_i}=\sum_{i=1}^nx_i

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