<p>As given in the picture, question ask us to transform dependent variable, by premultiplying with a weighting matrix W, (which will change the regression coefficient into C'β) such that the expression (given in question) has a minimum mean squared error.</p>

<p>So, I solved part a) by solving for the brackets of E[(w′y−c′β)(w′y−c′β)′] and differentiating with respect to w, I obtain the expression as</p>

<p><span class="math-container">$w′y=[c′ββ′x′(xββ′x′+σ^2.I)^{−1}]y$</span></p>

<p>My question is What does this espression signify or simply put what did we do/ what are we trying to achieve and why it is not practical to use above estimator? Any help or suggestion will be helpful. </p>

<p>PS: Is it impractical because <span class="math-container">$σ^2$</span> is unknown?</p>

<p>Expression can be rewritten as - </p>

<p><span class="math-container">$w′y=[c′ββ′x′(E(yy') )^{−1}]y$</span></p>

<p>Because Xβ = E(Y|X) and Var(y) =<span class="math-container">$σ^2$</span>I, By the formula </p>

<p>Var(Y) = E(YY') + E(Y|X)E(Y|X)'</p>

<p>Edit: Expressions for estimated β is given by (X'X)<span class="math-container">$^{-1}$</span>X'Y where, define A= (X'X)<span class="math-container">$^{-1}$</span>X'. This gives estimated β = AY. So, the expression W'Y be also denoting another expression for estimated β, an expression computed such that the discrepancy between the stylised expression for est_β ie W'Y and the weighted β ie C'β is minimum? Am I thinking on correct lines? </p>

<p><a href="https://i.sstatic.net/h9kuy.jpg" rel="nofollow noreferrer"><img src="https://i.sstatic.net/h9kuy.jpg" alt="Question"></a></p>

CC BY-SA 4.0
2019-09-08T13:56:30.960Z
Hi: I saw some version of this question earlier and still don't understand it. Is $\beta$ a vector and known ? If not, then what is $C^{\prime} \beta$, a new unknown scalar ?
mark leeds
CC BY-SA 4.0
2019-09-09T02:14:55.403Z
@markleeds Hi, $/beta$ is estimated using OLS and since we have transformed Y by a matrix W, this will change the beta coefficients denoted by C'$/beta$.
Elina Gilbert
CC BY-SA 4.0
2019-09-09T03:11:12.303Z
Hi: So, $\beta$ is a known vector and $W'y$ is still an known vector  since $W$ is known also ? $C$ is the only unknown ?
mark leeds
CC BY-SA 4.0
2019-09-09T04:30:02.507Z
@markleeds W is not known. We are trying to transform y (through premultiplying by W) in such a way that it minimizes the expression given in part a). C is a weighted vector and is assumed to be given. So, the question is if we have measured beta as a weighted measure then how should we transform Y so that it minimizes the expression.
Elina Gilbert
CC BY-SA 4.0
2019-09-09T19:52:56.893Z
Hi: I'm sorry for the confusion. That clarifies some of it but can you give the dimension of every variable in the model.  I'm confused between what's a vector and what's a matrix. Thanks.
mark leeds
CC BY-SA 4.0
2019-09-09T19:54:27.290Z
For example, is $C^{\prime} \beta$ a scalar ?
mark leeds
CC BY-SA 4.0
2019-09-09T21:20:31.320Z
@markleeds Yes C'beta is a scalar and so is W'y. Dimension of W is nx1 and dimension of Y is nx1. Dimension of C and Beta is given in the question. Thanks in advance for thinking through the question.
Elina Gilbert
CC BY-SA 4.0
2019-09-09T21:45:21.650Z
@markleeds Please look at the edited post. Does my line of thought make sense?
Elina Gilbert
CC BY-SA 4.0
2019-09-10T14:54:32.307Z
Elina: I think I understand better now ( but still not perfectly ).  Assuming your expression is correct  (the one where you say, we obtain the expression as ), I would leave it like that. The impracticailty of doing what you are doing might have to do with the fact that when you multiply the response by $W^{\prime}$, you are changing the noise term also. So, the expression you are minimizing is no longer minimizing the original noise term but some weighted version of it. This doesn't make sense to me from a practical perspective.
mark leeds
CC BY-SA 4.0
2019-09-10T15:03:27.647Z
@markleeds Thank you! This makes sense now.
Elina Gilbert
CC BY-SA 4.0
2019-09-11T02:56:41.250Z
Hi: I'm not clear on what helped ( and glad it did )  but I have a feeling this question is trying to prepare you-give you intuition for the topic of "generalized least squares" which is often referred to as GLS. If your course or whatever you're doing covers that, then you should go back to this question at that point because you'll understand the question more deeply once you cover GLS.  All the best.
mark leeds
CC BY-SA 4.0
2019-09-11T03:40:01.853Z
Yes it does. It hasn't been discussed yet in class but is part of syllabus. Thanks!
Elina Gilbert
CC BY-SA 4.0
2019-09-12T07:14:02.427Z
great. enjoy the class..
mark leeds
<econometrics><regression>
Minimizing the error: Transforming dependent variable
Elina Gilbert
# Minimizing the error: Transforming dependent variable

(Asked by: username_0 on 2019-09-07)

As given in the picture, question ask us to transform dependent variable, by premultiplying with a weighting matrix W, (which will change the regression coefficient into C'β) such that the expression (given in question) has a minimum mean squared error.

So, I solved part a) by solving for the brackets of E[(w′y−c′β)(w′y−c′β)′] and differentiating with respect to w, I obtain the expression as

$w′y=[c′ββ′x′(xββ′x′+σ^2.I)^{−1}]y$

My question is What does this espression signify or simply put what did we do/ what are we trying to achieve and why it is not practical to use above estimator? Any help or suggestion will be helpful.

PS: Is it impractical because $σ^2$ is unknown?

Expression can be rewritten as -

$w′y=[c′ββ′x′(E(yy') )^{−1}]y$

Because Xβ = E(Y|X) and Var(y) =$σ^2$I, By the formula

Var(Y) = E(YY') + E(Y|X)E(Y|X)'

Edit: Expressions for estimated β is given by (X'X)$^{-1}$X'Y where, define A= (X'X)$^{-1}$X'. This gives estimated β = AY. So, the expression W'Y be also denoting another expression for estimated β, an expression computed such that the discrepancy between the stylised expression for est\_β ie W'Y and the weighted β ie C'β is minimum? Am I thinking on correct lines?

[![Question](https://i.sstatic.net/h9kuy.jpg)](https://i.sstatic.net/h9kuy.jpg)

**Tags:** econometrics, regression
> **Comments:**
> **username_1**: Hi: I saw some version of this question earlier and still don't understand it. Is $\beta$ a vector and known ? If not, then what is $C^{\prime} \beta$, a new unknown scalar ?  
> **username_0**: @markleeds Hi, $/beta$ is estimated using OLS and since we have transformed Y by a matrix W, this will change the beta coefficients denoted by C'$/beta$.  
> **username_1**: Hi: So, $\beta$ is a known vector and $W'y$ is still an known vector  since $W$ is known also ? $C$ is the only unknown ?  
> **username_0**: @markleeds W is not known. We are trying to transform y (through premultiplying by W) in such a way that it minimizes the expression given in part a). C is a weighted vector and is assumed to be given. So, the question is if we have measured beta as a weighted measure then how should we transform Y so that it minimizes the expression.  
> **username_1**: Hi: I'm sorry for the confusion. That clarifies some of it but can you give the dimension of every variable in the model.  I'm confused between what's a vector and what's a matrix. Thanks.  
> **username_1**: For example, is $C^{\prime} \beta$ a scalar ?  
> **username_0**: @markleeds Yes C'beta is a scalar and so is W'y. Dimension of W is nx1 and dimension of Y is nx1. Dimension of C and Beta is given in the question. Thanks in advance for thinking through the question.  
> **username_0**: @markleeds Please look at the edited post. Does my line of thought make sense?  
> **username_1**: Elina: I think I understand better now ( but still not perfectly ).  Assuming your expression is correct  (the one where you say, we obtain the expression as ), I would leave it like that. The impracticailty of doing what you are doing might have to do with the fact that when you multiply the response by $W^{\prime}$, you are changing the noise term also. So, the expression you are minimizing is no longer minimizing the original noise term but some weighted version of it. This doesn't make sense to me from a practical perspective.  
> **username_0**: @markleeds Thank you! This makes sense now.  
> **username_1**: Hi: I'm not clear on what helped ( and glad it did )  but I have a feeling this question is trying to prepare you-give you intuition for the topic of "generalized least squares" which is often referred to as GLS. If your course or whatever you're doing covers that, then you should go back to this question at that point because you'll understand the question more deeply once you cover GLS.  All the best.  
> **username_0**: Yes it does. It hasn't been discussed yet in class but is part of syllabus. Thanks!  
> **username_1**: great. enjoy the class..  

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## 0 answers

