<p>Here's a little discussion that might help.</p>

<p>The population version has a solution because it is assumed to have a solution. You can see that this is the case from assumptions 3.1 (linearity), 3.3 (orthogonality conditions), and 3.4 (rank condition for identification). You can see the derivation of this fact from equations (3.3.3) and (3.3.4). </p>

<p>Now, you'll notice the population version having a solution implies that
$$
\text{rank}\left( \Sigma_{xz} \right ) 
= \text{rank}\left(\left [\Sigma_{xz} \mid \sigma_{xy} \right ]\right ),
$$
as discussed in exercise 5 of section 3.3. In the hint they note that the population version has a solution when
$$
\left [\Sigma_{xz} \mid \sigma_{xy} \right ]
$$
is of rank $L$. If it has rank $L+1$, it will not have a solution. In the hint, they say that this is a set of "equality conditions." This is referring to the fact that one of the columns must be a linear combination of the remaining columns. This is a very hard condition to satisfy. In many models, this condition will hold with probability zero. This is because the data has to come out so that the equality condition holds perfectly.</p>

<p>On the other hand, the condition that the sample version $S_{xy}$ of $\sum_{xz}$ be of full column rank is a set of "inequality" conditions on the matrix. Because they are inequality conditions on the matrix, they are much easier to satisfy by chance.</p>

<p>In conclusion, $S_{xy}$ will have full column rank for sufficiently large $n$ because the conditions are inequality conditions which occur with probability larger than zero. However, to have a solution we need $\left [S_{xz} \mid s_{xy} \right ]$ to have rank $L$. But because this is an equality condition on the matrix, it only occurs with probability zero.</p>

<p><em>Search Keywords: Solutions, Exercise 3, section 3.4 Hayashi Econometrics</em></p>

CC BY-SA 3.0
2015-01-19T16:56:17.247Z
2015-01-19T16:56:17.247Z
jmbejara
<p>In Hayashi Econometrics, page 207-8, ex3 (see hint), he says that even if a population moment conditions (GMM setting) system is overidentified it still has a solution, while the sample moment conditions system cannot.</p>
<p>I'm having a hard time understanding this.</p>
<p>Any help would be appreciated.</p>
<p>By the way, I've also posted this question on <a href="https://stats.stackexchange.com/questions/133578/possibility-of-solution-in-overdetermined-system-of-moment-conditions">CrossValidated</a>. I hope there's no problem.</p>
<p>Edit: Check Jayk answer in CV. ;)</p>

CC BY-SA 3.0
2016-11-25T12:47:49.407Z
Did you also try to solve Exercises 1 and 2 in the same section?
I am looking for potential ways of solving them.
For 2) I would suggest 0 as the solution.
Bonsaibubble
CC BY-SA 3.0
2015-01-15T22:50:15.123Z
Well the conditions of Weirestrass theorem are usually used to guarantee a solution. So informally I would say that the sample analogue could fail to have a solution because of lack of boundedness or closedness of the parameter space. Normally, this can be forced, what is problematic is uniqueness and thus identification. The population case is simple because is a quadratic objective minimized over a well behaved set of parameters (at least closed and bounded and usually compact).
user157623
CC BY-SA 3.0
2015-01-19T16:34:30.650Z
I've posted an answer to this on cross validated: http://stats.stackexchange.com/a/134056/52022
jayk
CC BY-SA 3.0
2015-01-19T17:12:08.117Z
Please don't cross-post
410 gone
<econometrics>
Solution of overdetermined system of moment conditions
An old man in the sea.
# Solution of overdetermined system of moment conditions

(Asked by: username_0 on 2015-01-15)

In Hayashi Econometrics, page 207-8, ex3 (see hint), he says that even if a population moment conditions (GMM setting) system is overidentified it still has a solution, while the sample moment conditions system cannot.

I'm having a hard time understanding this.

Any help would be appreciated.

By the way, I've also posted this question on [CrossValidated](https://stats.stackexchange.com/questions/133578/possibility-of-solution-in-overdetermined-system-of-moment-conditions). I hope there's no problem.

Edit: Check Jayk answer in CV. ;)

**Tags:** econometrics
> **Comments:**
> **username_2**: Well the conditions of Weirestrass theorem are usually used to guarantee a solution. So informally I would say that the sample analogue could fail to have a solution because of lack of boundedness or closedness of the parameter space. Normally, this can be forced, what is problematic is uniqueness and thus identification. The population case is simple because is a quadratic objective minimized over a well behaved set of parameters (at least closed and bounded and usually compact).  
> **username_3**: I've posted an answer to this on cross validated: http://stats.stackexchange.com/a/134056/52022  
> **username_4**: Please don't cross-post  
> **username_1**: Did you also try to solve Exercises 1 and 2 in the same section? I am looking for potential ways of solving them. For 2) I would suggest 0 as the solution.  

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## 1 answers

### Answer by username_5 on 2015-01-19. Score: 2

Here's a little discussion that might help.

The population version has a solution because it is assumed to have a solution. You can see that this is the case from assumptions 3.1 (linearity), 3.3 (orthogonality conditions), and 3.4 (rank condition for identification). You can see the derivation of this fact from equations (3.3.3) and (3.3.4).

Now, you'll notice the population version having a solution implies that
$$
\text{rank}\left( \Sigma\_{xz} \right )
= \text{rank}\left(\left [\Sigma\_{xz} \mid \sigma\_{xy} \right ]\right ),
$$
as discussed in exercise 5 of section 3.3. In the hint they note that the population version has a solution when
$$
\left [\Sigma\_{xz} \mid \sigma\_{xy} \right ]
$$
is of rank $L$. If it has rank $L+1$, it will not have a solution. In the hint, they say that this is a set of "equality conditions." This is referring to the fact that one of the columns must be a linear combination of the remaining columns. This is a very hard condition to satisfy. In many models, this condition will hold with probability zero. This is because the data has to come out so that the equality condition holds perfectly.

On the other hand, the condition that the sample version $S\_{xy}$ of $\sum\_{xz}$ be of full column rank is a set of "inequality" conditions on the matrix. Because they are inequality conditions on the matrix, they are much easier to satisfy by chance.

In conclusion, $S\_{xy}$ will have full column rank for sufficiently large $n$ because the conditions are inequality conditions which occur with probability larger than zero. However, to have a solution we need $\left [S\_{xz} \mid s\_{xy} \right ]$ to have rank $L$. But because this is an equality condition on the matrix, it only occurs with probability zero.

*Search Keywords: Solutions, Exercise 3, section 3.4 Hayashi Econometrics*


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