<p>I am struggling with understanding one result in <a href="http://www.econ.ucla.edu/fjbuera/papers/matFINAL2.pdf" rel="nofollow noreferrer">Buera and Nicolini (2004)</a> published in JME. In page 538, they write
<span class="math-container">$$A_t(h^t) b_{t-1}(h^{t-1}) = Z_t(h^{t-1})$$</span>
where <span class="math-container">$A_t(h^t)$</span> is a <span class="math-container">$N \times J$</span> matrix, <span class="math-container">$b_{t-1}(h^{t-1})$</span> is a <span class="math-container">$J\times 1$</span> vector, and <span class="math-container">$Z_t(h^{t-1})$</span> is a <span class="math-container">$N\times 1$</span> vector. <span class="math-container">$N$</span> is the number of states and <span class="math-container">$J$</span> is the number of maturities. Then, in <a href="https://www.econ.ucla.edu/fjbuera/papers/matFINAL2.pdf#page=11" rel="nofollow noreferrer">page 539</a>, they say that</p>
<p>&quot;a necessary condition for the Ramsey to be implementable is that the government can issue non-contingent bonds in at least as many maturities as possible realization of the stock, i.e., <span class="math-container">$J \ge N$</span>.''</p>
<p>For an ease of notation, write the above equation as <span class="math-container">$A b = Z$</span>. Suppose that <span class="math-container">$J &gt; N$</span>. Premultiplying a <span class="math-container">$J \times N$</span> matrix <span class="math-container">$D$</span>,
<span class="math-container">$$D Ab = D Z$$</span>
If <span class="math-container">$D$</span> is an inverse of <span class="math-container">$A$</span>, rank(<span class="math-container">$DA$</span>) = <span class="math-container">$J$</span>. But, <span class="math-container">$\text{rank}(DA) \le \min(\text{rank}(D), \text{rank}(A))\le N$</span>. Contradiction. I think I am missing something important. I would appreciate if someone helps me with this.</p>

CC BY-SA 4.0
2024-03-26T12:11:23.490Z
There is something I can't understand. $A$ is a $n\times j$ matrix, if $n\neq j$ $A$ is a non-square matrix, and can't have an inverse, the inverse of a matrix is defined for square matrices only.
BakerStreet
CC BY-SA 4.0
2024-03-26T16:40:44.853Z
@shk910 Is the whole first part of the question necessary, or could it be reduced to the part starting with "*For an ease of notation*..."?
Giskard
CC BY-SA 4.0
2024-03-26T16:41:13.840Z
@BakerStreet Perhaps it is a [generalized inverse](https://en.wikipedia.org/wiki/Generalized_inverse)? The article is "*an*", as in "$D$ is an inverse of $A$". In that case rank of $DA$ would be $N \leq J$ though.
Giskard
CC BY-SA 4.0
2024-03-26T17:01:58.663Z
@Giskard  maybe, but nothing can be said without reading the paper, it sholuld be specified.
BakerStreet
CC BY-SA 4.0
2024-03-26T17:10:02.090Z
@Giskard The pseudo inverse  also is unique,  https://en.wikipedia.org/wiki/Moore%E2%80%93Penrose_inverse, so it is difficult to say without more information what the paper of Buera and Nicolini is talking about.
BakerStreet
CC BY-SA 4.0
2024-03-26T17:10:45.503Z
Is this $D$ still in the paper though? I thought the papers part stops at the quotation marks at the middle.
Giskard
CC BY-SA 4.0
2024-03-26T17:13:04.060Z
@BakerStreet Here is the relevant page, in case you are interested http://www.econ.ucla.edu/fjbuera/papers/matFINAL2.pdf#page=12
Giskard
<macroeconomics><theory>
some question on Buera and Nicolini (2004)
shk910
# some question on Buera and Nicolini (2004)

(Asked by: username_0 on 2024-03-26)

I am struggling with understanding one result in [Buera and Nicolini (2004)](http://www.econ.ucla.edu/fjbuera/papers/matFINAL2.pdf) published in JME. In page 538, they write
$$A\_t(h^t) b\_{t-1}(h^{t-1}) = Z\_t(h^{t-1})$$
where $A\_t(h^t)$ is a $N \times J$ matrix, $b\_{t-1}(h^{t-1})$ is a $J\times 1$ vector, and $Z\_t(h^{t-1})$ is a $N\times 1$ vector. $N$ is the number of states and $J$ is the number of maturities. Then, in [page 539](https://www.econ.ucla.edu/fjbuera/papers/matFINAL2.pdf#page=11), they say that

"a necessary condition for the Ramsey to be implementable is that the government can issue non-contingent bonds in at least as many maturities as possible realization of the stock, i.e., $J \ge N$.''

For an ease of notation, write the above equation as $A b = Z$. Suppose that $J > N$. Premultiplying a $J \times N$ matrix $D$,
$$D Ab = D Z$$
If $D$ is an inverse of $A$, rank($DA$) = $J$. But, $\text{rank}(DA) \le \min(\text{rank}(D), \text{rank}(A))\le N$. Contradiction. I think I am missing something important. I would appreciate if someone helps me with this.

**Tags:** macroeconomics, theory
> **Comments:**
> **username_1**: There is something I can't understand. $A$ is a $n\times j$ matrix, if $n\neq j$ $A$ is a non-square matrix, and can't have an inverse, the inverse of a matrix is defined for square matrices only.  
> **username_2**: @shk910 Is the whole first part of the question necessary, or could it be reduced to the part starting with "*For an ease of notation*..."?  
> **username_2**: @BakerStreet Perhaps it is a [generalized inverse](https://en.wikipedia.org/wiki/Generalized_inverse)? The article is "*an*", as in "$D$ is an inverse of $A$". In that case rank of $DA$ would be $N \leq J$ though.  
> **username_1**: @Giskard  maybe, but nothing can be said without reading the paper, it sholuld be specified.  
> **username_1**: @Giskard The pseudo inverse  also is unique,  https://en.wikipedia.org/wiki/Moore%E2%80%93Penrose_inverse, so it is difficult to say without more information what the paper of Buera and Nicolini is talking about.  
> **username_2**: Is this $D$ still in the paper though? I thought the papers part stops at the quotation marks at the middle.  
> **username_2**: @BakerStreet Here is the relevant page, in case you are interested http://www.econ.ucla.edu/fjbuera/papers/matFINAL2.pdf#page=12  

---

## 0 answers

