3
$\begingroup$

I have so far discovered three different ways of utilizing the Cholesky decomposition for calculating the OIRFs of a VAR(k). The different methods seem contradictory so I would like some input on where I am making the mistake. First, all methods use the Cholesky decomposition, which decomposes the contemporaneous variance covariance matrix of the error term into PP'.

First method to solve the OIRFS is to first estimate the reduced VAR(1) equation (for simplicity):

$Y_t = A_1 Y_{t-1} + e_t $

Premultiply the equation by the P and the OIRFs can be solved recursively.

Second method is the same, but only the right side is multiplied by P. The following question contains an example and more details: SVAR, Cholesky decomposition and impulse-response function in R

The third method is to use the VMA form of the VAR (thanks to hejseb for the equation): $ y_t=\sum_{s=0}^\infty\Psi_se_{t-s}=\sum_{s=0}^\infty\Psi_sPP^{-1}e_{t-s}=\sum_{s=0}^\infty\Psi_s^*v_{t-s}. $

Where $\Psi_s^*$ is the matrix of orthogonalized impulse responses, $\Psi_s$ being the matrix of simple IRFs. I have personally checked that this method provides the correct result. More details on the method: How to calculate the impulse response function of a VAR(1)? (With example)

Two questions:

1) As is apparent, the equation used for the first and second methods are not the same, there is an added P on the left side in the first case. Yet they are supposed to provide the same result. How is this possible? To me the 2nd method seems just wrong as there doesn't even appear to be a contemporaneous effect in it.

2) In the VMA representation the newly created uncorrelated error term is $P^{-1}\epsilon_{t}$ In the VAR representations it is $P\epsilon_{t}$. Isn't this a contradiction? The VMA representation makes more sense as the inverse of P is what de-correlates correlated variables.

Still looking for answer on this.

$\endgroup$
2
$\begingroup$

I think the second method is wrong, both algebraically and conceptually. The third one is absolutely correct, while the first one presents a single mistake in the use of the Cholesky factor you report. To be clearer I try to explain it to you using an example. Say that we come up with a model that induces the following $SVAR$ (structural $VAR$) for $\mathbf{y}_{t} \in \rm I\!R^{n} $ \begin{equation} A\mathbf{y}_{t} = \Gamma \mathbf{y}_{t-1} + \mathbf{v}_{t}, \quad \mathbf{v}_{t} \sim WN(\mathbf{0}, \Sigma), \forall\, t \end{equation} and we want to be able to identify it using the usual reduced form $VAR$ that the data give us \begin{equation} \mathbf{y}_{t} = \Pi \mathbf{y}_{t-1} + \mathbf{u}_{t}, \quad \mathbf{u}_{t} \sim WN(\mathbf{0}, \Omega), \forall\, t \end{equation} Doing simple algebra, it is clear that $A^{-1}\Gamma = \Pi$ and $A^{-1}\mathbf{v}_{t}=\mathbf{u}_{t}$. From the latter, it is also true that $\Omega = A^{-1} \Sigma {A^{-1}}^{T}$. Now, using a Cholesky identification scheme (I suppose you were referring to the one proposed by Sims) requires assuming $\Sigma = I_{n}$ and $A$ lower triangular. Moreover, being $\Omega$ a p.d. matrix, you can always decompose it as $\Omega = GG'$ with $G$ lower triangular. It is easy to see now that $G=A^{-1}$ thanks to the assumption we made. More is actually true: we now have a bijection between our $SVAR$ and the rf $VAR$, indeed premultiplying the latter for the inverse of the Cholesky factor (and not for the Cholesky factor as you said in the first method) \begin{equation} \mathbf{y}_{t} = \Pi \mathbf{y}_{t-1} + \mathbf{u}_{t} \Rightarrow G^{-1}\mathbf{y}_{t} = G^{-1}\Pi \mathbf{y}_{t-1} + G^{-1}\mathbf{u}_{t} \Rightarrow A\mathbf{y}_{t} = \Gamma \mathbf{y}_{t-1} + \mathbf{v}_{t} \end{equation} using the relations described above. Notice now that this method is nothing more than the initial step you have to do to compute $IRFs$. Indeed, what people basically do it is using your third method as the final step to get $IRFs$. The $VMA(\infty)$ representation is the most used one because it simplifies notation and computation a lot, especially in higher order $VARs$. Therefore, assuming $A$ to have eigenvalues smaller than one in absolute value, you can rewrite your rf $VAR$ as \begin{equation} \mathbf{y}_{t}= \Psi(L)\mathbf{u}_{t}= \sum_{i=0}^{\infty}A^{i}\mathbf{u}_{t}= \sum_{i=0}^{\infty}A^{i}G\mathbf{v}_{t}=\Psi^{\star}(L)\mathbf{v}_{t} \end{equation} For what regards your second question, you just have to pay attention to what it is claimed to be structural and what is not. Using my notation, $\mathbf{v}_{t}$ is structural, $\mathbf{u}_{t}$ not. The mapping between the two is always the same just written in different ways \begin{equation} A^{-1}\mathbf{v}_{t}=\mathbf{u}_{t} \Rightarrow G\mathbf{v}_{t}=\mathbf{u}_{t} \Rightarrow \mathbf{v}_{t} = G^{-1}\mathbf{u}_{t} \Rightarrow \mathbf{v}_{t} = A \mathbf{u}_{t} \end{equation} In the initial step you go from reduced form to structural form so you use one direction, in the $VMA$ representation, instead, you go back to the structural model from the reduced form so you need to use the other direction. Everything is perfectly coherent, but I know that it is quite easy to get lost in it if you do not pay too much attention in what you are doing.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.