<p><span class="math-container">$\epsilon_t$</span> here is the structural shock. <span class="math-container">$\nu_t$</span> are the reduced form residuals. <span class="math-container">$A_0^{-1}$</span> is a rotation matrix providing the mapping between reduced form and structural shocks. <span class="math-container">$\tilde A_0^{-1}$</span> is essentially the same, but normalizing the structural shock variances to 1. In the paper, we are interested in the IRFs to a structural shock, which implies we need to identify the vector <span class="math-container">$\nu_t$</span> corresponding to the shock of interest. Generally, for a one-standard deviation shock that would be something like <span class="math-container">$\tilde A_0^{-1} e_1'$</span>, because, loosely speaking, <span class="math-container">$\tilde A_0^{-1}$</span> has the standard deviation of the shocks on the diagonal and <span class="math-container">$e_1$</span> is a unit shock. However, we are interested in a 0.1 unit shock instead of a one-standard deviation one. That is where the <span class="math-container">$\xi$</span> comes in. The numerator uses 0.1 times a one-standard deviation shock, while the denominator divides this by the standard deviation. The result is a 0.1 unit shock to the first structural shock.</p>

CC BY-SA 4.0
2024-11-08T14:48:32.450Z
I understand your answer but I think $\xi$ is a column vector of $0.1$ standard deviation shock normalized by one-standard deviation(since we need normalized matrix $A_0^{-1}$ to calculate impulse-response function). Then the dimension of impulse-response result $\Psi_j^k=e_j \Lambda^k \xi$ is $(1 \times n)*(n \times n)*(n \times 1)= 1\times 1$ which means it is a scalar(note that $n$ is a random number that I choose).
guest
CC BY-SA 4.0
2024-11-09T08:40:12.713Z
Yes, that is correct because it is only the IRF for variable $j$. If you drop the selection vector $e_j$ you will get the full vector of IRFs of dimension $n\times 1$.
jpfeifer
CC BY-SA 4.0
2024-11-07T13:19:41.040Z
2024-11-07T13:19:41.040Z
jpfeifer
<p>Counterfactual analysis is used when we want to compare the actual IRF with counterfactual senario IRF. For example, if you want to examine inflation response to oil price shocks just ignoring the second round effects, you can use counterfactual IRF(no second round effects) to compare with the baseline IRF(with second round effects) in SVAR model.<br />
Counterfactual IRF is produced from baseline IRF in SVAR using additional restrictions. This methodology has been used by Bernanke et al(1997), Sims and Zha(2006), Kilian and Lewis(2011) etc. In here I mainly refered the <a href="https://onlinelibrary.wiley.com/doi/full/10.1111/jmcb.12288" rel="nofollow noreferrer"><strong>Wong(2015)</strong></a>.</p>
<p>Consider the following VAR model.<br />
<span class="math-container">$A_0 y_t =\sum_{i=1}^{p} A_i y_{t-i} +\epsilon_t $</span><br />
where <span class="math-container">$y_t =[oil_t, \pi^e_t, \pi_t ]$</span> and <span class="math-container">$\epsilon_t$</span> is a vector of orthogonal structural shock.<br />
<br />
Define the above equation in companion form.<br />
<span class="math-container">$A_0 y_t =\sum_{i=1}^{p} A_i y_{t-i} +\epsilon_t $</span><br />
<span class="math-container">$y_t =\sum_{i=1}^{p} A^{-1}_0A_i y_{t-i} +A^{-1}_0\epsilon_t $</span><br />
<span class="math-container">$X_t =\Lambda X_{t-1}+v_t$</span></p>
<p>Let <span class="math-container">$E(v_t' v_t)=\Omega$</span> , <span class="math-container">$\tilde{A_0^{-1}}=chol(\Omega)$</span> and <span class="math-container">$e_j$</span> as a row vector with 1 as the <span class="math-container">$j$</span>th element and 0 elsewhere.<br />
Then the impluse response function of variable <span class="math-container">$j$</span> to <span class="math-container">$10\%$</span> oil shock at horizon <span class="math-container">$k$</span>, <span class="math-container">$\Psi_j^k$</span> is<br />
<span class="math-container">$\Psi_j^k=e_j \Lambda^k \xi$</span><br />
where <span class="math-container">$\xi=\frac{0.1 \times \tilde{A_0^{-1}}e_1' }{e_1 \tilde{A_0^{-1}} e_1'}$</span></p>
<p>Usually the response of variable <span class="math-container">$j$</span> to shock <span class="math-container">$i$</span> at horizon <span class="math-container">$k$</span> is  <span class="math-container">$ \Lambda^k_{i,j} $</span> <br />
so <span class="math-container">$\Lambda^k_{j,i}=e_j\Lambda^k e_i'$</span> <br />
Here I can't understand why we use <span class="math-container">$\xi$</span> and the meaning of its numerator and denominator.</p>
<p>[Note] The difference between <span class="math-container">$\tilde{A_0^{-1}}$</span> and <span class="math-container">$A_0^{-1}$</span> is just normalization. <span class="math-container">$A_0^{-1}$</span> is normalized to have 1 in its diagnal.</p>

CC BY-SA 4.0
2024-11-05T15:19:15.150Z
What is the paper in question? It seems there are additional unstated assumptions.
jpfeifer
CC BY-SA 4.0
2024-11-06T00:29:28.680Z
@jpfeifer The paper is "Do Inflation Expectations Propagate the Inflationary Impact of Real Oil Price Shocks?: Evidence from the Michigan Survey", and I added link(https://onlinelibrary.wiley.com/doi/full/10.1111/jmcb.12288) in the text.
guest
<macroeconomics><svar>
Counterfactual analysis in SVAR
guest
# Counterfactual analysis in SVAR

(Asked by: username_0 on 2024-11-05)

Counterfactual analysis is used when we want to compare the actual IRF with counterfactual senario IRF. For example, if you want to examine inflation response to oil price shocks just ignoring the second round effects, you can use counterfactual IRF(no second round effects) to compare with the baseline IRF(with second round effects) in SVAR model.  
Counterfactual IRF is produced from baseline IRF in SVAR using additional restrictions. This methodology has been used by Bernanke et al(1997), Sims and Zha(2006), Kilian and Lewis(2011) etc. In here I mainly refered the [**Wong(2015)**](https://onlinelibrary.wiley.com/doi/full/10.1111/jmcb.12288).

Consider the following VAR model.  
$A\_0 y\_t =\sum\_{i=1}^{p} A\_i y\_{t-i} +\epsilon\_t $  
where $y\_t =[oil\_t, \pi^e\_t, \pi\_t ]$ and $\epsilon\_t$ is a vector of orthogonal structural shock.  
  
Define the above equation in companion form.  
$A\_0 y\_t =\sum\_{i=1}^{p} A\_i y\_{t-i} +\epsilon\_t $  
$y\_t =\sum\_{i=1}^{p} A^{-1}\_0A\_i y\_{t-i} +A^{-1}\_0\epsilon\_t $  
$X\_t =\Lambda X\_{t-1}+v\_t$

Let $E(v\_t' v\_t)=\Omega$ , $\tilde{A\_0^{-1}}=chol(\Omega)$ and $e\_j$ as a row vector with 1 as the $j$th element and 0 elsewhere.  
Then the impluse response function of variable $j$ to $10\%$ oil shock at horizon $k$, $\Psi\_j^k$ is  
$\Psi\_j^k=e\_j \Lambda^k \xi$  
where $\xi=\frac{0.1 \times \tilde{A\_0^{-1}}e\_1' }{e\_1 \tilde{A\_0^{-1}} e\_1'}$

Usually the response of variable $j$ to shock $i$ at horizon $k$ is $ \Lambda^k\_{i,j} $   
so $\Lambda^k\_{j,i}=e\_j\Lambda^k e\_i'$   
Here I can't understand why we use $\xi$ and the meaning of its numerator and denominator.

[Note] The difference between $\tilde{A\_0^{-1}}$ and $A\_0^{-1}$ is just normalization. $A\_0^{-1}$ is normalized to have 1 in its diagnal.

**Tags:** macroeconomics, svar

> **Comments:**
> **username_1**: What is the paper in question? It seems there are additional unstated assumptions.  
> **username_0**: @username_1 The paper is "Do Inflation Expectations Propagate the Inflationary Impact of Real Oil Price Shocks?: Evidence from the Michigan Survey", and I added link(https://onlinelibrary.wiley.com/doi/full/10.1111/jmcb.12288) in the text.  

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## 1 answers

### Answer by username_1 on 2024-11-07. Score: 0 | Accepted answer

$\epsilon\_t$ here is the structural shock. $\nu\_t$ are the reduced form residuals. $A\_0^{-1}$ is a rotation matrix providing the mapping between reduced form and structural shocks. $\tilde A\_0^{-1}$ is essentially the same, but normalizing the structural shock variances to 1. In the paper, we are interested in the IRFs to a structural shock, which implies we need to identify the vector $\nu\_t$ corresponding to the shock of interest. Generally, for a one-standard deviation shock that would be something like $\tilde A\_0^{-1} e\_1'$, because, loosely speaking, $\tilde A\_0^{-1}$ has the standard deviation of the shocks on the diagonal and $e\_1$ is a unit shock. However, we are interested in a 0.1 unit shock instead of a one-standard deviation one. That is where the $\xi$ comes in. The numerator uses 0.1 times a one-standard deviation shock, while the denominator divides this by the standard deviation. The result is a 0.1 unit shock to the first structural shock.

> **Comments:**
> **username_0**: I understand your answer but I think $\xi$ is a column vector of $0.1$ standard deviation shock normalized by one-standard deviation(since we need normalized matrix $A_0^{-1}$ to calculate impulse-response function). Then the dimension of impulse-response result $\Psi_j^k=e_j \Lambda^k \xi$ is $(1 \times n)*(n \times n)*(n \times 1)= 1\times 1$ which means it is a scalar(note that $n$ is a random number that I choose).  
> **username_1**: Yes, that is correct because it is only the IRF for variable $j$. If you drop the selection vector $e_j$ you will get the full vector of IRFs of dimension $n\times 1$.  

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