TransmissionChannelAnalysis.jl

Calculate the transmission effect along a collection of paths in an SVAR. See paper for description.

Information for Users

Variables follow the convention that they start with "x" followed by a number. The number corresponds to the variable number in the rewritten structural form in the paper. For example, for a SVAR with $k$ variables, the following mapping applies

• $y_{1, t} \to x_1$
• $y_{2, t} \to x_2$
• ...
• $y_{k, t} \to x_k$
• $y_{1, t+1} \to x_{k+1}$
• ...

Transmission mechanisms are defined in terms of Boolean statements. To create a transmission condition, first create a string corresponding to the desired Boolean statement. For example, if the interest is in the transmission along the paths that go through $x_2$ but not through $x_3$ or $x_4$, then the string is of the form

s = "x1 & !x3 & !x4".

This could equivalently be represented as x1 & !(x3 | x4), however, the above representation is often more efficient.

String representations of Boolean statements can then be transformed into transmission conditions/queries using make_condition as follows

cond = make_condition(s)

cond is now of type Q, which represents a transmission condition or Qery. The effect of this transmission query can be calculated using transmission as follows

from_shock = 1
effect = transmission(from_shock, B, Qbb, cond)
effect = transmission(from_shock, B, Qbb, cond; method = :BQbb)  # equivalent to above.

where B and Qbb correspond to the structural matrices in the rewritten structural form (also see the documentation for transmission). To obtain B and Qbb from an estimated and (partially) identified SVAR, use to_structural_transmission_model. If only IRFs are available, or if the IRF method is preferred, then effect can be calculated by using transmission and setting the keyword argument method = :irfs as follows

from_shock = 1
effect_irfs = transmission(from_shock, irfs, irfs_ortho, cond; method = :irfs)

where the irfs and irfs_ortho are the structural IRFs (with possibly NaN columns indicating unidentified shocks) and irfs_ortho are the orthogonalised IRFs. Both are in transmission form obtained using to_transmission_irfs.

The :BQbb method is often the more efficient one, and as implemented here can be used for all Boolean statements. Both the :BQbb and the :irfs method return a vector with index i being the transmission effect on $x_i$. If $x_k$ is the variable with the highest subscript involved in the Boolean statement, then the first k entries in the returned vector are NaN since interpretation of those effects is nonsensical.

Internals

The internals of TransmissionChannelAnalysis.jl all revolve around the type Q which represents a transmission condition or query. It has two fields. The vars field is a Vector{String} field with each element corresponding to a term in a transmission condition. The second field is the multiplier field which is a Vector{Number} and holds the multiplier for each term. For example

• The condition $Q[x_1]$ results in a single element in vars corresponding to "x1" and a single element in multiplier equal to 1.
• The condition $Q[x_1 \land x_2]$ results in a single element in vars corresponding to "x2 & x1" and a single element in multiplier corresponding to 1.
• The condition $Q[x_1 \lor x_2] = Q[x_1] + Q[x_2] - Q[x_1 \land x_2]$ results in three elements in vars corresponding to "x1", "x2", and "x2 & x1" and three elements in multiplier corresponding to 1, 1, and -1 respectively.

Starting with simple queries of the form Q("x1"), and Q("x2"), the overloaded operators yield simplifications consistent with the algebra in the paper. This consistency cannot be guaranteed if one is not starting with these "atomic" queries.

Q follows three rules:

1. Assume $Q[b] = \sum_{i=1}^{N_1}m_i Q[b_i]$ and $Q[b^] =\sum_{j=1}^{N_2}m^jQ[b^_j]$. Then $Q[b \land b^] = \sum{i=1}^{N_1}\sum_{j=1}^{N_2}m_im^_jQ[b_i \land b^_j]$
2. $Q[b \lor b^] = Q[b] + Q[b^] - Q[b \land b^*]$
3. $Q[b \land \neg b^] = Q[b] - Q[b \land b^]$

Rule (2) and (3) are included in the rules in the paper. Rule (1), on the other hand, requires some more explanation. For this, I will first show that if $Q[b] = \sum_{i = 1}^{N_1} m_i Q[b_i]$ then $Q[b \land b^] = \sum_{i = 1}^{N_1} m_i Q[b_i \land b^]$.

In order for $Q[b]$ to simplify to $\sum_{i =1}^{N_1}Q[b_i]$, either rule (2) or rule (3) must be applied at some point, because if $b$ was only to consist of conjunctions, then the term would never be split (follows from the algebra). Thus, if $b$ was only consisting of conjunctions, then the statement above is trivially true. Now assume that simplification of $Q[b]$ to the sum of terms takes $I$ iterations of applying rules (2) or (3) to some terms. At iteration zero, where $Q[b]$ has not yet been split into other terms, $Q[b \land b^] = Q[b \land b^]$ is trivially true. Thus, assume that the statement is true at the beginning of some iteration $k$. We then must show that it is true at the end of iteration $k$ and thus at the beginning of iteration $k+1$. The only two operations that would further split a term are rules (2) and (3). Note though, that

1. $Q[(b_j \lor b_k) \land b^] = Q[(b_j \land b^) \lor (b_k \land b^)] = Q[b_j \land b^] + Q[b_k \land b^] - Q[b_j \land b_k \land b^]$
2. $Q[(b_j \land \neg b_k) \land b^] = Q[b_j \land b^] - Q[b_j \land b_k \land b^*]$

Thus, no-matter which rule is applied, the statement above would still hold for the split term, and thus for the entire sum. This proofs the correctness by induction and shows that $Q[b \land b^] = \sum_{i=1}^{N_1}m_iQ[b_i \land b^]$.

To obtain the full statement, use the same arguments as above, but this time, $b^$ takes the role that $b$ took above, and $b_i$ takes the role that $b^$ took above. This results in $Q[b_i \land b^] = \sum_{j=1}^{N_2}m^jQ[b_i \land b_j^]$. Thus, $Q[b \land b^] = \sum{i=1}^{N_1}\sum_{j=1}^{N_2}m_im^_jQ[b_i \land b^_j]$

We can thus implement the algebra in the paper by overwriting the AND (&), the OR (|), and the NOT (!) operator for the type of Q, and implement the rules above. This is what we internally do with the single exception that we do not further simplify negations of single variables. Reason for this is that it is faster to calculate the effect including the negation than it is to simplify the negated statements to a state in which each term is only a conjunction of ANDs.

There are obviously some more internal details. Please ask if you are interested.