# Linearly Transforming Variables in the VAR Model, How Does it Change the Impulse Response?

• Peter Reusens and Christophe Croux

## Abstract

This paper analyzes the impulse response function of vector autoregression models for variables that are linearly transformed. The impulse response is equal to the linear transformation of the original impulse response if and only if the shock is equal to the linear transformation of the original shock. In particular, we consider shocks in one error term only, generalized shocks, structural shocks identified by short-run recursive restrictions and structural shocks identified by long-run recursive restrictions. A vector autoregression model with expected inflation, the overnight rate and a long term ex-ante real interest rate that replaces the corresponding long term nominal interest rate, illustrates our results.

JEL Classification: C30; C32

## Acknowledgments

We are grateful to the Editor and two anonymous referees for their helpful comments and suggestions. Furthermore, we thank Ferre De Graeve, Geert Dhaene, Gerdie Everaert, Michele Lenza and Frank Smets for comments on an earlier version of the manuscript. Finally, financial support from the Agency for Innovation by Science and Technology (IWT) in Flanders is kindly acknowledged.

## Appendix

In the proofs below, the subscripts “.j”, “j.” and “jj” of a matrix respectively denote the jth column, the jth row and the element of the jth row and jth column of that matrix.

### Proof of Proposition 1

1. Let us assume that δ* and δ are economically equivalent, i.e. δ* = Aδ. Then, it follows from (3), (4) and (5) that, for each s,

Is=Ψsδ=AΨsA1δ=AΨsA1Aδ=AIs,

which proves that the impulse responses Is and Is are economically equivalent □.

2. Assume that for a given s > 0, that Is and Is are economically equivalent and that Ψs is invertible. Then, Ψs is also invertible since A is assumed to be invertible. It follows from (3), (4) and (5) that

δ=Ψs1Is=(AΨsA1)1AIs=AΨs1Is=Aδ,

which proves that the shocks δ* and δ are economically equivalent. □

### Proof of Proposition 2

Take, without loss of generalization, j = 1. The shocks can be written as

δ=(δ1,00)Tandδ=(δ1,00)T.
1. For a one standard deviation shock, δ1 and δ1 are set the standard deviation of the first error terms, denoted by Σ11 and Σ11, respectively. We have that Aδ is equal to A.1Σ11. Let us assume that the conditions of Proposition 2a are satisfied for j = 1. Since A1. has zero elements except for the first position, it follows from (2) that A11Σ11=Σ11 and hence (Aδ)1=δ1, where the subscript 1 denotes the first element of the vector. Since A.1 has zero elements except for the first position, Aδ = δ*. □

2. For a one unit shock, both δ1 and δ1 are set to one. Under the conditions of Proposition 2b, Aδ simplifies to the first column of A, which equals δ*. Hence, by definition (6), the shocks δ and δ* are economically equivalent. □

### Proof of Proposition 3

Let us assume that the jth row of A contains zeros for all elements except for the jth element which is equal to a strictly positive number c. Using (2), (8) and the assumed structure on the jth row of A, δ* can be written as

δ=Σ.jΣjj=[AΣAT].j[AΣAT]jj=AΣ[AT].j[AΣAT]jj=AcΣ.jc2Σjj=AΣ.jΣjj=Aδ,

which proves the economic equivalence of the shocks δ and δ*. □

### Proof of Proposition 4

The proof is analogous to the proof of Proposition 3. □

### Proof of Proposition 5a

Let us assume that A is lower triangular. From (2) and (9), it follows that Σ* = AΣAT =APPTAT = (AP)(AP)T. As both A and P are lower triangular, AP is also lower triangular and therefore, AP is equal to the unique lower triangular matrix P* of the Cholesky decomposition of Σ*. Therefore, the shock δ* is equal to

δ=P.j=(AP).j=A(P).j=Aδ.

This proves that the shocks δ* and δ are economically equivalent. □

Before proving Proposition 5b, an intermediate result is needed.

### Lemma 1

Let B be the identity matrix of dimension n with rows i and k swapped, with either i, k < j or i, k > j and where j is a given number with 1 ≤ j ≤ n. Let P and PB be the lower triangular matrices of the Cholesky decompositions of the n × n dimensional symmetric matrices Σ and BΣBT, respectively. Then,

P.jB=BP.j.

### Proof of Lemma 1

Let j be a given number with 1 ≤ jn. B can be partitioned as

(14)B=(B10001000B2),

where the scalar 1 is at position (j, j) and B1 and B2 are square matrices respectively of size j − 1 and nj. Either B1 or B2 is an identity matrix and either B2 or B1 is an identity matrix where two rows are interchanged. Next, partition P in the same way as B:

(15)P=(P100P2p0P3P4P5),

where p is a scalar and P1 and P5 are lower triangular matrices, respectively of size (j − 1) and (nj). Note that we have a smaller number of sub-matrices in the partition of B and P if j = 1 or j = n.

Let LA and LB be the lower triangular matrices and QA and QB the orthogonal matrices of decompositions B1P1 = LAQA and B2P5 =LBQB. Then, BP can be written as

(16)BP=(B1P100P2p0B2P3B2P4B2P5)=L~Q,

with L~ the lower triangular matrix

(17)L~=(LA00P2QA1p0B2P3QA1B2P4LB),

and Q the orthogonal matrix

Q=(QA0001000QB).

From (9), (16) and the fact that Q is an orthogonal matrix, it follows that

BΣBT=BPPTBT=L~QQTL~T=L~L~T.

As L~ is lower triangular, it is equal to the unique lower triangular matrix PB of the Cholesky decomposition of the symmetric matrix BΣ BT. Hence, using (14), (15) and (17), it follows that

P.jB=L~.j=(0pB2P4)=B(0pP4)=BP.j,

which proves Lemma 1. □

### Proof of Proposition 5b

Let AB be the lower triangular matrix that is obtained by swapping the rows i and k and/or the columns i and k of A. Let BR be the identity matrix with swapping of the same rows as to obtain AB and let BC be the identity matrix with swapping of the same columns as to obtain AB. Note that both BR and BC are symmetric matrices. Then,

(18)AB=BRABCT

is lower triangular.

Denote

(19)ΣBC=BCΣBCT
(20)ΣBR=BRΣBRT.

Applying Lemma 1 yields

(21)P.jBC=BCP.j
(22)P.jBR=BRP.j,

where PBC and P*BR are the lower triangular matrices of the Cholesky decompositions of ΣBC and Σ*BR, respectively.

Since BR and BC are either “identity matrices” or “swapping matrices”, BRT=BR1 and BCT=BC1 and using (2), (18), (19) and (20), it follows that

(23)ΣBR=BRAΣATBRT=BRABCTBCΣBCTBCATBRT=ABPBC(ABPBC)T.

As lower triangularity of AB and PBC implies that also ABPBC is lower triangular, it follows from (23) that

(24)PBR=ABPBC.

From (18), (21) and (24) and using BCTBC=I, it then follows that

(25)P.jBR=ABP.jBC=BRAP.j.

Finally, equaling the expressions for P.jBR in (22) and (25) and premultiplying both sides of the equation by BR1 gives

P.j=AP.j,

which proves that shocks δ and δ* are economically equivalent. □

### Proof of Proposition 6a

The proof is similar to the proof of Proposition 5a. Let us assume that A is lower triangular with the jth diagonal element equal to 1. Next, define X as a diagonal matrix with diagonal elements equal to those of AL, which is a lower triangular matrix with the jth diagonal element equal to 1. Note that since A is invertible, X−1 exists. Using (2) and (10) it then follows that

Σ=AΣAT=ALDLTAT=(ALX1)(XDX)(ALX1)T.

Hence, ALX−1 and XDX are respectively equal to L* and D* of the unique decomposition Σ* = L*D*L*T. Since X is a diagonal matrix with jth diagonal element equal to one, it follows that

L.j=(ALX1).j=AL(X1).j=AL.j,

which proves the economic equivalence of the shocks δ and δ*. □

Before proving Proposition 6b, an intermediate result is needed.

### Lemma 2

Let B be the identity matrix of dimension n with rows i and k swapped, with either i, k < j or i, k > j and where j is a given number with 1 ≤ j ≤ n. Let L and LB be the lower unitriangular matrices and D and DB be the diagonal matrices of the decompositions Σ = LDLT and BΣ BT = LBDB(LB)T, where Σ is a n × n dimensional symmetric matrix. Then,

L.jB=BL.j.

### Proof of Lemma 2

Let j be a given number with 1 ≤ jn. B can be partitioned as

(26)B=(B10001000B2),

where the scalar 1 is at position (j, j) and B1 and B2 are square matrices respectively of size j − 1 and nj. Either B1 or B2 is an identity matrix and either B2 or B1 is an identity matrix where two rows are interchanged. Next, partition L and D in the same way as B

(27)L=(L100L210L3L4L5)
D=(D1000d000D2),

where L1 and L5 are lower unitriangular matrices, respectively of size j − 1 and nj, d is a scalar and D1 and D2 are diagonal matrices, respectively of size j − 1 and nj. Note that we have a smaller number of sub-matrices in the partition of B, L and D if j = 1 or j = n.

Let D1/2, D11/2 and D21/2 be the principal square root of the diagonal matrices D, D1 and D2 and let LA and LB be the lower triangular matrices and QA and QB the orthogonal matrices of the decompositions B1L1D11/2=LAQA and B2L5D21/2=LBQB. Then,

(28)BLD1/2=(B1L1D11/200L2D11/2d1/20B2L3D11/2B2L4d1/2B2L5D21/2)=L~Q,

with L~ the lower triangular matrix

(29)L~=(LA00L2D11/2QA1d1/20B2L3D11/2QA1B2L4d1/2LB),

and Q the orthogonal matrix

Q=(QA0001000QB).

Define X~ as a diagonal matrix with diagonal elements equal to those of L~. Since these diagonal elements are strictly greater than zero, X~−1 exists. From (26), (28) and the fact that Q is an orthogonal matrix, it follows that

BΣBT=BLDLTBT=L~QQTL~T=L~L~T=(L~X~1)(X~X~)(L~X~1)T,

where L~X~−1 is a lower unitriangular matrix and X~X~ is a diagonal matrix. Hence, L~X~−1 is equal to the lower unitriangular matrix LB of the decomposition ΣB = LBDB(LB)T. Therefore, using (26), (27) and (29), it follows that

L.jB=(L~X~1).j=(01B2L4)=B(01L4)=BL.j,

which proves Lemma 2. □

### Proof of Proposition 6b

The proof is similar to the proof of Proposition 5b. Let AB be the lower triangular matrix with the jth diagonal element equal to 1 that is obtained by swapping the rows i and k and/or the columns i and k of A. Let BR be the identity matrix with swapping of the same rows as to obtain AB and let BC be the identity matrix with swapping of the same columns as to obtain AB. Note that both BR and BC are symmetric matrices. Then,

(30)AB=BRABCT

is lower triangular with the jth diagonal element equal to 1.

Denote

(31)ΣBC=BCΣBCT
(32)ΣBR=BRΣBRT.

Applying Lemma 2 yields

(33)L.jBC=BCL.j
(34)L.jBR=BRL.j,

where LBC and L*BR are the lower unitriangular matrices of the decompositions ΣBC = LBCDBC(LBC)T and Σ*BR = L*BRD*BR(L*BR)T, where DBC and D*BR are diagonal matrices.

Since BR and BC are either “identity matrices” or “swapping matrices”, BRT=BR1 and BCT=BC1 and using (2), (30), (31) and (32), it follows that

ΣBR=BRAΣATBRT=BRABCTBCΣBCTBCATBRT=ABΣBC(AB)T.

Using the assumption that AB is lower triangular with the jth diagonal element equal to 1, a similar derivation as in the proof of Proposition 6a can show that

(35)L.jBR=ABL.jBC.

From (30), (33) and (35) and using BCTBC=I, it then follows that

(36)L.jBR=ABL.jBC=BRABC1BCL.j=BRAL.j.

Finally, equaling the expressions for L.jBR in (34) and (36) and premultiplying both sides of the equation by BR1 gives

L.j=AL.j,

which proves that shocks δ and δ* are economically equivalent.

### Proof of Proposition 7a

Let us assume that A is lower triangular. From (1), (2), (12) and (13), it follows that

Φ(1)1ΣΦ(1)T=AΦ(1)1ΣΦ(1)TAT=(AΞ)(AΞ)T.

As both A and Ξ are lower triangular, is also lower triangular and therefore, is equal to the unique lower triangular matrix Ξ* of the Cholesky decomposition (13).

Therefore, the shock δ* is equal to

δ=(Φ(1)Ξ).j=Φ(1)(Ξ).j=AΦ(1)A1AΞ.j=A(Φ(1)Ξ).j=Aδ,

which proves that shocks δ and δ* are economically equivalent.

### Proof of Proposition 7b

The proof is similar in spirit to the proof of Proposition 5b. Let AB be the lower triangular matrix that is obtained by swapping the rows i and k and/or the columns i and k of A. Let BR be the identity matrix with swapping of the same rows as to obtain AB and let BC be the identity matrix with swapping of the same columns as to obtain AB. Note that both BR and BC are symmetric matrices. Then,

(37)AB=BRABCT

is lower triangular.

Denote

(38)ΣBC=BCΦ(1)1ΣΦ(1)TBCT
(39)ΣBR=BRΦ(1)1ΣΦ(1)TBRT.

Given that Φ(1)−1Σ Φ(1)T and Φ*(1)−1Σ*Φ*(1)T are symmetric matrices, we can apply Lemma 1 such that

(40)Ξ.jBC=BCΞ.j
(41)Ξ.jBR=BRΞ.j,

where ΞBC and Ξ*BR are the lower triangular matrices of the Cholesky decompositions of ΣBC and Σ*BR, respectively.

Since BR and BC are either “identity matrices” or “swapping matrices”, BRT=BR1 and BCT=BC1 and using (1), (2), (37), (38) and (39), it follows that

(42)ΣBR=BRΦ(1)1ΣΦ(1)TBRT=BRAΦ(1)1A1AΣATATΦ(1)TATBRT=BRABCTBCΦ(1)1ΣΦ(1)TBCTBCATBRT=ABΞBC(ABΞBC)T.

As lower triangularity of AB and ΞBC implies that also ABΞBC is lower triangular, it follows from (42) that

(43)ΞBR=ABΞBC.

From (37), (40) and (43) and using BCTBC=I, it then follows that

(44)Ξ.jBR=ABΞ.jBC=BRAΞ.j.

Finally, equaling the expressions for Ξ.jBR in (41) and (44) and premultiplying both sides of the equation by BR1 gives

Ξ.j=AΞ.j.

Therefore, the shock δ* is equal to

δ=[Φ(1)1Ξ].j=AΦ(1)A1AΞ.j=A[Φ(1)Ξ].j=Aδ,

which proves that shocks δ and δ* are economically equivalent. □

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Published Online: 2017-7-6

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