Act I · The equations are real

What changes when both variables respond?

Where we are: give the predictor an equation

C02 · Module 8’s ARDL → a connected system

yt=α+ϕ1yt−1+β0xt+β1xt−1+ϵty_t=\alpha+\phi_1y_{t-1}+\beta_0x_t+\beta_1x_{t-1}+\epsilon_t

Mixing-board channel Today
Own memory Keep one lag
Cross-variable history Let both outcomes depend on it
Current connections Allow reciprocal dependence
Equations Give each outcome its own equation
Module 10 let two variables adjust. What if we model both current outcomes together?

One disturbance, a connected pond

C03 · Disturbance location ≠ observed movement

A stone disturbs a connected body of water.

A disturbance starts somewhere.

Connections transmit it elsewhere.

Does movement at a second point tell us where the stone entered?

Think about inflation and the policy rate.

The ARDL already had a current connection

C04 · One conditional outcome

Dated graph: past y and x and current x enter current y; current x has no supplied equation.

yt=α+ϕ1yt−1y_t=\alpha+\phi_1y_{t-1} +β0xt+β1xt−1+ϵt\quad+\beta_0x_t+\beta_1x_{t-1}+\epsilon_t

Which current variable still lacks an equation?

Give current x its own equation

C05 · A second outcome, the same history

Both current outcomes now receive arrows from yesterday’s y and x; current x still enters y.

yt=αy+byxxty_t=\alpha_y+b_{yx}x_t +ϕyyyt−1+ϕyxxt−1+ηy,t\quad+\phi_{yy}y_{t-1}+\phi_{yx}x_{t-1}+\eta_{y,t} xt=αx+ϕxyyt−1x_t=\alpha_x+\phi_{xy}y_{t-1} +ϕxxxt−1+ηx,t\quad+\phi_{xx}x_{t-1}+\eta_{x,t}

Which terms record the new arrows?

Turn on reciprocal current dependence

C06 · The new term creates a current cycle

Separate current arrows x to y and y to x form a directed cycle.

yt=αy+byxxty_t=\alpha_y+b_{yx}x_t +ϕyyyt−1+ϕyxxt−1+ηy,t\quad+\phi_{yy}y_{t-1}+\phi_{yx}x_{t-1}+\eta_{y,t} xt=αx+bxyyt⏟new current connectionx_t=\alpha_x+\underbrace{b_{xy}y_t}_{\text{new current connection}} +ϕxyyt−1+ϕxxxt−1+ηx,t\quad+\phi_{xy}y_{t-1}+\phi_{xx}x_{t-1}+\eta_{x,t}

Can we calculate either current value without the other?

Read the structural pair

C07 · First subscript: receiving equation · second: entering variable

yt=αy+byxxt+ϕyyyt−1+ϕyxxt−1+ηy,t,xt=αx+bxyyt+ϕxyyt−1+ϕxxxt−1+ηx,t.\begin{aligned} y_t&=\alpha_y+b_{yx}x_t+\phi_{yy}y_{t-1}+\phi_{yx}x_{t-1}+\eta_{y,t},\\[.5em] x_t&=\alpha_x+b_{xy}y_t+\phi_{xy}y_{t-1}+\phi_{xx}x_{t-1}+\eta_{x,t}. \end{aligned}

b: current connectionsφ: lagged connectionsη: structural shocks

Our shocks have zero means, positive variances, no serial/past dependence, and zero covariance with each other.

Must the unexpected movements in y and x also be uncorrelated?

Act II · Where did that coefficient come from?

Solve for today’s values using yesterday’s information.

Track the x-lag coefficient

C08 · Our question through the algebra

yt=αy+byxxt+ϕyyyt−1+ϕyxxt−1⏟direct x-lag term+ηy,t,xt=αx+bxyyt+ϕxyyt−1+ϕxxxt−1+ηx,t.\begin{aligned} y_t&=\alpha_y+b_{yx}x_t+\phi_{yy}y_{t-1} +\underbrace{\phi_{yx}x_{t-1}}_{\text{direct x-lag term}}+\eta_{y,t},\\[.5em] x_t&=\alpha_x+b_{xy}y_t+\phi_{xy}y_{t-1}+\phi_{xx}x_{t-1}+\eta_{x,t}. \end{aligned}

If yesterday’s x has no direct connection to today’s y, could it still help predict today’s y?

First, xt and yt are determined together. Each appears as a regressor in the other equation and can be correlated with that equation’s structural shock.

Substitution gives us equations using past values plus a current innovation—the form we can estimate with OLS under our shock assumptions.

Substitute for current x

C09 · Find the coefficient on yesterday’s x in today’s y

yt=αy+byxxt+ϕyyyt−1+ϕyxxt−1+ηy,ty_t=\alpha_y+b_{yx}x_t+\phi_{yy}y_{t-1}+\phi_{yx}x_{t-1}+\eta_{y,t}

Replace xtx_t by the entire right-hand side of its equation:

yt=αy+byx[αx+bxyyt+ϕxyyt−1+ϕxxxt−1+ηx,t]+ϕyyyt−1+ϕyxxt−1+ηy,t.\begin{aligned} y_t={}&\alpha_y+b_{yx}\Big[\alpha_x+b_{xy}y_t +\phi_{xy}y_{t-1}+\phi_{xx}x_{t-1}+\eta_{x,t}\Big]\\ &+\phi_{yy}y_{t-1}+\phi_{yx}x_{t-1}+\eta_{y,t}. \end{aligned}

What did we replace? Which current variable remains?

Multiply every term by byx

C10 · The current variable comes back

yt=αy+byxαx+byxbxyyt⏟returning current y+byxϕxyyt−1+byxϕxxxt−1+byxηx,t+ϕyyyt−1+ϕyxxt−1+ηy,t.\begin{aligned} y_t={}&\alpha_y+b_{yx}\alpha_x +\underbrace{b_{yx}b_{xy}y_t}_{\text{returning current y}}\\ &+b_{yx}\phi_{xy}y_{t-1}+b_{yx}\phi_{xx}x_{t-1}+b_{yx}\eta_{x,t}\\ &+\phi_{yy}y_{t-1}+\phi_{yx}x_{t-1}+\eta_{y,t}. \end{aligned}

Find both contributions multiplying xt−1.

Collect the current variable and the lags

C11 · Four groups survive the substitution

(1−byxbxy)yt=αy+byxαx⏟intercept+(ϕyy+byxϕxy)yt−1⏟y lag+(ϕyx+byxϕxx)xt−1⏟target: x lag+ηy,t+byxηx,t⏟shocks.\begin{aligned} (1-b_{yx}b_{xy})y_t={}&\underbrace{\alpha_y+b_{yx}\alpha_x}_{\text{intercept}}\\[.35em] &+\underbrace{(\phi_{yy}+b_{yx}\phi_{xy})y_{t-1}}_{\text{y lag}}\\[.35em] &+\underbrace{(\phi_{yx}+b_{yx}\phi_{xx})x_{t-1}}_{\text{target: x lag}}\\[.35em] &+\underbrace{\eta_{y,t}+b_{yx}\eta_{x,t}}_{\text{shocks}}. \end{aligned}

Why do φyx and byxφxx belong together?

Divide by the common factor

C12 · Define D = 1 − byxbxy ≠ 0

yt=αy+byxαxD+ϕyy+byxϕxyDyt−1+ϕyx+byxϕxxDxt−1+ηy,t+byxηx,tD.\begin{aligned} y_t={}&\frac{\alpha_y+b_{yx}\alpha_x}{D} +\frac{\phi_{yy}+b_{yx}\phi_{xy}}{D}y_{t-1}\\[.6em] &+\frac{\phi_{yx}+b_{yx}\phi_{xx}}{D}x_{t-1} +\frac{\eta_{y,t}+b_{yx}\eta_{x,t}}{D}. \end{aligned}

Where did current x go?

Into the transformed intercept, lag coefficients, and innovation.

Read the transformed x-lag coefficient

C13 · One coefficient combines several connections

ayx=ϕyx⏞direct lag+byxϕxx⏞through current xDa_{yx}=\frac{\overbrace{\phi_{yx}}^{\text{direct lag}} +\overbrace{b_{yx}\phi_{xx}}^{\text{through current x}}}{D}

xt−1→ϕxxxt→byxytx_{t-1}\ \xrightarrow{\ \phi_{xx}\ }\ x_t \ \xrightarrow{\ b_{yx}\ }\ y_t

D accounts for solving the reciprocal current system.
If the direct lag disappears, does the other route disappear too?

Now substitute for current y

C14 · Solve the second equation just as carefully

xt=αx+bxyyt+ϕxyyt−1+ϕxxxt−1+ηx,tx_t=\alpha_x+b_{xy}y_t+\phi_{xy}y_{t-1}+\phi_{xx}x_{t-1}+\eta_{x,t}

Replace yty_t by the entire right-hand side of its structural equation:

xt=αx+bxy[αy+byxxt+ϕyyyt−1+ϕyxxt−1+ηy,t]+ϕxyyt−1+ϕxxxt−1+ηx,t.\begin{aligned} x_t={}&\alpha_x+b_{xy}\Big[\alpha_y+b_{yx}x_t +\phi_{yy}y_{t-1}+\phi_{yx}x_{t-1}+\eta_{y,t}\Big]\\ &+\phi_{xy}y_{t-1}+\phi_{xx}x_{t-1}+\eta_{x,t}. \end{aligned}

Which current variable will return this time?

Expand the second equation

C15 · Keep the subscripts attached to their equations

xt=αx+bxyαy+bxybyxxt⏟returning current x+bxyϕyyyt−1+bxyϕyxxt−1+bxyηy,t+ϕxyyt−1+ϕxxxt−1+ηx,t.\begin{aligned} x_t={}&\alpha_x+b_{xy}\alpha_y +\underbrace{b_{xy}b_{yx}x_t}_{\text{returning current x}}\\ &+b_{xy}\phi_{yy}y_{t-1}+b_{xy}\phi_{yx}x_{t-1}+b_{xy}\eta_{y,t}\\ &+\phi_{xy}y_{t-1}+\phi_{xx}x_{t-1}+\eta_{x,t}. \end{aligned}

Name the two y-lag contributions and the two x-lag contributions.

Collect the second equation

C16 · The same reciprocal product returns

(1−bxybyx)xt=αx+bxyαy⏟intercept+(ϕxy+bxyϕyy)yt−1⏟y lag+(ϕxx+bxyϕyx)xt−1⏟x lag+ηx,t+bxyηy,t⏟shocks.\begin{aligned} (1-b_{xy}b_{yx})x_t={}&\underbrace{\alpha_x+b_{xy}\alpha_y}_{\text{intercept}}\\[.35em] &+\underbrace{(\phi_{xy}+b_{xy}\phi_{yy})y_{t-1}}_{\text{y lag}}\\[.35em] &+\underbrace{(\phi_{xx}+b_{xy}\phi_{yx})x_{t-1}}_{\text{x lag}}\\[.35em] &+\underbrace{\eta_{x,t}+b_{xy}\eta_{y,t}}_{\text{shocks}}. \end{aligned}

Why is the factor on the left still D?

Solve for current x

C17 · Every group is divided by D

xt=αx+bxyαyD+ϕxy+bxyϕyyDyt−1+ϕxx+bxyϕyxDxt−1+ηx,t+bxyηy,tD.\begin{aligned} x_t={}&\frac{\alpha_x+b_{xy}\alpha_y}{D} +\frac{\phi_{xy}+b_{xy}\phi_{yy}}{D}y_{t-1}\\[.6em] &+\frac{\phi_{xx}+b_{xy}\phi_{yx}}{D}x_{t-1} +\frac{\eta_{x,t}+b_{xy}\eta_{y,t}}{D}. \end{aligned}

Which structural shock appears in both solved equations?
The current variables are now on the left; the current shocks still enter on the right.

Give the transformed quantities names

C18 · New letters name expressions we already derived

yt=ay0+ayyyt−1+ayxxt−1+ϵy,t,xt=ax0+axyyt−1+axxxt−1+ϵx,t.\begin{aligned} y_t&=a_{y0}+a_{yy}y_{t-1}+a_{yx}x_{t-1}+\epsilon_{y,t},\\[.5em] x_t&=a_{x0}+a_{xy}y_{t-1}+a_{xx}x_{t-1}+\epsilon_{x,t}. \end{aligned}

ayx=ϕyx+byxϕxxD,ϵy,t=ηy,t+byxηx,tDa_{yx}=\frac{\phi_{yx}+b_{yx}\phi_{xx}}D, \qquad \epsilon_{y,t}=\frac{\eta_{y,t}+b_{yx}\eta_{x,t}}D

Which coefficient will a regression on the two lags estimate?

Zero direct cross-lag, nonzero reduced cross-lag

C19 · Return to our prediction

yt=0.2+0.5xt+0.3yt−1+0xt−1⏟ϕyx=0+ηy,t,xt=0.1+0.25yt+0.1yt−1+0.4xt−1+ηx,t.\begin{aligned} y_t&=0.2+0.5x_t+0.3y_{t-1}+\underbrace{0x_{t-1}}_{\phi_{yx}=0}+\eta_{y,t},\\[.35em] x_t&=0.1+0.25y_t+0.1y_{t-1}+0.4x_{t-1}+\eta_{x,t}. \end{aligned}

D=1−(0.5)(0.25)=78D=1-(0.5)(0.25)=\frac78

ayx=0+(0.5)(0.4)0.875=835≈0.229a_{yx}=\frac{0+(0.5)(0.4)}{0.875}=\frac8{35}\approx0.229

The equations are real. That reduced coefficient is a mixture of their connections.

Act III · Record the same calculation with matrices

The compact notation must reproduce our scalar answer.

A vector is an ordered column

C20 · Two rows, one column · always y then x

𝐘t=(ytxt),𝚪0=(αyαx),𝛈t=(ηy,tηx,t)\mathbf Y_t=\begin{pmatrix}y_t\\x_t\end{pmatrix},\qquad \boldsymbol\Gamma_0=\begin{pmatrix}\alpha_y\\\alpha_x\end{pmatrix},\qquad \boldsymbol\eta_t=\begin{pmatrix}\eta_{y,t}\\\eta_{x,t}\end{pmatrix}

Each object is 2 × 1.

Row 1 belongs to the yy equation. Row 2 belongs to the xx equation.

What breaks if one column silently reverses the order?

Move current terms to the left

C21 · B records current connections

yt−byxxt=αy+ϕyyyt−1+ϕyxxt−1+ηy,t,−bxyyt+xt=αx+ϕxyyt−1+ϕxxxt−1+ηx,t.\begin{aligned} y_t-b_{yx}x_t&=\alpha_y+\phi_{yy}y_{t-1}+\phi_{yx}x_{t-1}+\eta_{y,t},\\[.35em] -b_{xy}y_t+x_t&=\alpha_x+\phi_{xy}y_{t-1}+\phi_{xx}x_{t-1}+\eta_{x,t}. \end{aligned}

𝐁=(1−byx−bxy1),𝚪1=(ϕyyϕyxϕxyϕxx)\mathbf B=\begin{pmatrix}1&-b_{yx}\\-b_{xy}&1\end{pmatrix},\qquad \mathbf\Gamma_1=\begin{pmatrix}\phi_{yy}&\phi_{yx}\\\phi_{xy}&\phi_{xx}\end{pmatrix}

Rows receive Columns enter
𝐁\mathbf B y equation, x equation current y, x
𝚪1\mathbf\Gamma_1 y equation, x equation lagged y, x

One row times one column

C22 · Multiply matching entries, then add

(1−byx−bxy1)⏟2×2(ytxt)⏟2×1=(yt−byxxt−bxyyt+xt)⏟2×1\underbrace{\begin{pmatrix}1&-b_{yx}\\-b_{xy}&1\end{pmatrix}}_{2\times2} \underbrace{\begin{pmatrix}y_t\\x_t\end{pmatrix}}_{2\times1} =\underbrace{\begin{pmatrix}y_t-b_{yx}x_t\\-b_{xy}y_t+x_t\end{pmatrix}}_{2\times1}

What is the first component of Γ1Yt−1?

(ϕyyϕyxϕxyϕxx)(yt−1xt−1)=(ϕyyyt−1+ϕyxxt−1ϕxyyt−1+ϕxxxt−1)\begin{pmatrix}\phi_{yy}&\phi_{yx}\\\phi_{xy}&\phi_{xx}\end{pmatrix} \begin{pmatrix}y_{t-1}\\x_{t-1}\end{pmatrix} =\begin{pmatrix}\phi_{yy}y_{t-1}+\phi_{yx}x_{t-1}\\\phi_{xy}y_{t-1}+\phi_{xx}x_{t-1}\end{pmatrix}

Two scalar equations, one vector equation

C23 · The notation changed; the model did not

𝐁𝐘t=𝚪0+𝚪1𝐘t−1+𝛈t\mathbf B\mathbf Y_t=\boldsymbol\Gamma_0+\mathbf\Gamma_1\mathbf Y_{t-1}+\boldsymbol\eta_t

Piece Meaning Resulting dimension
𝐁𝐘t\mathbf B\mathbf Y_t Current values and their connections 2×12\times1
𝚪0\boldsymbol\Gamma_0 Structural intercepts 2×12\times1
𝚪1𝐘t−1\mathbf\Gamma_1\mathbf Y_{t-1} Structural lag terms 2×12\times1
𝛈t\boldsymbol\eta_t Structural shocks 2×12\times1
Read the first row back as a scalar equation.

What would undo multiplication by B?

C24 · Start with the operation we want

𝐈=(1001),𝐈𝐘t=𝐘t\mathbf I=\begin{pmatrix}1&0\\0&1\end{pmatrix},\qquad \mathbf I\mathbf Y_t=\mathbf Y_t

An inverse satisfies

𝐁−1𝐁=𝐈.\mathbf B^{-1}\mathbf B=\mathbf I.

It undoes multiplication by 𝐁\mathbf B.

The inverse is a matrix. It is not the reciprocal of every entry.

Verify the two-by-two inverse

C25 · Our scalar denominator reappears

𝐁−1=1D(1byxbxy1),D=1−byxbxy≠0\mathbf B^{-1}=\frac1D\begin{pmatrix}1&b_{yx}\\b_{xy}&1\end{pmatrix}, \qquad D=1-b_{yx}b_{xy}\ne0

1D(1byxbxy1)(1−byx−bxy1)=1D(D00D)=𝐈\frac1D\begin{pmatrix}1&b_{yx}\\b_{xy}&1\end{pmatrix} \begin{pmatrix}1&-b_{yx}\\-b_{xy}&1\end{pmatrix} =\frac1D\begin{pmatrix}D&0\\0&D\end{pmatrix}=\mathbf I

Top left:1−byxbxy=DTop right:−byx+byx=0\begin{aligned} \text{Top left:}&\quad 1-b_{yx}b_{xy}=D\\ \text{Top right:}&\quad -b_{yx}+b_{yx}=0 \end{aligned}

Supply the bottom row. Where have we seen D before?

Apply the inverse to the whole equation

C26 · Multiply every term on the left by B−1

𝐁−1𝐁⏟𝐈𝐘t=𝐁−1𝚪0+𝐁−1𝚪1𝐘t−1+𝐁−1𝛈t\underbrace{\mathbf B^{-1}\mathbf B}_{\mathbf I}\mathbf Y_t =\mathbf B^{-1}\boldsymbol\Gamma_0 +\mathbf B^{-1}\mathbf\Gamma_1\mathbf Y_{t-1} +\mathbf B^{-1}\boldsymbol\eta_t

𝐀0=𝐁−1𝚪0intercepts𝐀1=𝐁−1𝚪1lag coefficients𝛜t=𝐁−1𝛈tinnovations\begin{aligned} \mathbf A_0&=\mathbf B^{-1}\boldsymbol\Gamma_0&&\text{intercepts}\\ \mathbf A_1&=\mathbf B^{-1}\mathbf\Gamma_1&&\text{lag coefficients}\\ \boldsymbol\epsilon_t&=\mathbf B^{-1}\boldsymbol\eta_t&&\text{innovations} \end{aligned}

𝐘t=𝐀0+𝐀1𝐘t−1+𝛜t\mathbf Y_t=\mathbf A_0+\mathbf A_1\mathbf Y_{t-1}+\boldsymbol\epsilon_t

A0 is the reduced intercept vector. B is the current-coefficient matrix.

Find our coefficient inside the matrix product

C27 · Predict the row and column, then reveal their product

1D(1byx)(ϕyxϕxx)=ϕyx+byxϕxxD=ayx\frac1D\begin{pmatrix}1&b_{yx}\end{pmatrix} \begin{pmatrix}\phi_{yx}\\\phi_{xx}\end{pmatrix} =\frac{\phi_{yx}+b_{yx}\phi_{xx}}D=a_{yx}

Row 1 receives in the y equation; column 2 enters as lagged x.

1(0)+0.5(0.4)0.875=835\frac{1(0)+0.5(0.4)}{0.875}=\frac8{35}

Static fallback

Row 1 of B−1 times column 2 of Γ1:

(1/D)φyx + (byx/D)φxx = ayx.

Receiving y equation; entering lagged x. For the fixed example: (1×0 + 0.5×0.4)/0.875 = 8/35.

Let R confirm the calculation

C28 · Live R · Let the computer do the arithmetic 
B <- matrix(c(1, -0.5, -0.25, 1), 2, byrow = TRUE)
Gamma1 <- matrix(c(0.3, 0, 0.1, 0.4), 2, byrow = TRUE)
solve(B, Gamma1)
y lag x lag
y equation 0.4 0.2286
x equation 0.2 0.4571

ayx=8/35≈0.2286a_{yx}=8/35\approx0.2286

Use the solver for more equations and lags. Keep the row/column interpretation.

Solving today and stability over time

C29 · Two different questions

Can we solve today’s system?

D=1−byxbxy≠0D=1-b_{yx}b_{xy}\ne0

𝐁\mathbf B is invertible.

Does propagation die away?

eigen(A1)$values

Eigenvalues: 0.644 and 0.213.

Both have modulus below one.
If we change the lag coefficients, must B change?

Act IV · What did OLS estimate?

Generate the system we know. Inspect the output we would usually read.

Independent disturbances can mix

C30 · Return to our prediction about unexpected movements

ϵy,t=ηy,t+byxηx,tD\epsilon_{y,t}=\frac{\eta_{y,t}+b_{yx}\eta_{x,t}}D

ϵx,t=bxyηy,t+ηx,tD\epsilon_{x,t}=\frac{b_{xy}\eta_{y,t}+\eta_{x,t}}D

The two innovations share structural-shock components.

Why can εy,t and εx,t covary even when the two η shocks do not?

Read the covariance matrix before its fractions

C31 · Two variances and one distinct covariance

𝚺=(Var(ϵy,t)Cov(ϵy,t,ϵx,t)Cov(ϵy,t,ϵx,t)Var(ϵx,t))\boldsymbol\Sigma=\begin{pmatrix} \operatorname{Var}(\epsilon_{y,t})&\operatorname{Cov}(\epsilon_{y,t},\epsilon_{x,t})\\ \operatorname{Cov}(\epsilon_{y,t},\epsilon_{x,t})&\operatorname{Var}(\epsilon_{x,t}) \end{pmatrix}

Cov(ϵy,t,ϵx,t)=bxyωy⏟shared y shock+byxωx⏟shared x shockD2\operatorname{Cov}(\epsilon_{y,t},\epsilon_{x,t}) =\frac{\underbrace{b_{xy}\omega_y}_{\text{shared y shock}} +\underbrace{b_{yx}\omega_x}_{\text{shared x shock}}}{D^2}

𝚺=(80/4948/4948/4968/49)\boldsymbol\Sigma=\begin{pmatrix}80/49&48/49\\48/49&68/49\end{pmatrix}

Why does the off-diagonal appear twice but supply only one number?

The loop is our reduced system

C32 · Live R · Seed 637611 · n = 2000 · burn-in = 250 
for (tt in seq_len(nrow(eta))) {
  Y[tt + 1, ] <- A0 + A1 %*% Y[tt, ] +
                B_inverse %*% eta[tt, ]
}

fit_y <- lm(y ~ y_lag + x_lag, data = dat)
fit_x <- lm(x ~ y_lag + x_lag, data = dat)

𝐘t=𝐀0+𝐀1𝐘t−1+𝛜t\mathbf Y_t=\mathbf A_0+\mathbf A_1\mathbf Y_{t-1}+\boldsymbol\epsilon_t

Both current values come from the joint solution, using the same past.

Three columns, three different objects

C33 · Which column should OLS approach?
Entry Structural Population reduced Estimated reduced
y intercept 0.2 0.2857 0.2550
y: y lag 0.3 0.4000 0.3756
y: x lag 0.0 0.2286 0.2742
x intercept 0.1 0.1714 0.1577
x: y lag 0.1 0.2000 0.1852
x: x lag 0.4 0.4571 0.4838

ϕyx=0⏟structuralayx=8/35⏟reducedâyx=0.274⏟estimated reduced\underbrace{\phi_{yx}=0}_{\text{structural}} \qquad \underbrace{a_{yx}=8/35}_{\text{reduced}} \qquad \underbrace{\hat a_{yx}=0.274}_{\text{estimated reduced}}

A useful estimate of the reduced coefficient does not recover the primitive zero.

What does the coefficient test?

C34 · An estimated reduced coefficient answers a reduced-form question
Estimate Std. Error t value Pr(>|t|)
x_lag 0.2742 0.0287 9.5477 <0.001

yt=ây0+âyyyt−1+âyxxt−1+ey,ty_t=\hat a_{y0}+\hat a_{yy}y_{t-1}+\hat a_{yx}x_{t-1}+e_{y,t}

ayx=0⇔ϕyx+byxϕxx=0(D≠0)a_{yx}=0\quad\Longleftrightarrow\quad\phi_{yx}+b_{yx}\phi_{xx}=0 \qquad(D\ne0)

Is that the same null as φyx = 0?

OLS uses innovations with conditional mean zero given the history. Cross-equation innovation covariance can remain.

Put current x back: does that recover its effect?

C35 · Same sample · Break the structural interpretation 
naive_y <- lm(y ~ x + y_lag + x_lag, data = dat)
naive_x <- lm(x ~ y + y_lag + x_lag, data = dat)
Connection True structural Naive OLS
current x into y 0.50 0.6841
current y into x 0.25 0.6143

Cov(xt,ηy,t∣ℱt−1)=bxyωyD≠0\operatorname{Cov}(x_t,\eta_{y,t}\mid\mathcal F_{t-1}) =\frac{b_{xy}\omega_y}{D}\ne0

Current x contains part of the structural disturbance in the y equation.

Act V · Can we recover the original system?

Now work backward. Put the generating coefficients out of view.

The analyst receives A0, A1, and Σ. The simulator’s truth is withheld.

What the reduced form gives us

C36 · Separate available information from the sought structure
Available to the analyst Sought under identifying assumptions
𝐀0\mathbf A_0: reduced intercepts 𝚪0\boldsymbol\Gamma_0: structural intercepts
𝐀1\mathbf A_1: reduced lags 𝚪1\mathbf\Gamma_1: structural lags
𝚺\boldsymbol\Sigma: innovation covariance 𝐁\mathbf B: current links; 𝛀\boldsymbol\Omega: shock variances

Σyy=ωy+byx2ωxD2,Σxx=bxy2ωy+ωxD2,Σyx=bxyωy+byxωxD2\Sigma_{yy}=\frac{\omega_y+b_{yx}^2\omega_x}{D^2},\quad \Sigma_{xx}=\frac{b_{xy}^2\omega_y+\omega_x}{D^2},\quad \Sigma_{yx}=\frac{b_{xy}\omega_y+b_{yx}\omega_x}{D^2}

𝚺=𝐁−1𝛀(𝐁−1)′,𝛀=(ωy00ωx)\boldsymbol\Sigma=\mathbf B^{-1}\boldsymbol\Omega(\mathbf B^{-1})', \qquad \boldsymbol\Omega=\begin{pmatrix}\omega_y&0\\0&\omega_x\end{pmatrix}

Three covariance numbers, four unknowns

C37 · Counting locates the missing information

(ΣyyΣyxΣyxΣxx)↔byx,bxy,ωy,ωx\begin{pmatrix}\Sigma_{yy}&\Sigma_{yx}\\\Sigma_{yx}&\Sigma_{xx}\end{pmatrix} \quad\longleftrightarrow\quad b_{yx},\ b_{xy},\ \omega_y,\ \omega_x

For a proposed compatible invertible 𝐁\mathbf B, set

𝚪0=𝐁𝐀0,𝚪1=𝐁𝐀1.\boldsymbol\Gamma_0=\mathbf B\mathbf A_0,\qquad \mathbf\Gamma_1=\mathbf B\mathbf A_1.

The unrestricted structural intercepts and lags can adapt to that choice.

What additional assumption would let us actually solve backward?

Remove one current arrow

C38 · Assume y first: byx = 0

yt=αy+ϕyyyt−1+ϕyxxt−1+ηy,ty_t=\alpha_y+\phi_{yy}y_{t-1}+\phi_{yx}x_{t-1}+\eta_{y,t} xt=αx+bxyyt+ϕxyyt−1+ϕxxxt−1+ηx,tx_t=\alpha_x+b_{xy}y_t+\phi_{xy}y_{t-1}+\phi_{xx}x_{t-1}+\eta_{x,t}

yt→bxyxty_t\ \xrightarrow{\ b_{xy}\ }\ x_t

ϵy,t=ηy,t,ϵx,t=bxyηy,t+ηx,t\epsilon_{y,t}=\eta_{y,t},\qquad \epsilon_{x,t}=b_{xy}\eta_{y,t}+\eta_{x,t}

Which connection did we exclude? Can lagged feedback remain?

Recover one quantity at a time

C39 · Variance → covariance ratio → remaining variance

Σyy=ωy⇒ωy=ΣyyΣyx=bxyωy⇒bxy=ΣyxΣyyΣxx=bxy2ωy+ωx⇒ωx=Σxx−Σyx2Σyy\begin{aligned} \Sigma_{yy}&=\omega_y &&\Rightarrow\quad\omega_y=\Sigma_{yy}\\[.7em] \Sigma_{yx}&=b_{xy}\omega_y &&\Rightarrow\quad b_{xy}=\frac{\Sigma_{yx}}{\Sigma_{yy}}\\[.7em] \Sigma_{xx}&=b_{xy}^2\omega_y+\omega_x &&\Rightarrow\quad \omega_x=\Sigma_{xx}-\frac{\Sigma_{yx}^2}{\Sigma_{yy}} \end{aligned}

What component are we subtracting from the second innovation’s variance?
A positive-definite Σ gives positive recovered variances.

A recovered structure under the chosen restriction

C40 · Apply y first to the fixed reduced covariance

bxy=48/4980/49=0.6,ωy=8049,ωx=0.8b_{xy}=\frac{48/49}{80/49}=0.6,\qquad \omega_y=\frac{80}{49},\qquad\omega_x=0.8

Have we recovered the original simultaneous generator?
Imposed/recovered y first Instructor-known generator
byxb_{yx} 0 assumed 0.5
bxyb_{xy} 0.6 recovered 0.25

The imposed zero is false in this simulation. The recursive structure still reproduces the reduced form.

Another restriction, the same reduced form

C41 · Choose an economic restriction; follow its coefficient through the system
Parameter Generating y first x first
byxb_{yx} 1/21/2 00 12/1712/17
bxyb_{xy} 1/41/4 3/53/5 00
ϕyx\phi_{yx} 00 8/358/35 −8/85-8/85
ωy\omega_y 11 80/4980/49 16/1716/17
ωx\omega_x 11 4/54/5 68/4968/49

All three reproduce the same 𝐀0\mathbf A_0, 𝐀1\mathbf A_1, and 𝚺\boldsymbol\Sigma.

Neither recursive zero is true for the generating system.

Static fallback
Parameter Generating y first x first
byx 1/2 0 12/17
bxy 1/4 3/5 0
φyx 0 8/35 −8/85
ωy 1 80/49 16/17
ωx 1 4/5 68/49

Both recursive restrictions reproduce the same reduced intercepts, lag matrix, and covariance. Neither zero matches the instructor-known generator.

Give three outcomes their own equations

C42 · Quarterly illustration · unemployment u, inflation π, federal funds rate i

ut=qu,t−1+buππt+buiit+ηu,t,πt=qπ,t−1+bπuut+bπiit+ηπ,t,it=qi,t−1+biuut+biππt+ηi,t.\begin{aligned} u_t&=q_{u,t-1}+b_{u\pi}\pi_t+b_{ui}i_t+\eta_{u,t},\\[.25em] \pi_t&=q_{\pi,t-1}+b_{\pi u}u_t+b_{\pi i}i_t+\eta_{\pi,t},\\[.25em] i_t&=q_{i,t-1}+b_{iu}u_t+b_{i\pi}\pi_t+\eta_{i,t}. \end{aligned}

𝐙t=(utπtit),𝐁3=(1−buπ−bui−bπu1−bπi−biu−biπ1)\mathbf Z_t=\begin{pmatrix}u_t\\\pi_t\\i_t\end{pmatrix},\qquad \mathbf B_3=\begin{pmatrix} 1&-b_{u\pi}&-b_{ui}\\ -b_{\pi u}&1&-b_{\pi i}\\ -b_{iu}&-b_{i\pi}&1 \end{pmatrix}

qj,t−1q_{j,t-1} collects the intercept and all three one-period lags. They remain unrestricted.

Nine unknowns, six covariance entries

C43 · Keep the normalization and shock assumptions fixed

diag(𝐁3)=𝟏,𝛀3=diag(ωu,ωπ,ωi),ωj>0\operatorname{diag}(\mathbf B_3)=\mathbf1,\qquad \boldsymbol\Omega_3=\operatorname{diag}(\omega_u,\omega_\pi,\omega_i),\qquad\omega_j>0

Current/shock unknowns Count
Current cross-variable coefficients 3×2=63\times2=6
Structural shock variances 33
Total 9

𝚺3=(ΣuuΣuπΣuiΣuπΣππΣπiΣuiΣπiΣii)\boldsymbol\Sigma_3=\begin{pmatrix} \Sigma_{uu}&\Sigma_{u\pi}&\Sigma_{ui}\\ \Sigma_{u\pi}&\Sigma_{\pi\pi}&\Sigma_{\pi i}\\ \Sigma_{ui}&\Sigma_{\pi i}&\Sigma_{ii} \end{pmatrix}

Three variances + three distinct covariances = 6.

The gap is 9 − 6 = 3 additional restrictions for the whole system.

Σ describes reduced-form innovations after conditioning on lags, not the raw macroeconomic series.

A recursive ordering removes three current arrows

C44 · Assume unemployment → inflation → policy within the quarter

𝐁3rec=(100−bπu10−biu−biπ1)\mathbf B_3^{\mathrm{rec}}=\begin{pmatrix} 1&0&0\\ -b_{\pi u}&1&0\\ -b_{iu}&-b_{i\pi}&1 \end{pmatrix}

buπ=bui=bπi=0b_{u\pi}=b_{ui}=b_{\pi i}=0

Allowed contemporaneous arrows under the illustrative ordering: unemployment to inflation, unemployment to policy, and inflation to policy.

Receiving equation Assumed current response within the quarter
Unemployment Does not respond to current inflation or policy
Inflation May respond to unemployment; does not respond to current policy
Federal funds rate May respond to both current unemployment and inflation
Which timing assumptions would you challenge? All lagged connections remain allowed.

Recover the third equation using the same logic

C45 · Work with the reduced innovations, after removing predictable history

ϵu,t=ηu,t,ϵπ,t=bπuϵu,t+ηπ,t,ϵi,t=biuϵu,t+biπϵπ,t+ηi,t\epsilon_{u,t}=\eta_{u,t},\quad \epsilon_{\pi,t}=b_{\pi u}\epsilon_{u,t}+\eta_{\pi,t},\quad \epsilon_{i,t}=b_{iu}\epsilon_{u,t}+b_{i\pi}\epsilon_{\pi,t}+\eta_{i,t}

The first two equations repeat our bivariate recovery:

ωu=Σuu,bπu=ΣuπΣuu,ωπ=Σππ−Σuπ2Σuu.\omega_u=\Sigma_{uu},\qquad b_{\pi u}=\frac{\Sigma_{u\pi}}{\Sigma_{uu}},\qquad \omega_\pi=\Sigma_{\pi\pi}-\frac{\Sigma_{u\pi}^2}{\Sigma_{uu}}.

The two policy coefficients must be solved together:

(ΣuuΣuπΣuπΣππ)(biubiπ)=(ΣuiΣπi)\begin{pmatrix}\Sigma_{uu}&\Sigma_{u\pi}\\\Sigma_{u\pi}&\Sigma_{\pi\pi}\end{pmatrix} \begin{pmatrix}b_{iu}\\b_{i\pi}\end{pmatrix} =\begin{pmatrix}\Sigma_{ui}\\\Sigma_{\pi i}\end{pmatrix}

Why isn’t Σui/Σuu alone the direct unemployment coefficient in the policy equation?

Add GDP growth ahead of the original three

C46 · Same quarterly illustration · new order (g, u, π, i)

gt=100Δlog(realGDPt)g_t=100\,\Delta\log(\mathrm{real\ GDP}_t)

Four-variable recursive current matrix. The three orange outlined zeros in the GDP-growth row are new; the three blue outlined zeros retain the unemployment, inflation, policy ordering.

Three new exclusions:

bgu=bgπ=bgi=0b_{gu}=b_{g\pi}=b_{gi}=0

GDP growth does not respond to current unemployment, inflation, or policy within the quarter.

The later equations may respond to current GDP growth.

Why should GDP come first? That is an assumption to defend.

Four equations need six additional restrictions

C47 · Same normalization, same diagonal positive shock covariance
Three equations Four equations
Current cross-variable coefficients 3×2=63\times2=6 4×3=124\times3=12
Structural shock variances 33 44
Current/shock unknowns 9 16
Distinct reduced covariance entries 3+3=63+3=6 4+6=104+6=10
Additional restrictions needed by the count 3 6

3⏟original exclusions+3⏟new GDP exclusions=6⏟four-equation recursive set\underbrace{3}_{\text{original exclusions}} +\underbrace{3}_{\text{new GDP exclusions}} =\underbrace{6}_{\text{four-equation recursive set}}

Adding one variable creates three new variable pairs. Where is each new exclusion in B4?

Count the general m-equation system

C48 · m is the number of outcomes and equations

Maintain unit diagonal B, uncorrelated positive-variance structural shocks, and unrestricted structural intercepts and lags.

Current/shock structure Distinct reduced innovation moments
m(m−1)m(m-1) off-diagonal current coefficients mm innovation variances
mm structural shock variances m(m−1)/2m(m-1)/2 innovation covariances
m2m^2 unknowns m(m+1)/2m(m+1)/2 entries of Σ

m2−m(m+1)2=m(m−1)2m^2-\frac{m(m+1)}2=\frac{m(m-1)}2

The order count requires m(m−1)/2 additional, suitably independent restrictions for the whole current/shock structure.

The normalizations and shock-uncorrelatedness assumptions are already in the baseline.

The upper triangle supplies the count—and a recovery

C49 · A consistent ordering is one workable construction

𝐁mrec=(100⋯0*10⋯0**1⋯0⋮⋮⋮⋱⋮***⋯1)\mathbf B_m^{\mathrm{rec}}=\begin{pmatrix} 1&0&0&\cdots&0\\ *&1&0&\cdots&0\\ *&*&1&\cdots&0\\ \vdots&\vdots&\vdots&\ddots&\vdots\\ *&*&*&\cdots&1 \end{pmatrix}

The stars are unrestricted current entries.

First equation: exclude m − 1 later current variables.

Second: exclude m − 2.

Continue down to 1, then 0.

1+2+⋯+(m−1)=m(m−1)21+2+\cdots+(m-1)=\frac{m(m-1)}2

A positive-definite Σ and a specified recursive ordering give a unique normalized recovery.
The right number alone is insufficient. Do the proposed restrictions actually permit unique recovery—and make economic sense?

Cholesky implements the ordering

C50 · A recognizable software operation · shown for our two-variable example

𝐏=(p110p21p22),𝚺=𝐏𝐏′\mathbf P=\begin{pmatrix}p_{11}&0\\p_{21}&p_{22}\end{pmatrix},\qquad \boldsymbol\Sigma=\mathbf P\mathbf P'

P_y_first <- t(chol(Sigma))

R returns the upper factor; transpose it for the lower factor.

The zero implements the y-first restriction. The factorization does not choose the economic timing.

Act VI · Defend the interpretation

The economic argument chooses the restriction.

What can happen within a month?

C51 · Return to the monthly two-variable example · y = inflation · x = policy rate

Assume inflation first

byx=0:yt→xtb_{yx}=0:\quad y_t\longrightarrow x_t

Policy may respond to current inflation news. Measured inflation does not respond to an unexpected policy disturbance within the month.

Assume policy first

bxy=0:xt→ytb_{xy}=0:\quad x_t\longrightarrow y_t

Inflation may respond to current policy. Policy does not respond to current inflation within the month.

Which timing story can you defend? How do publication dates, aggregation, and frequency affect it?

The equations are real. Now ask about the system.

C52 · Key takeaways + next time

The connected pond, revisited.

Coefficients are transformed. Recall ϕyx=0\phi_{yx}=0 but ayx=8/35a_{yx}=8/35.

Structure needs assumptions. A valid decomposition does not prove its timing story.

Disturbances move through the system. One coefficient is only one part of that account.

Next question Later module
How does each variable respond over time to an identified disturbance? IRF · 12
Which disturbances account for forecast-error variance at a given horizon? FEVD · 13