Who Heard the Alarm?

Module 8 · ARDL, information, and exogeneity

Gary Cornwall

Econ 6376 · The George Washington University

Act I · Letting another variable into the room

One series’ memory becomes a relationship across series.

ARMA: one series remembers outcomes and surprises

$y_t=\alpha+\phi_1y_{t-1}+\theta_1\epsilon_{t-1}+\epsilon_t$

A time-unrolled DAG: each date is a different node. Innovations ε are unobserved DGP quantities.

ARDL: another series brings its own history

$y_t=\alpha+\phi_1y_{t-1}+\beta_0x_t+\beta_1x_{t-1}+\epsilon_t$

Observed x replaces the explicit MA block. Drawing an arrow does not identify a causal effect.

Where we are: expand the mixing board

$y_t=\alpha+\delta t+\sum_{j=1}^{p}\phi_jy_{t-j} +\sum_{\ell=1}^{q}\theta_\ell\epsilon_{t-\ell}+\epsilon_t$

Channel	Module 8 baseline
Intercept and outcome memory	On: $\alpha$ , $\phi_jy_{t-j}$
Trend and explicit MA memory	Off
Observed predictor history	New: $\sum_{\ell=0}^{q}\beta_\ell x_{t-\ell}$
Current disturbance	Still present; its relation to $x$ now matters

Module 7 combined forecasts. Today we add information absent from that forecast pool.

One conditional equation, many possible lags

$y_t=\alpha+\sum_{j=1}^{p}\phi_jy_{t-j} +\sum_{\ell=0}^{q}\beta_\ell x_{t-\ell}+\epsilon_t$

Restriction	Model left on the board
All $\beta_\ell=0$	AR $(p)$
All $\phi_j=0$	Distributed-lag regression
$q=0$	Outcome lags and current $x$
$\beta_0=0$	Lag-only predictor block

What does ARDL(1,2) include that ARDL(1,1) does not?

Here q means the maximum x lag. It is not the MA order.

The DGP and our fitted equation remain different

DGP: parameters and disturbances we know only in a simulation

$y_t=\alpha+\phi_1y_{t-1}+\beta_0x_t+\beta_1x_{t-1}+\epsilon_t$

Estimated equation: hats and an observed residual

$y_t=\hat\alpha+\hat\phi_1y_{t-1}+\hat\beta_0x_t+\hat\beta_1x_{t-1}+e_t$

$e_t=y_t-\hat y_t$

Fitted values: ŷ. Forecasts from Module 7 onward:

\tilde y_{t+h\mid t}

Keep the stationary baseline visible

$\Phi(L)=1-\phi_1L-\cdots-\phi_pL^p$

Stationary inputs + stable dynamics + suitable disturbances.

AR lag-polynomial roots lie outside the unit circle.
For one outcome lag: $|\phi_1|<1$ .
If $x$ responds to past $y$ , check stability of the joint system.

Trending levels need additional reasoning. Cointegration is a later route, not a blanket exemption.

[2 minutes. Source: notes §8.2.] Enders’ characteristic roots inside are the reciprocal convention. No detailed root calculation on the deck. A regression with ARMA disturbances is also a different specification: the notes show the restricted ARDL produced by an AR(1) disturbance. Do not translate an xreg coefficient mechanically into this model.

Student reference — the root condition. For one outcome lag,

$1-\phi_1 L=0\quad\Longrightarrow\quad L=1/\phi_1.$

For $\phi_1\ne0$ , $|L|>1$ is equivalent to $|\phi_1|<1$ . If $\phi_1=0$ , there is no autoregressive propagation and no finite AR root to check. This is the same condition expressed using reciprocal roots, not two competing definitions of stability.

Student reference — why AR(1) disturbances impose a restriction. Start with the alternative model in notes §8.2:

$\begin{aligned} y_t&=\alpha+\beta x_t+u_t,\\ u_t&=\phi_1u_{t-1}+\epsilon_t. \end{aligned}$

Here $u_t$ is the disturbance around the static regression, while $\epsilon_t$ is its innovation. The preceding regression equation gives

$u_{t-1}=y_{t-1}-\alpha-\beta x_{t-1}.$

Substitute and collect terms:

$\begin{aligned} y_t&=\alpha+\beta x_t +\phi_1(y_{t-1}-\alpha-\beta x_{t-1})+\epsilon_t\\ &=(1-\phi_1)\alpha+\phi_1y_{t-1}\\ &\qquad+\beta x_t-\phi_1\beta x_{t-1}+\epsilon_t. \end{aligned}$

Thus the resulting ARDL intercept is $(1-\phi_1)\alpha$ , the current predictor coefficient is $\beta$ , and the lag coefficient is restricted to $-\phi_1\beta$ . The $\alpha$ in this calculation belongs to the original static regression; it is not equal to the resulting ARDL intercept. An unrestricted ARDL does not impose this coefficient restriction. Under stability its long-run step multiplier in this restricted case is

$\frac{\beta-\phi_1\beta}{1-\phi_1}=\beta.$

This is why a coefficient from a regression with AR disturbances should not be read as the unrestricted ARDL’s impact multiplier without checking the model.

Act II · Who else heard the news?

Exogeneity is the central question.

The Office: everyone reacts to the same emergency

If claims and unemployment rise in the same month, which story has the regression established?

Optional clip: The Office · Fire Drill (official). The diagram remains usable offline.

[4 minutes. Source: notes §8.3 common current news; original teaching diagram.] Invite competing stories: layoffs affect both; claims provide timely measurement; there may be feedback; the definitions cover different populations. Neither co-movement nor an arrow settles which intervention would move the outcome. Optional instructor hook: the opening fire drill in The Office, Stress Relief, Part1 (S5E14). Official source and episode description: https://www.peacocktv.com/watch-online/tv/the-office/4902514835143843112/seasons/5/episodes/stress-relief-part-1-episode-14/194b509e-14fb-377a-ac23-c1ec8a412703 Official clip linked by Peacock’s twentieth-anniversary episode article: https://www.youtube.com/watch?v=gO8N3L_aERg If played, use only the short setup needed for the common-event analogy; no internet dependency is required to teach this slide. The drill is initiated by Dwight, so do not imply it is itself an exogenous real-world experiment. It is an analogy for shared news, not a proposed identification strategy.

Exogeneity asks what remains in the disturbance

Let $\mathcal F_{t-1}$ contain the relevant histories of $y$ and $x$ .

$\boxed{E[\epsilon_t\mid\mathcal F_{t-1},x_t]=0}$

After using the modeled history and current x, no predictable mean component remains.

This covers all included regressors, including lagged $y$ .
With stability, moments, and adequate regressor variation, it supports large-sample OLS consistency.
It does not promise exact recovery in one finite sample.

[5 minutes. Source: notes §8.3.] Read the conditioning bar aloud. The regressor vector is (1, y(t−1), x(t), x(t−1)) in our ARDL(1,1). The sufficient condition implies orthogonality for that vector. Do not reduce the condition to Cov(x_t,epsilon_t)=0 alone. We have deliberately not imported exact textbook unbiasedness or t/F distributions from a strictly exogenous Gaussian regression. Formal conditions stay in the notes Technical Note.

Student reference — from the conditional mean to the OLS moments. Let $z_t=(1,y_{t-1},x_t,x_{t-1})'$ for this ARDL(1,1). Every component is known given $\mathcal F_{t-1}$ and $x_t$ . Assuming the products are integrable, the law of iterated expectations gives

$\begin{aligned} E[z_t\epsilon_t] &=E\!\left[E[z_t\epsilon_t\mid\mathcal F_{t-1},x_t]\right]\\ &=E\!\left[z_t E[\epsilon_t\mid\mathcal F_{t-1},x_t]\right]\\ &=E[z_t\,0]=0. \end{aligned}$

The same argument applies component by component to every included outcome lag and predictor lag. Merely showing that one covariance is zero would not prove this conditional-mean statement.

Optional technical reference — where consistency enters. Write $y_t=z_t'\vartheta+\epsilon_t$ . For a nonsingular sample regressor matrix, substitute that equation into the OLS formula:

$\begin{aligned} \hat\vartheta &=\left(\sum_t z_tz_t'\right)^{-1}\sum_t z_ty_t\\ &=\vartheta+ \left(\sum_t z_tz_t'\right)^{-1}\sum_tz_t\epsilon_t,\\ \hat\vartheta-\vartheta &=\left(T^{-1}\sum_tz_tz_t'\right)^{-1} \left(T^{-1}\sum_tz_t\epsilon_t\right). \end{aligned}$

Under stationary/ergodic sampling and suitable moments, the sample averages converge to $E[z_tz_t']$ and zero. If $E[z_tz_t']$ is positive definite, the estimation error converges to zero. This is the argument in the notes’ Technical Note; it supplies neither exact finite-sample unbiasedness nor exact $t/F$ distributions. Inference needs its own variance and central-limit conditions.

Three claims that sound similar

x and y are relatedOften why we included x. This is not the exogeneity restriction.

ε has no serial correlationA statement about the disturbance across dates.

$E[\epsilon_t\mid\mathcal F_{t-1},x_t]=0$ A statement about the disturbance relative to the regressor information.

Could two white-noise shocks be correlated with each other today?

Yes. No memory is not independence from every other process.

Common current news creates an omitted route

If the outcome disturbance contains shared news, current x may reveal part of that disturbance.

Lagged feedback is a different issue

$x_t=0.5x_{t-1}+0.2y_{t-1}+\epsilon_{x,t}$

With independent current shocks, independent of the past, this feedback can coexist with our conditional-mean restriction.

[5 minutes. Source: notes §8.3 and simulation §8.4.] The graph isolates an across-date route present in the model; the complete equations have other arrows too. Today’s y disturbance enters y_t, which can affect tomorrow’s x. It need not predict today’s x innovation. Sequential conditioning permits this distinction. Joint stability still matters and is checked for the simulation. Ask students to explain why ‘feedback’ alone is too vague a verdict.

Student reference — current exogeneity and future dependence can coexist. Use the feedback-only case: $c=0.2$ , independent current shocks, and shock pairs independent across dates. Conditional on $\mathcal F_{t-1}$ and $x_t$ , the outcome’s predictable component is

$g_t=0.2+0.6y_{t-1}+0.4x_t+0.2x_{t-1},$

so $y_t=g_t+\epsilon_t$ . Because current $x_t$ depends on the past and $\epsilon_{x,t}$ , it does not reveal the independent $\epsilon_t$ :

$E[\epsilon_t\mid\mathcal F_{t-1},x_t]=0.$

But the next predictor value is

$\begin{aligned} x_{t+1}&=0.5x_t+0.2y_t+\epsilon_{x,t+1}\\ &=0.5x_t+0.2g_t+0.2\epsilon_t+\epsilon_{x,t+1}. \end{aligned}$

Holding the current information fixed, the two remaining shocks are independent and have unit variance. Therefore

$\begin{aligned} \operatorname{Cov}(\epsilon_t,x_{t+1} \mid\mathcal F_{t-1},x_t) &=0.2\operatorname{Var}(\epsilon_t)=0.2. \end{aligned}$

The route is $\epsilon_t\rightarrow y_t\rightarrow x_{t+1}$ . The unconditional covariance is also 0.2 here: $\epsilon_t$ is mean independent of the current information. Thus a zero conditional mean given the entire predictor path cannot hold, even though the sequential restriction does hold. The calculation depends on the independent-shock construction; it is not a universal formula for feedback.

Optional technical reference — checking joint stability. Substitute the predictor equation into the outcome equation:

$\begin{aligned} y_t&=0.2+(0.6+0.4c)y_{t-1}+0.4x_{t-1}\\ &\qquad+0.4\epsilon_{x,t}+\epsilon_t,\\ x_t&=c\,y_{t-1}+0.5x_{t-1}+\epsilon_{x,t}. \end{aligned}$

For $c=0.2$ , the transition matrix for $(y_t,x_t)'$ is

$A=\begin{pmatrix}0.68&0.4\\0.2&0.5\end{pmatrix}.$

Its characteristic roots solve

$\begin{aligned} 0&=(\lambda-0.68)(\lambda-0.5)-0.4(0.2)\\ &=\lambda^2-1.18\lambda+0.26,\\ \lambda&=\frac{1.18\pm\sqrt{1.18^2-4(0.26)}}{2}\\ &\approx0.887,\;0.293. \end{aligned}$

Both characteristic roots are inside the unit circle, equivalently the joint lag-polynomial roots are outside. For the two no-feedback cases the transition matrix is triangular, with characteristic roots 0.6 and 0.5. Changing current shock correlation changes the innovation covariance, not these transition roots. This verifies the simulation’s stability claim; the matrix machinery is optional reference and is developed in Module 11.

Stronger assumptions and narrower diagnostics

Strict exogeneity asks today’s disturbance to be unrelated to the entire $x$ path—including future $x$ .

Lagged feedback may violate that stronger requirement while the sequential conditional mean remains valid.

Proposed fix	What it cannot establish
“Use a lag”	Earlier does not guarantee exogenous
“The residual ACF looks clean”	No proof of structural exogeneity
“Use HAC standard errors”	No repair for invalid coefficient moments

Are we making a prediction claim, a conditional-path claim, or a causal claim?

Act III · Build it, then break it

Same structural coefficients. Different information.

Two equations we control

$\begin{aligned} x_t&=0.5x_{t-1}+c\,y_{t-1}+\epsilon_{x,t},\\ y_t&=0.2+0.6y_{t-1}+0.4x_t+0.2x_{t-1}+\epsilon_t. \end{aligned}$

Current shocks have variance one and correlation $r$ . Shock pairs are independent across dates.

Case	Feedback $c$	Shock correlation $r$
Baseline	0	0
Shared current shocks	0	0.7
Lagged feedback only	0.2	0

Live code: the for-loop is the model

LIVE CODE · build, fit, then change one assumption

# Excerpt from ardl_simulator(); initialization is in the full helper.
epsilon_x <- rnorm(total)
epsilon <- shock_corr * epsilon_x +
  sqrt(1 - shock_corr^2) * rnorm(total)
for (t in 2:total) {
  x[t] <- 0.5*x[t-1] + feedback*y[t-1] + epsilon_x[t]
  y[t] <- 0.2 + 0.6*y[t-1] + 0.4*x[t] +
    0.2*x[t-1] + epsilon[t]
}

Which change should move OLS away from the structural coefficient 0.4?

[7 minutes. Source: notes §8.4, canonical helpers/ardl_utils.R.] This is a visible excerpt from the canonical simulator, not a new function. Complete runnable live setup from the project root (or use the helper directly):

set.seed(637608)
n <- 400L; burn_in <- 200L; total <- n + burn_in + 1L
shock_corr <- 0; feedback <- 0
epsilon_x <- rnorm(total)
epsilon <- shock_corr * epsilon_x +
  sqrt(1 - shock_corr^2) * rnorm(total)
x <- y <- numeric(total)
for (t in 2:total) {
  x[t] <- 0.5*x[t-1] + feedback*y[t-1] + epsilon_x[t]
  y[t] <- 0.2 + 0.6*y[t-1] + 0.4*x[t] +
    0.2*x[t-1] + epsilon[t]
}
keep <- (burn_in + 2L):total
d <- data.frame(y=y[keep], x=x[keep])
d$y_lag1 <- c(NA, head(d$y, -1))
d$x_lag1 <- c(NA, head(d$x, -1))
coef(lm(y ~ y_lag1 + x + x_lag1, data=d))

Show the equation’s order: x_t is generated before y_t. Change shock_corr to .7, then instead set feedback=.2, shock_corr=0. Use the same underlying draws by resetting the seed for the three single samples. Larger n does not cure a wrong target.

One sample is not the distribution

Monte Carlo distributions of the estimated contemporaneous coefficient under the baseline, shared-current-shock, and lagged-feedback designs.

More observations reduce sampling noise. They cannot identify the structural coefficient from an invalid restriction.

The wrong structural answer can have clean residuals

Residual ACF from the shared-current-shock regression.

What did the ACF check—and what did it leave unanswered?

OLS is answering a different conditional question

For the Gaussian shared-shock case,

$E[\epsilon_t\mid\epsilon_{x,t}]=0.7\epsilon_{x,t}, \qquad \epsilon_{x,t}=x_t-0.5x_{t-1}.$

Therefore,

$E[y_t\mid\mathcal F_{t-1},x_t] =0.2+0.6y_{t-1}+1.1x_t-0.15x_{t-1}.$

Coefficient	Structural DGP	Conditional regression
Current $x$	0.4	1.1
Lagged $x$	0.2	−0.15

A useful conditional predictor need not recover the causal mechanism we wanted.

[5 minutes. Source: notes §8.4, Why the Shared-Shock Regression Can Still Look Clean.] Present the two coefficient adjustments aloud: .4+.7 and .2−.7*.5; the complete calculation below is available for student reference. The residual after projection is epsilon_t−.7epsilon_x,t, independent of current x innovations and serially independent in this construction. Thus clean residuals are entirely compatible with the observed coefficient pattern. Pause for a ten-minute break after this slide. Application bridge: claims may signal omitted labor news.

Student reference — the complete conditional-mean calculation. In this case there is no feedback ( $c=0$ ). Construct independent standard normal variables $\epsilon_{x,t}$ and $\eta_t$ , independently across dates, and set

$\begin{aligned} \epsilon_t&=0.7\epsilon_{x,t}+\sqrt{1-0.7^2}\,\eta_t\\ &=0.7\epsilon_{x,t}+\sqrt{0.51}\,\eta_t. \end{aligned}$

The structural disturbance has variance $0.7^2+0.51=1$ and covariance 0.7 with the unit-variance predictor innovation. Their correlation is therefore 0.7. Independence and the zero mean of $\eta_t$ imply

$\begin{aligned} E[\epsilon_t\mid\epsilon_{x,t}] &=0.7\epsilon_{x,t}+\sqrt{0.51}\,E[\eta_t\mid\epsilon_{x,t}]\\ &=0.7\epsilon_{x,t}. \end{aligned}$

Since the predictor follows $x_t=0.5x_{t-1}+\epsilon_{x,t}$ , observing current $x_t$ and its history reveals

$\epsilon_{x,t}=x_t-0.5x_{t-1}.$

The current $\eta_t$ remains independent of that information and of the past. Take the conditional expectation of the structural outcome equation and then substitute for the predictor innovation:

$\begin{aligned} E[y_t\mid\mathcal F_{t-1},x_t] &=0.2+0.6y_{t-1}+0.4x_t+0.2x_{t-1}\\ &\qquad+E[\epsilon_t\mid\mathcal F_{t-1},x_t]\\ &=0.2+0.6y_{t-1}+0.4x_t+0.2x_{t-1}\\ &\qquad+0.7(x_t-0.5x_{t-1})\\ &=0.2+0.6y_{t-1}+(0.4+0.7)x_t\\ &\qquad+(0.2-0.7\times0.5)x_{t-1}\\ &=0.2+0.6y_{t-1}+1.1x_t-0.15x_{t-1}. \end{aligned}$

Call the remaining population conditional disturbance $\epsilon_t^c$ :

$\begin{aligned} \epsilon_t^c &=y_t-E[y_t\mid\mathcal F_{t-1},x_t]\\ &=\epsilon_t-0.7\epsilon_{x,t}\\ &=\sqrt{0.51}\,\eta_t,\\ \operatorname{Var}(\epsilon_t^c)&=0.51. \end{aligned}$

It is independent of current $\epsilon_{x,t}$ and of the past, and is iid across dates in this construction. It is not the original structural disturbance $\epsilon_t$ , and it is not the finite-sample fitted residual $e_t$ . Estimated residuals approximate this conditional disturbance; estimation still induces finite-sample restrictions on them. Thus a residual ACF can look clean while OLS approaches 1.1 and $-0.15$ instead of the structural 0.4 and 0.2. The shared news has entered the conditional coefficients, not disappeared from the mechanism.

Act IV · Follow every path

A coefficient is one route. A response follows all routes.

First specify the input experiment

$y_t=0.6y_{t-1}+0.4x_t+0.2x_{t-1}+\epsilon_t$

Imposed change in $x$	Today	Next period	Later
Pulse	+1	0	0
Permanent step	+1	+1	+1

Hold initial outcome history and outcome shocks fixed; compare the two $y$ paths.

Should a one-period pulse and a permanent step have the same long-run response?

[3 minutes. Source: notes §8.5.] The intercept cancels between paths. Say conditional experiment: these are imposed x paths, not automatically paths generated by an innovation to x’s equation. Under stable dynamics our pulse decays, whereas the permanent step has a nonzero limit.

Student reference — subtracting two paths. Label the baseline path $(0)$ and the altered-input path $(1)$ . Both satisfy the same ARDL with the same parameters and outcome shocks:

$\begin{aligned} y_{t+h}^{(a)}&=\alpha+\phi_1y_{t+h-1}^{(a)} +\beta_0x_{t+h}^{(a)}\\ &\qquad+\beta_1x_{t+h-1}^{(a)}+\epsilon_{t+h}, \qquad a\in\{0,1\}. \end{aligned}$

Define cross-path deviations $d_{y,h}=y_{t+h}^{(1)}-y_{t+h}^{(0)}$ and $d_{x,h}=x_{t+h}^{(1)}-x_{t+h}^{(0)}$ . These compare two paths at the same date; they are not the time difference $\Delta y_t=y_t-y_{t-1}$ . Subtracting yields

$\begin{aligned} d_{y,h}&=(\alpha-\alpha)+\phi_1d_{y,h-1} +\beta_0d_{x,h}\\ &\qquad+\beta_1d_{x,h-1}+(\epsilon_{t+h}-\epsilon_{t+h})\\ &=\phi_1d_{y,h-1}+\beta_0d_{x,h}+\beta_1d_{x,h-1}. \end{aligned}$

Identical histories imply $d_{y,h}=d_{x,h}=0$ before $h=0$ . For an imposed pulse, $d_{x,0}=1$ and $d_{x,h}=0$ for $h\ne0$ . For a permanent step, $d_{x,h}=1$ for every $h\ge0$ . The same recurrence handles both experiments; only the supplied predictor path changes. This is a conditional calculation holding the model and shocks fixed, not by itself an identified causal claim.

Tomorrow has two routes from today’s x

$m_0=\beta_0=0.4$

$m_1=\underbrace{0.6(0.4)}_{\text{inherited}}+\underbrace{0.2}_{\text{direct}}=0.44$

β₁ alone is not the total next-period response.

[4 minutes. Source: notes §8.5.] This graph is the opening DAG extended one date. The path calculation is the one derivation that earns its time: next-period response follows all routes. At horizon2, the direct lag has run out, so m2=.6*.44=.264. The optional general recursion is recorded below for reference.

Student reference — each horizon of the pulse. Use the cross-path recurrence from the preceding slide, with $m_h=d_{y,h}$ for the pulse and $m_{-1}=0$ :

$\begin{aligned} m_0&=\phi_1(0)+\beta_0(1)+\beta_1(0)=\beta_0,\\ m_1&=\phi_1m_0+\beta_0(0)+\beta_1(1) =\phi_1\beta_0+\beta_1,\\ m_2&=\phi_1m_1+\beta_0(0)+\beta_1(0) =\phi_1m_1. \end{aligned}$

Numerically,

$\begin{aligned} m_0&=0.4,\\ m_1&=0.6(0.4)+0.2=0.24+0.2=0.44,\\ m_2&=0.6(0.44)=0.264,\\ m_3&=0.6(0.264)=0.1584. \end{aligned}$

At horizon one, the direct route contributes $\beta_1$ and the inherited route contributes $\phi_1\beta_0$ . At horizons $h\ge2$ there is no remaining direct input from this one-period pulse, so repeated substitution gives

$m_h=\phi_1^{h-1}(\phi_1\beta_0+\beta_1),\qquad h\ge1.$

Optional generalization from the notes’ Deeper Dive. The same subtraction for an ARDL $(p,q)$ gives

$m_h=\beta_h+\sum_{j=1}^{p}\phi_jm_{h-j},$

where $m_h=0$ for $h<0$ and $\beta_h=0$ for $h>q$ . Each lagged outcome carries its own earlier response forward. The direct predictor coefficient at a horizon is only one term of the total response.

Pulse decays; repeated pulses accumulate

Pulse and permanent-step response paths for phi1 equal .6, beta0 equal .4, and beta1 equal .2.

$m_2=0.264,\qquad \text{step at }h=2=0.4+0.44+0.264=1.104.$

[4 minutes. Source: notes §8.5.] The permanent step is a sequence of pulses; linear time-invariant dynamics make its response equal the cumulative pulse response. Say which experiment each curve depicts. One example only; notes keep the negative-phi cancellation case.

Student reference — why the step equals the cumulative pulse response. A unit step is a unit pulse at date $t$ , another at $t+1$ , and another at every later date. Linearity permits adding their responses; time-invariant coefficients make a pulse introduced at $t+j$ contribute $m_{h-j}$ at $t+h$ . Denote the step response at horizon $h$ by $s_h$ . Then

$\begin{aligned} s_h&=m_h+m_{h-1}+\cdots+m_0\\ &=\sum_{j=0}^{h}m_{h-j}=\sum_{r=0}^{h}m_r. \end{aligned}$

For our coefficients this gives

$\begin{aligned} s_0&=0.4,\\ s_1&=0.4+0.44=0.84,\\ s_2&=0.4+0.44+0.264=1.104,\\ s_3&=1.104+0.1584=1.2624. \end{aligned}$

The result also follows directly from the step input in the path recurrence: $s_0=0.4$ and $s_h=0.6s_{h-1}+0.4+0.2$ for $h\ge1$ . For example, $s_2=0.6(0.84)+0.6=1.104$ . This is accumulation of the full pulse responses, including outcome persistence, not just a sum of direct $\beta$ coefficients.

The long-run multiplier needs stability

$\text{Long-run step multiplier} =\frac{\sum_{\ell=0}^{q}\beta_\ell}{1-\sum_{j=1}^{p}\phi_j}$

$\frac{0.4+0.2}{1-0.6}=1.5$

A permanent +1 in $x$ eventually changes $y$ by +1.5.
The one-period pulse itself converges to zero.
Estimated multipliers carry hats; a denominator near zero makes estimates sensitive.

If φ₁ = 1.05, does the nonzero denominator make −12 an eventual response?

[3 minutes. Source: notes §8.5.] Use the path picture during presentation; the full algebra below supports student reference after class. This is a change, not the unconditional mean. Units must be y-units per x-unit. Joint coefficient uncertainty matters, but delta method stays in the Technical Note. The next slide answers the question by examining the path.

Student reference — the general steady-state difference. Let $s_h$ denote the response to a permanent unit increase in the imposed predictor path. For $h\ge q$ , every included predictor lag is one unit above baseline:

$s_h=\sum_{j=1}^{p}\phi_js_{h-j}+\sum_{\ell=0}^{q}\beta_\ell.$

Stability makes the transient responses vanish, so $s_h$ converges to a common limit $s_\infty$ . Taking that limit and collecting terms gives

$\begin{aligned} s_\infty&=\left(\sum_{j=1}^{p}\phi_j\right)s_\infty +\sum_{\ell=0}^{q}\beta_\ell,\\ \left(1-\sum_{j=1}^{p}\phi_j\right)s_\infty &=\sum_{\ell=0}^{q}\beta_\ell,\\ s_\infty&=\frac{\sum_{\ell=0}^{q}\beta_\ell} {1-\sum_{j=1}^{p}\phi_j}. \end{aligned}$

The intercept has already canceled between the two paths. For a predictor level $\bar x$ , the deterministic equilibrium level is instead $\bar y=[\alpha+(\sum_\ell\beta_\ell)\bar x]/[1-\sum_j\phi_j]$ . The multiplier describes a change between equilibria, not that level.

Student reference — summing our ARDL(1,1) responses. Since $m_0=\beta_0$ and $m_h=\phi_1^{h-1}(\phi_1\beta_0+\beta_1)$ for $h\ge1$ ,

$\begin{aligned} \sum_{h=0}^{\infty}m_h &=\beta_0+(\phi_1\beta_0+\beta_1) \sum_{k=0}^{\infty}\phi_1^k\\ &=\beta_0+\frac{\phi_1\beta_0+\beta_1}{1-\phi_1}\\ &=\frac{\beta_0+\beta_1}{1-\phi_1},\qquad |\phi_1|<1. \end{aligned}$

Numerically, $0.4+0.44/(1-0.6)=0.4+1.1=1.5$ . Meanwhile $m_h=0.44(0.6)^{h-1}\rightarrow0$ : the pulse fades even though the step converges to 1.5. For finite $h\ge1$ , the step response is

$s_h=0.4+0.44\frac{1-0.6^h}{1-0.6}=1.5-1.1(0.6)^h.$

Optional generalization from the notes’ Deeper Dive. Stability with finite predictor lags makes the pulse responses absolutely summable. Sum the general pulse recurrence and shift indices, using $m_h=0$ at negative horizons:

$\begin{aligned} \sum_{h=0}^{\infty}m_h &=\sum_{\ell=0}^{q}\beta_\ell +\sum_{j=1}^{p}\phi_j\sum_{h=0}^{\infty}m_{h-j}\\ &=\sum_{\ell=0}^{q}\beta_\ell +\left(\sum_{j=1}^{p}\phi_j\right) \sum_{h=0}^{\infty}m_h. \end{aligned}$

Rearranging recovers the same ratio. Absolute summability justifies these infinite sums; a nonzero denominator alone does not justify taking a long-run limit.

Break it: a finite ratio can describe no limit

Permanent-step paths under stable, unit-root, and explosive outcome persistence.

At φ₁ = 1.05 the path grows. The algebraic ratio −12 is not its limit.

[4 minutes. Source: notes §8.5, Break Stability.] Live if time: ardl_multipliers(1.05,c(.4,.2),20). At phi1=1, increments from successive imposed pulses accumulate; phi1=1.05 amplifies them. An algebraic steady-state solution is not necessarily an attracting limit. The chart is the assumption-break for the multiplier concept.

Student reference — the paths that invalidate the limiting interpretation. Keep $\beta_0=0.4$ and $\beta_1=0.2$ , and change only $\phi_1$ . The pulse recursion always gives

$m_0=0.4,\qquad m_h=\phi_1^{h-1}(0.4\phi_1+0.2),\quad h\ge1.$

At $\phi_1=1$ , every later pulse response equals 0.6. Hence

$s_h=0.4+\sum_{j=1}^{h}0.6=0.4+0.6h\longrightarrow+\infty.$

At $\phi_1=1.05$ , $m_1=1.05(0.4)+0.2=0.62$ . For $h\ge1$ ,

$\begin{aligned} s_h&=0.4+0.62\sum_{k=0}^{h-1}1.05^k\\ &=0.4+0.62\frac{1.05^h-1}{1.05-1}\\ &=12.4(1.05)^h-12\longrightarrow+\infty. \end{aligned}$

The equation $s=1.05s+0.6$ has the algebraic fixed point $s=-12$ , but it is repelling. In fact, $s_h+12=1.05(s_{h-1}+12)$ ; the actual initial response $s_0=0.4$ is 12.4 above that fixed point, and the gap grows. A fixed-point calculation finds a candidate equilibrium, not proof that the observed path approaches it. Neither the unit-root case nor this explosive case has a finite long-run step response from the stated initial history.

An x innovation can create a whole input path

In the independent-shock, no-feedback baseline,

$x_t=0.5x_{t-1}+\epsilon_{x,t}.$

A unit innovation creates an x-path deviation of $(1,0.5,0.25,\ldots)$ , not an imposed pulse.

$\text{Outcome response at }h=1=0.44+0.4(0.5)=0.64.$

Conditional path response, response to an innovation, and causal intervention are distinct objects.

What else must be followed if x responds to past y?

[4 minutes. Source: notes §8.5.] Hold the outcome shock path fixed for this comparison. Shared current shocks could also change the expected outcome disturbance; .64 is the mechanical input path contribution only outside the independent-shock baseline. With feedback, follow both equations. That motivates the later VAR—not an automatic causal solution. Application bridge: a step in claims growth is not a step in the level of claims.

Student reference — trace the predictor innovation before the outcome. In the independent-shock, no-feedback baseline, compare paths with identical initial histories and shocks except for a one-unit increase in $\epsilon_{x,t}$ . Keep every outcome innovation fixed. The predictor-path deviation obeys

$d_{x,0}=1,\qquad d_{x,h}=0.5d_{x,h-1}=(0.5)^h.$

Thus there is an effective input of one at $t$ , 0.5 at $t+1$ , 0.25 at $t+2$ , and so on. Add the pulse responses to those inputs:

$d_{y,h}=\sum_{j=0}^{h}m_{h-j}(0.5)^j.$

The first three responses are

$\begin{aligned} d_{y,0}&=m_0=0.4,\\ d_{y,1}&=m_1+m_0(0.5)\\ &=0.44+0.4(0.5)=0.64,\\ d_{y,2}&=m_2+m_1(0.5)+m_0(0.25)\\ &=0.264+0.22+0.10=0.584. \end{aligned}$

Direct substitution into the same outcome-path recurrence verifies horizon one:

$\begin{aligned} d_{y,1}&=0.6d_{y,0}+0.4d_{x,1}+0.2d_{x,0}\\ &=0.6(0.4)+0.4(0.5)+0.2(1)=0.64. \end{aligned}$

The extra 0.20 relative to the imposed pulse’s 0.44 comes from tomorrow’s predictor still being 0.5 above baseline. The notation $d$ denotes a difference between paths, not a temporal difference $\Delta$ . With shared shocks, learning about $\epsilon_{x,t}$ also changes the expected outcome disturbance; with feedback, $d_{x,h}$ depends on the outcome path and is no longer simply $(0.5)^h$ . Those experiments require following the corresponding joint mechanism.

Act V · A predictor needs a calendar

Initial claims and unemployment, with an honest forecast origin.

The familiar workflow gets two new questions

Inspect → specifyUnits, transformations, lag lengths, release dates.

Estimate → diagnoseCommon estimation sample; residual plots and regression diagnostics.

Revise → evaluateTraining-only choices; identical holdout dates and benchmarks.

When was x available?
What interpretation do the assumptions support?

A future outcome may need a future predictor

$\tilde y_{t+1\mid t}=\hat\alpha+\hat\phi_1y_t +\hat\beta_0\tilde x_{t+1\mid t}+\hat\beta_1x_t$

Information about future $x$	Claim we can make
Forecast from its own model	Forecast with predictor uncertainty
Stipulated path	Conditional / scenario forecast
Released before the target outcome	Nowcast using that information

If an evaluation substitutes the future realized x, what advantage has it given the model?

[3 minutes. Source: notes §8.6.] The notes’ feasible-versus-oracle simulation is the optional live extension if students need to see leakage. No new chart needed. The error can include beta0 times x forecast error plus the outcome innovation; predictor uncertainty and any covariance matter. A scenario band intentionally conditions on x. Lag-only models avoid future x at one step only when lagged data are released.

Student reference — derive the one-step forecast. Start with the true ARDL(1,1) at the target date:

$y_{t+1}=\alpha+\phi_1y_t+\beta_0x_{t+1}+\beta_1x_t+\epsilon_{t+1}.$

Assume $y_t,x_t$ are available in $\mathcal F_t$ , and the sequential restriction holds at $t+1$ . By iterated expectations,

$\begin{aligned} E[\epsilon_{t+1}\mid\mathcal F_t] &=E\!\left[E[\epsilon_{t+1}\mid\mathcal F_t,x_{t+1}] \mid\mathcal F_t\right]=0. \end{aligned}$

Taking the conditional expectation of the target equation gives

$\begin{aligned} E[y_{t+1}\mid\mathcal F_t] &=\alpha+\phi_1y_t+\beta_1x_t\\ &\qquad+\beta_0E[x_{t+1}\mid\mathcal F_t]. \end{aligned}$

Replace the unknown coefficients by estimates and the predictor conditional mean by its feasible forecast to obtain the slide’s plug-in expression. If $x_t$ or $y_t$ has not yet been released, it cannot simply be inserted as known; the information set and forecasting problem must change accordingly.

Student reference — separate predictor uncertainty from outcome noise. First hold parameters known and use the true predictor conditional mean. Define $v_{t+1}=x_{t+1}-E[x_{t+1}\mid\mathcal F_t]$ . Subtraction gives

$\begin{aligned} y_{t+1}-E[y_{t+1}\mid\mathcal F_t] &=\beta_0v_{t+1}+\epsilon_{t+1}. \end{aligned}$

In general, the variance of this sum includes $2\beta_0\operatorname{Cov}(v_{t+1},\epsilon_{t+1}\mid\mathcal F_t)$ . Under the maintained sequential conditional-mean restriction that covariance is zero, because $v_{t+1}$ is known given $\mathcal F_t,x_{t+1}$ and $E[v_{t+1}\epsilon_{t+1}\mid\mathcal F_t]=0$ by iterated expectations. Thus in this maintained known-parameter case,

$\begin{aligned} \operatorname{Var}(y_{t+1}\mid\mathcal F_t) &=\beta_0^2\operatorname{Var}(x_{t+1}\mid\mathcal F_t)\\ &\qquad+\operatorname{Var}(\epsilon_{t+1}\mid\mathcal F_t). \end{aligned}$

This formula does not include parameter-estimation uncertainty. A conditional scenario calculation with a fixed predictor path intentionally removes the predictor uncertainty; it answers a different question.

Student reference — feasible versus oracle with the same fitted coefficients. Let the oracle insert realized $x_{t+1}$ while the feasible forecast inserts $\tilde x_{t+1\mid t}$ . Their difference is exactly

$\begin{aligned} \tilde y_{t+1\mid t}^{\mathrm{oracle}} -\tilde y_{t+1\mid t}^{\mathrm{feasible}} &=\hat\beta_0(x_{t+1}-\tilde x_{t+1\mid t}). \end{aligned}$

Since forecast error is actual minus forecast,

$\begin{aligned} e_{t+1\mid t}^{\mathrm{feasible}} &=e_{t+1\mid t}^{\mathrm{oracle}} +\hat\beta_0(x_{t+1}-\tilde x_{t+1\mid t}),\\ e_{t+1\mid t}^{\mathrm{oracle}} &=\epsilon_{t+1}+(\alpha-\hat\alpha) +(\phi_1-\hat\phi_1)y_t\\ &\qquad+(\beta_0-\hat\beta_0)x_{t+1} +(\beta_1-\hat\beta_1)x_t. \end{aligned}$

The first identity is exact even with estimated coefficients. The second shows why an oracle error is not just the structural innovation when coefficients are estimated. Observing future $x$ removes one information gap; it does not remove estimation error or establish structural exogeneity. The resulting score is a feasible nowcast score only if that predictor was actually released at the forecast origin.

Reference month is not release date

One step ahead in the monthly index; usually a nowcast of the current calendar month.

Schematic timing, not an exact historical release calendar. Use

\mathcal I_t^{\mathrm{release}}

for this origin.

What are we actually modeling?

$y_t=\Delta U_t,\qquad x_t=100\Delta\log C_t$

U: unemployment rateMonthly percent. y is its change in percentage points.

C: initial claimsMean of weekly counts whose week-ending Saturdays fall in the month.

x: claims growth100 × log change, approximately percent growth in that monthly mean.

A monthly mean of weekly claims is not a monthly total.

[2 minutes. Source: notes §8.7; FRED UNRATE and ICSA cached under data/.] Both source series seasonally adjusted. Household-survey unemployment and administrative benefit claims cover different populations. A permanent step in x here means faster claims growth forever—not one additional claimant. Data checks reject missing weeks or months before aggregation and lag construction.

Student reference — transformations and units. If $n_t$ weekly observations have week-ending Saturdays in reference month $t$ , the monthly predictor level is

$C_t=\frac{1}{n_t}\sum_{w\,\text{in month }t}\mathrm{ICSA}_w.$

It remains a mean weekly claims count. It is not the sum of all claims during a calendar month. The modeled predictor is

$\begin{aligned} x_t&=100(\log C_t-\log C_{t-1})\\ &=100\log(C_t/C_{t-1}),\\ 100\left(\frac{C_t}{C_{t-1}}-1\right) &=100\big[\exp(x_t/100)-1\big]\approx x_t. \end{aligned}$

The approximation uses $\exp(a)\approx1+a$ for a small log change. Meanwhile $y_t=U_t-U_{t-1}$ is a difference of rates recorded in percent, so its unit is percentage points, not percent growth.

To interpret a permanent one-unit increase in $x$ , compare paths with the same initial $C_{t-1}$ and add one to claims log growth at each date from $t$ onward. Summing those log-growth differences through horizon $h$ gives

$\begin{aligned} \log C_{t+h}^{(1)}-\log C_{t+h}^{(0)}&=(h+1)/100,\\ \frac{C_{t+h}^{(1)}}{C_{t+h}^{(0)}}&=\exp((h+1)/100). \end{aligned}$

Thus a step in growth makes the claims-level ratio keep rising. A one-period pulse in growth would instead leave a constant level ratio $\exp(1/100)$ from that date onward. These are translations of the imposed input experiment, not causal estimates from the observational application.

Inspect the transformed training data

Training paths of unemployment-rate changes and growth in monthly mean weekly claims.

1990–2017 only. ADF rejects a unit root; KPSS does not reject level stationarity for either transformation at 5%.

Fit on a common sample; keep the holdout untouched

$y_t=\alpha+\sum_{j=1}^{p}\phi_jy_{t-j} +\sum_{\ell=1}^{q}\beta_\ell x_{t-\ell}+\epsilon_t$

TRAIN · Jan 1990–Dec 2017336 common rows · choose by BIC

HOLDOUT2018–2019 · 24 rows

Lag-only claims: $\beta_0=0$ . No complete target-month claims average required.

First menu: $p=1,2,3$ ; add one to three claims lags. Compare with AR benchmarks.

Does using yesterday’s claims prove that omitted labor-market news is irrelevant?

Diagnose, revise once, and report what remains

Training decision	What happened
Small menu	AR(3) and lag-only ARDL(3,3) selected
Residual serial-correlation check	BG(12) $p\approx0.0000085$
One justified expansion	Up to 12 outcome lags for both families
Revised selection	AR(6), lag-only ARDL(6,3)
Revised diagnostics	Stable AR roots; BG(12) $p\approx0.0031$

Residual dependence remains. Stop this illustration and disclose it; do not search until the diagnostic passes.

Roll the information, keep the coefficients fixed

Observed unemployment-rate changes and fixed-coefficient predictions on the 2018–2019 holdout.

Observed lags update after release. Coefficients remain estimated on 1990–2017. No-change predicts ΔU = 0.

[3 minutes. Source: notes §8.7 app-forecast and app-holdout-plot.] The plot shows changes rather than reconstructed levels for slide clarity; errors are identical because each level forecast adds observed U(t−1). These are24rolling one-step-index predictions, not a single24-step path fromDecember2017. The origin is the post-release information set, and all caches are revised historical data.

Student reference — reconstruct levels without changing one-step errors. At each post-release origin for target month $t$ , let $\tilde y_{t\mid \mathcal I_t^{\mathrm{release}}}$ denote the predicted change using only available lags. The level prediction adds that change to the latest observed level:

$\tilde U_{t\mid\mathcal I_t^{\mathrm{release}}} =U_{t-1}+\tilde y_{t\mid\mathcal I_t^{\mathrm{release}}}.$

Since $U_t=U_{t-1}+y_t$ , subtracting gives

$\begin{aligned} U_t-\tilde U_{t\mid\mathcal I_t^{\mathrm{release}}} &=(U_{t-1}+y_t) -(U_{t-1}+\tilde y_{t\mid\mathcal I_t^{\mathrm{release}}})\\ &=y_t-\tilde y_{t\mid\mathcal I_t^{\mathrm{release}}}. \end{aligned}$

The level and change errors are identical date by date, hence have identical RMSE and MAE. A no-change level forecast sets $\tilde U_{t\mid\mathcal I_t^{\mathrm{release}}}=U_{t-1}$ , equivalently $\tilde y_{t\mid\mathcal I_t^{\mathrm{release}}}=0$ . At the next origin the newly released $U_t$ becomes the observed lag. Accumulating all predicted changes from December 2017 would replace those observed lags with a different recursive forecasting experiment.

The claims model improves this score—not every claim

Model	RMSE	MAE
AR(6)	0.1129	0.0929
Lag-only ARDL(6,3)	0.1057	0.0831
No change in unemployment	0.1099	0.0875

Unemployment percentage points · identical 24 dates · revised-data pseudo-out-of-sample evaluation.

Which statement is supported: “claims helped this forecast” or “raising claims causes unemployment to rise”?

A short-window ranking is descriptive evidence. Residual dependence and real-time-vintage limitations remain.

[3 minutes. Source: notes §8.7 app-scores, exact values verified by make_figures.R.] No significance claim or DM test;24dates is a narrow window. Retuning on these errors would consume the holdout. No causal intervention interpretation. The complete application deliberately shows imperfect diagnostics rather than manufacturing an all-green result. That is the practical exogeneity lesson.

Student reference — scores and their units. For each model $M$ and each of the same 24 holdout dates, let $e_{M,t}$ be observed minus predicted unemployment change. The preceding slide shows that it is also the corresponding level error. Compute

$\begin{aligned} \mathrm{MSE}_M&=\frac{1}{24}\sum_{t\,\text{in holdout}}e_{M,t}^2,\\ \mathrm{RMSE}_M&=\sqrt{\mathrm{MSE}_M},\\ \mathrm{MAE}_M&=\frac{1}{24}\sum_{t\,\text{in holdout}}|e_{M,t}|. \end{aligned}$

The errors are percentage points, MSE is squared percentage points, and taking the square root returns RMSE to percentage points. The application code applies these formulas to the fixed-coefficient predictions: ARDL RMSE rounds to 0.1057 and MAE to 0.0831. For no change, the errors are simply the realized $\Delta U_t$ , giving RMSE 0.1099 and MAE 0.0875 after rounding. No denominator or evaluation date changes between models. The inequalities in this table describe this window; the arithmetic itself supplies neither a significance test nor a causal claim.

Act VI · Three doors out of ARDL

Brief orientation to the next modules.

Looking Ahead: does x’s past add predictive information?

MODULE 9 · GRANGER CAUSALITY · OPTIONAL PREVIEW

For a lag-only linear model and a stated conditioning set,

$H_0:\beta_1=\beta_2=\cdots=\beta_q=0.$

A joint question: do these $x$ lags add information beyond the history already included?

Predictive precedence can arise from an early signal of a common driver. It does not identify an intervention effect.

Would adding another predictor potentially change this answer?

Looking Ahead: can wandering levels share a stable gap?

MODULE 10 · COINTEGRATION · OPTIONAL PREVIEW

$y_t\sim I(1),\qquad x_t\sim I(1),\qquad y_t-a-bx_t\sim I(0).$

Harry and his dog, Mouse: both wander, but their separation can remain stationary.

Cointegration concerns a stationary combination of nonstationary levels.
An error-correction model links short-run changes to last period’s deviation.
Rewriting an ARDL as an ECM is algebra, not a cointegration test.

What would differ if they wandered independently with no stable separation?

[2 minutes. Source: notes §8.9, LookingAhead.] Keep the existing Harry-and-Mouse leash analogy for Module10; no large cartoon sequence or new simulation here. Ordinary ADF critical values do not automatically apply to estimated cointegrating residuals. Bounds testing/Johansen stay outside this introduction. Alpha remains the intercept; the optional reference below writes the adjustment coefficient explicitly as phi1−1.

Optional preview reference — the full error-correction rearrangement. This is the algebra in notes §8.9, not additional Module 8 assessable content. Start with the level ARDL(1,1):

$y_t=\alpha+\phi_1y_{t-1}+\beta_0x_t+\beta_1x_{t-1}+\epsilon_t.$

Subtract $y_{t-1}$ from both sides:

$\begin{aligned} \Delta y_t&=\alpha+(\phi_1-1)y_{t-1}\\ &\qquad+\beta_0x_t+\beta_1x_{t-1}+\epsilon_t. \end{aligned}$

Substitute $x_t=x_{t-1}+\Delta x_t$ and collect the two lagged-level terms:

$\begin{aligned} \Delta y_t&=\alpha+(\phi_1-1)y_{t-1}\\ &\qquad+\beta_0(x_{t-1}+\Delta x_t) +\beta_1x_{t-1}+\epsilon_t\\ &=\alpha+(\phi_1-1)y_{t-1}\\ &\qquad+(\beta_0+\beta_1)x_{t-1} +\beta_0\Delta x_t+\epsilon_t. \end{aligned}$

When $\phi_1\ne1$ , define the level-relation coefficients

$a=\frac{\alpha}{1-\phi_1},\qquad b=\frac{\beta_0+\beta_1}{1-\phi_1}.$

Then the first three terms factor as

$\begin{aligned} &\alpha+(\phi_1-1)y_{t-1} +(\beta_0+\beta_1)x_{t-1}\\ &\quad=-(1-\phi_1)[y_{t-1}-a-bx_{t-1}]. \end{aligned}$

To check the signs, expand the bracket: the three terms are $-(1-\phi_1)y_{t-1}$ , $+(1-\phi_1)a=\alpha$ , and $+(1-\phi_1)b x_{t-1}=(\beta_0+\beta_1)x_{t-1}$ . Therefore

$\begin{aligned} \Delta y_t&=-(1-\phi_1)[y_{t-1}-a-bx_{t-1}]\\ &\qquad+\beta_0\Delta x_t+\epsilon_t. \end{aligned}$

The bracket is the preceding period’s deviation from the level relation. When $\phi_1<1$ , a positive deviation contributes negatively to the next change, other inputs held fixed; this sign alone is not a full stability condition. The adjustment coefficient is $\phi_1-1$ , while $\alpha$ remains the original intercept. The coefficient $b$ reproduces the familiar long-run ratio.

This rewriting also works for stationary variables and therefore cannot prove cointegration. For $I(1)$ variables, one must additionally establish that the level combination is $I(0)$ and justify the conditioning and adjustment model. At $\phi_1=1$ the unfactored difference equation is still valid, but division by $1-\phi_1$ is unavailable. Module 10 supplies the statistical arguments and appropriate cointegration tests; no test is supplied by this algebra.

Looking Ahead: give every variable an equation

MODULES 11–12 · VAR AND SHOCK INTERPRETATION

$\begin{aligned} y_t&=a_y+b_{yy}y_{t-1}+b_{yx}x_{t-1}+\epsilon_{y,t},\\ x_t&=a_x+b_{xy}y_{t-1}+b_{xx}x_{t-1}+\epsilon_{x,t}. \end{aligned}$

The ensemble cast: both histories can help explain both outcomes.

A system models joint evolution and forecast feedback. It still needs assumptions to identify causal shocks.

Key takeaways + next time

ARDL adds observed predictor history to outcome memory.

Exogeneity concerns the disturbance and the whole information set.

A response follows every path; state the input experiment first.

A forecast needs an honest calendar and a common benchmark.

Next: does one variable’s past improve prediction beyond the information already being used?

Depth and runnable code: Module 8 notes §§8.1–8.7. Enders ch. 5 §2; Hamilton §8.2. Optional previews: Modules 9–10.