```{r} #| label: setup #| include: false set.seed(2026) library(ggplot2) theme_set(theme_minimal(base_size = 18)) ``` ## 🎬 The Idea in 3 Minutes ::: {style="text-align:center"} ```{=html} ``` [⬅ Previous lecture: 9 · Trends, Unit Roots and Brownian Motion](../09-trends-unit-root-asymptotics/trends_unit_root_asymptotics_lecture9.html){style="font-size:28px"} [A short film with the intuition. Watch it before the slides, or after. Captions: CC button.]{style="font-size:28px"} ::: --- ## 🎯 Learning Objectives ::: {style="font-size: 32px"} By the end of this lecture, you will be able to: - **Derive** the Dickey–Fuller limit of $T(\hat\rho-1)$ from the FCLT, and explain why it is pivotal - **Carry out** the DF test with the right deterministic terms and the right critical values - **Correct** for serially correlated errors, nonparametrically (Phillips–Perron) or parametrically (ADF) - **Interpret** a unit-root verdict as a statement about the persistence of shocks, and its limits ::: --- ## 🔁 Warm-up from last time {.quiz-question} Lecture 9: $y_t=y_{t-1}+\varepsilon_t$, $y_0=0$, $\varepsilon_t$ i.i.d.$(0,\sigma^2)$. What is the limit of $\frac1T\sum_{t=1}^T y_{t-1}\varepsilon_t$? - [$\frac{\sigma^2}{2}\{\chi^2(1)-1\}$]{.correct data-explanation="✅ Telescoping gives Σ y_{t−1}ε_t = ½{y_T² − Σε_t²}. Divided by T: y_T²/T → σ²χ²(1) and Σε_t²/T → σ². In Brownian form this is σ²∫B dB = (σ²/2)(B(1)² − 1)."} - $N(0,\sigma^4)$, as in the stationary case - $0$, because $E(y_{t-1}\varepsilon_t)=0$ - $\frac{\sigma^2}{2}\chi^2(1)$ --- ## 🗺️ Where We Are ::: {style="font-size: 32px"} Lecture 9 ended on a limit: $$T(\hat\rho-1)\ \xrightarrow{d}\ \frac{\int_0^1 \mathcal B\,d\mathcal B}{\int_0^1 \mathcal B(r)^2dr}.$$ It is non-normal, skewed to the left, and free of $\sigma^2$. ::: {.fragment} Today that limit becomes a **test**. Three questions decide whether the test is usable in practice: which deterministic terms enter the regression, what to do when the errors are serially correlated, and what a rejection or non-rejection actually tells us. ::: ::: --- ## ❓ The Question ::: {.callout-important} ## Are recessions forgotten? US real GDP fell about 4% in 2008–09. Does log GDP return to its old trend line, or is the loss carried forward for ever? ::: ::: {style="font-size: 30px"} If log GDP is **trend-stationary**, a shock dies out and the old path returns. If it has a **unit root**, every shock shifts the whole future path, and "returning to trend" has no meaning. The two views give the same short-run forecasts and very different long-run ones. ::: --- ## 📐 Random-Walk Moments (FCLT) ::: {style="font-size: 28px"} Let $\varepsilon_t$ be white noise with variance $\sigma^2$ and $x_t=x_{t-1}+\varepsilon_t$. With $S_T(r)=\frac{1}{\sigma\sqrt T}\,x_{[Tr]}$, $$S_T(r)\ \Rightarrow\ \mathcal B(r)\quad\text{(standard Brownian motion on }[0,1]),$$ $$\frac1T\sum_{t=2}^{T}x_{t-1}\varepsilon_t\ \xrightarrow{d}\ \sigma^2\!\int_0^1\!\mathcal B\,d\mathcal B,\qquad \frac1{T^2}\sum_{t=2}^{T}x_{t-1}^2\ \xrightarrow{d}\ \sigma^2\!\int_0^1\!\mathcal B(r)^2dr,$$ jointly. - **Remark 1.** The same holds for weak white noise that is a martingale difference sequence; the limit still has independent increments. - **Remark 2.** It holds for $x_0\neq0$, even random, provided $x_0$ is bounded in probability. ::: --- ## 📐 OLS Under a Unit Root ::: {style="font-size: 28px"} Model $x_t=\rho x_{t-1}+\varepsilon_t$ with $\rho=1$. The OLS estimator is $$\hat\rho=\frac{\sum_{t=2}^{T}x_tx_{t-1}}{\sum_{t=2}^{T}x_{t-1}^2}=1+\frac{\sum_{t=2}^{T}x_{t-1}\varepsilon_t}{\sum_{t=2}^{T}x_{t-1}^2}.$$ ::: {.fragment} Scale the numerator by $T$ and the denominator by $T^2$: $$T(\hat\rho-1)=\frac{\frac1T\sum x_{t-1}\varepsilon_t}{\frac1{T^2}\sum x_{t-1}^2}\ \xrightarrow{d}\ \underbrace{\frac{\int_0^1\mathcal B\,d\mathcal B}{\int_0^1\mathcal B(r)^2\,dr}}_{DF\ \text{distribution}}.$$ The $\sigma^2$ cancels. The rate is $T$, not $\sqrt T$: $\hat\rho$ is **superconsistent**. ::: ::: --- ## 📐 Pivotal, and Skewed to the Left ::: {style="font-size: 28px"} Compare the stationary case, $|\rho|<1$: $$\sqrt T(\hat\rho-\rho)\ \xrightarrow{d}\ N(0,\,1-\rho^2).$$ This limit depends on $\rho$: it is **non-pivotal**, and we must standardise before tabulating. ::: {.fragment} The DF distribution contains no unknown parameter: it is **pivotal**, and can be tabulated once, by simulation. ::: ::: {.fragment} Its shape: the numerator is $\int_0^1\mathcal B\,d\mathcal B=\tfrac12\{\mathcal B(1)^2-1\}$, and the denominator is positive, so $$P\big(T(\hat\rho-1)<0\big)\to P\big(\chi^2(1)<1\big)\approx0.68.$$ Two thirds of the mass lies below zero: $\hat\rho$ is biased downwards. ::: ::: --- ## 📐 The Dickey–Fuller Test ::: {style="font-size: 28px"} $$H_0:\ \rho=1\ \text{(unit root)}\qquad\text{vs}\qquad H_A:\ \rho<1\ \text{(stationarity)}$$ The test is one-sided and **left-tailed**. Two statistics: - the coefficient statistic: reject $H_0$ if $T(\hat\rho-1)-21.8$ and $-2.17>-3.41$. **Do not reject** the unit root at 5%. ::: ::: {.fragment} **The trap.** Against $N(0,1)$, $\tau=-2.17<-1.645$ would "reject". The normal table overstates the evidence against a unit root. ::: ::: --- ## 📐 Serially Correlated Errors ::: {style="font-size: 28px"} Return to $x_t=\rho x_{t-1}+u_t$, $\rho=1$. If the true process is not a random walk, $u_t=\Delta x_t$ is **serially correlated**. Write $$\Gamma_j=E(u_tu_{t-j}),\qquad \omega^2=\sum_{j=-\infty}^{\infty}\Gamma_j,\qquad \lambda=\sum_{j=1}^{\infty}\Gamma_j=\frac{\omega^2-\Gamma_0}{2}.$$ ::: {.fragment} With $\mathcal B$ now the Brownian motion from the FCLT normalised by $\omega$, $$T(\hat\rho-1)\ \xrightarrow{d}\ \frac{\int_0^1\mathcal B\,d\mathcal B+\lambda/\omega^2}{\int_0^1\mathcal B(r)^2\,dr}.$$ The extra term depends on nuisance parameters, so the DF tables no longer apply. ::: ::: --- ## 📐 Two Remedies ::: {style="font-size: 28px"} **Nonparametric: Phillips–Perron.** Subtract the bias directly: $$Z=T(\hat\rho-1)-\frac{\hat\lambda}{\frac1{T^2}\sum x_{t-1}^2}\ \xrightarrow{d}\ DF .$$ $\hat\lambda=(\hat\omega^2-\hat\Gamma_0)/2$ uses a Newey–West estimate $\hat\omega^2$. A one-sided sum need not be positive, so no positive-definite estimate of $\lambda$ is required. ::: {.fragment} **Parametric: augmented Dickey–Fuller.** Add lagged differences: $$x_t=\rho x_{t-1}+\sum_{j=1}^{p}\alpha_j\,\underbrace{\Delta x_{t-j}}_{\text{stationary}}+\underbrace{\varepsilon_t}_{\text{pre-whitened}} .$$ Choose $p$ by BIC. Then $ADF=\dfrac{T(\hat\rho-1)}{1-\sum_j\hat\alpha_j}\xrightarrow{d}DF$, and the $t$-ratio on $\rho$ $\xrightarrow{d}DF_\tau$. ::: ::: --- ## 💻 Does the Correction Work? ```{r} #| label: pp-adf #| code-summary: "📦 Random walk with MA(1) increments: rejection rates at 5%" nw_lrv <- function(u, L = floor(4 * (length(u) / 100)^(2 / 9))) { n <- length(u); g <- sapply(0:L, function(j) sum(u[(j + 1):n] * u[1:(n - j)]) / n) g[1] + 2 * sum((1 - (1:L) / (L + 1)) * g[-1]) # Bartlett kernel } stats3 <- function(x, pmax = 8) { # all with a constant y <- x[-1]; xl <- x[-length(x)]; n <- length(y) f <- .lm.fit(cbind(1, xl), y); dfz <- n * (f$coefficients[2] - 1) lam <- (nw_lrv(f$residuals) - mean(f$residuals^2)) / 2 pp <- dfz - lam / (sum((xl - mean(xl))^2) / n^2) dx <- diff(x); fitp <- function(p, i) { X <- cbind(x[i - 1], 1) if (p > 0) for (j in 1:p) X <- cbind(X, dx[i - 1 - j]); .lm.fit(X, x[i]) } i0 <- (pmax + 2):length(x) bic <- sapply(0:pmax, function(p) length(i0) * log(mean(fitp(p, i0)$residuals^2)) + (p + 2) * log(length(i0))) p <- which.min(bic) - 1; i <- (p + 2):length(x); b <- fitp(p, i)$coefficients adf <- length(i) * (b[1] - 1) / (1 - if (p > 0) sum(b[3:(p + 2)]) else 0) c(DF = dfz, PP = pp, ADF = adf) } rate <- function(theta, R = 2000, T = 200) { r <- t(replicate(R, { e <- rnorm(T + 1); stats3(cumsum(e[-1] + theta * e[-(T + 1)])) })) round(colMeans(r < -14.1), 3) } # -14.1 = 5% point of DF-bar rbind(`theta = -0.5` = rate(-0.5), `theta = 0` = rate(0), `theta = +0.5` = rate(0.5)) ``` ::: {style="font-size: 28px"} Under $H_0$ every entry should be 0.05. Uncorrected DF fails badly; ADF comes closest. ::: --- ## 🧠 Think–Pair–Share ```{r} #| label: tps-timer #| echo: false countdown::countdown(minutes = 3, seconds = 0, top = 0, right = 0, font_size = "2em", warn_when = 30) ``` ::: {style="font-size: 30px"} Two processes, driven by the **same** shocks, $T=300$: $$\text{(A)}\ x_t=x_{t-1}+\varepsilon_t,\qquad \text{(B)}\ x_t=0.999\,x_{t-1}+\varepsilon_t .$$ 1. After a unit shock, how much remains in (B) after 10 periods? After 700? 2. Would a DF test at 5% tell (A) and (B) apart? Guess the rejection rate under (B). 3. Which of the two do you want for a 4-quarter GDP forecast? For a 50-year pension forecast? ::: --- ## ✅ Think–Pair–Share: Solution ::: {style="font-size: 28px"} 1. In (B) the impulse response is $0.999^h$: $0.999^{10}=0.990$, $0.999^{700}=0.496$. The half-life is $\ln 0.5/\ln 0.999\approx693$ periods. In (A) the whole shock remains for ever. ::: {.fragment} 2. No. With $T=300$ the two are practically indistinguishable; the rejection rate under (B) is barely above the 5% size (next slide). ::: ::: {.fragment} 3. For four quarters it does not matter: short-run forecasts coincide. For fifty years it matters entirely: (B) reverts to its mean, (A) never does. ::: ::: --- ## 📈 The Twins: ρ = 1 and ρ = 0.999 ```{r} #| label: twins #| echo: false #| fig-width: 11 #| fig-height: 3.9 set.seed(2026) n <- 300; e <- rnorm(n) xa <- cumsum(e); xb <- as.numeric(stats::filter(e, 0.999, method = "recursive")) H <- 3000; h <- 1:H tw <- rbind( data.frame(t = 1:n, x = xa, model = "rho = 1", panel = "sample, t = 1..300"), data.frame(t = 1:n, x = xb, model = "rho = 0.999", panel = "sample, t = 1..300"), data.frame(t = h, x = rep(xa[n], H), model = "rho = 1", panel = "forecast, h = 1..3000"), data.frame(t = h, x = 0.999^h * xb[n], model = "rho = 0.999", panel = "forecast, h = 1..3000")) tw$panel <- factor(tw$panel, levels = unique(tw$panel)) ggplot(tw, aes(t, x, colour = model)) + geom_line(linewidth = 1.1) + geom_hline(yintercept = 0, colour = "grey60") + facet_wrap(~ panel, scales = "free_x") + scale_colour_manual(values = c("#8b2635", "#2f4b7c"), name = NULL) + labs(x = NULL, y = expression(x[t])) + theme_minimal(base_size = 20) + theme(legend.position = "top") ``` ::: {style="font-size: 28px"} Same shocks. In the sample the two paths are one line; only the forecasts part company. ::: --- ## 📊 Can a Test Tell Them Apart? ```{r} #| label: twins-power #| code-summary: "📦 DF τ test (constant), 5%: rejection rates, T = 300" rej <- function(rho, R = 2000) mean(replicate(R, { x <- as.numeric(stats::filter(rnorm(300), rho, method = "recursive")) df_stats(x, "const")[2] < -2.86 })) c(`rho = 1` = rej(1), `rho = 0.999` = rej(0.999), `rho = 0.95` = rej(0.95)) ``` ::: {style="font-size: 28px"} At $\rho=0.999$ the test rejects about as often as under the null: its power is essentially its size. Even $\rho=0.95$, a half-life of 14 periods, is missed four times in ten. ::: --- ## 🎛️ Explore: How Persistent Is ρ? ```{ojs} //| echo: false viewof rho = Inputs.range([0.80, 1.00], {step: 0.005, value: 0.95, label: "ρ"}) ``` ```{ojs} //| echo: false shocks = { let s = 2026 >>> 0; const u = () => { s = (s + 0x6D2B79F5) >>> 0; let t = s; t = Math.imul(t ^ (t >>> 15), t | 1); t ^= t + Math.imul(t ^ (t >>> 7), t | 61); return ((t ^ (t >>> 14)) >>> 0) / 4294967296; }; return Array.from({length: 300}, () => Math.sqrt(-2 * Math.log(1 - u())) * Math.cos(2 * Math.PI * u())); } pathData = { const out = []; let x = 0; shocks.forEach((e, i) => { x = rho * x + e; out.push({t: i + 1, x, part: "sample"}); }); const xT = x; for (let h = 1; h <= 300; h++) out.push({t: 300 + h, x: Math.pow(rho, h) * xT, part: "forecast"}); return out; } Plot.plot({ width: 1100, height: 330, marginLeft: 70, marginBottom: 50, style: {fontSize: "18px"}, x: {label: "t"}, y: {label: "x_t"}, color: {domain: ["sample", "forecast"], range: ["#2f4b7c", "#8b2635"]}, marks: [ Plot.ruleY([0], {stroke: "#999"}), Plot.ruleX([300], {stroke: "#999", strokeDasharray: "4,4"}), Plot.line(pathData, {x: "t", y: "x", stroke: "part", strokeWidth: 2}) ] }) ``` ::: {style="font-size: 28px"} Blue: the sample, same shocks at every $\rho$. Red: the forecast $\rho^h x_T$. As $\rho\to1$ the path stops crossing zero and the forecast stops coming back. ::: --- ## 📐 What a Unit-Root Test Means ::: {style="font-size: 30px"} - The test is about the **persistence** of shocks. Under a unit root, as for log GNP on this view, a shock lasts for ever. - In a finite sample, $x_t=x_{t-1}+\varepsilon_t$ and $x_t=0.999x_{t-1}+\varepsilon_t$ cannot be told apart. Their short-run predictions are the same. - They differ only at long horizons, where the stationary forecast returns to the mean. - Low power against near-unit-root alternatives is therefore not a defect of a particular test: it is in the problem. - "Do not reject" means "the data are consistent with very persistent shocks", not "$\rho$ is exactly 1". ::: --- ## 💰 Case Study: Three US Series :::: columns ::: {.column width="50%"} ::: {.callout-note} ## 📈 The question 1. Does log real GDP have a unit root around a linear trend? 2. Does the unemployment rate wander, or return to a natural rate? 3. Is the euro–dollar rate a random walk? ::: ::: ::: {.column width="50%"} ::: {.callout-tip} ## 📊 The data **Series**: FRED `GDPC1` (quarterly, 1947–2026, logs, constant + trend); `UNRATE` (monthly, 1948–2019, constant); `DEXUSEU` (daily, 1999–2026, monthly averages, logs, constant) **Test**: ADF, lag $p$ by BIC ($p\le12$) ::: ::: :::: ```{.r .display-only} # How the CSVs were obtained (not run at render time): for (id in c("GDPC1", "UNRATE", "DEXUSEU")) download.file(paste0("https://fred.stlouisfed.org/graph/fredgraph.csv?id=", id), paste0("data/", id, ".csv")) ``` --- ## 💰 Case Study: ADF by Hand ```{r} #| label: case-adf #| code-summary: "📦 ADF regression with BIC lag choice, on vendored FRED data" rd <- function(f) { d <- read.csv(f); d[[2]] <- suppressWarnings(as.numeric(d[[2]])) d <- na.omit(d); d$date <- as.Date(d$observation_date); d } gdp <- rd("data/GDPC1.csv"); un <- rd("data/UNRATE.csv"); eu <- rd("data/DEXUSEU.csv") eum <- aggregate(eu$DEXUSEU, list(m = format(eu$date, "%Y-%m")), mean) adf <- function(x, det, pmax = 12) { dx <- diff(x); fitp <- function(p, i) { X <- cbind(x[i - 1], 1); if (det == "trend") X <- cbind(X, i) if (p > 0) for (j in 1:p) X <- cbind(X, dx[i - 1 - j]) f <- lm.fit(X, x[i]); s2 <- sum(f$residuals^2) / (length(i) - ncol(X)) list(f = f, k = ncol(X), se = sqrt(s2 * solve(crossprod(X))[1, 1])) } i0 <- (pmax + 2):length(x) bic <- sapply(0:pmax, function(p) { r <- fitp(p, i0) length(i0) * log(mean(r$f$residuals^2)) + r$k * log(length(i0)) }) p <- which.min(bic) - 1; r <- fitp(p, (p + 2):length(x)); b <- r$f$coefficients[1] c(p = p, T = length(x) - p - 1, rho = round(b, 4), tau = round((b - 1) / r$se, 2), crit_5pct = if (det == "trend") -3.41 else -2.86) } rbind(log_GDP = adf(log(gdp$GDPC1), "trend"), unemployment = adf(un$UNRATE, "const"), log_EURUSD = adf(log(eum$x), "const")) ``` ::: {style="font-size: 28px"} Read the `tau` column against `crit_5pct`: reject the unit root only when `tau` is further left. ::: --- ## 💰 Case Study: What the Data Say ```{r} #| label: case-fig #| echo: false #| fig-width: 10 #| fig-height: 3.6 lv <- c("log real GDP, OLS trend", "unemployment rate, %", "log USD per EUR") cs <- rbind( data.frame(date = gdp$date, v = log(gdp$GDPC1), s = "log real GDP, OLS trend"), data.frame(date = un$date, v = un$UNRATE, s = "unemployment rate, %"), data.frame(date = as.Date(paste0(eum$m, "-01")), v = log(eum$x), s = "log USD per EUR")) tr <- data.frame(date = gdp$date, v = fitted(lm(log(gdp$GDPC1) ~ seq_along(gdp$date))), s = "log real GDP, OLS trend") cs$s <- factor(cs$s, levels = lv); tr$s <- factor(tr$s, levels = lv) ggplot(cs, aes(date, v)) + geom_line(colour = "#2f4b7c", linewidth = 0.8) + geom_line(data = tr, colour = "#8b2635", linetype = "dashed") + facet_wrap(~ s, scales = "free") + labs(x = NULL, y = NULL) + theme_minimal(base_size = 20) + theme(strip.text = element_text(size = 17)) ``` ::: {style="font-size: 28px"} **What the data say:** log GDP ($\tau=-1.25$) and the euro ($\tau=-2.08$) do not reject the unit root: GDP stayed below its pre-2008 trend. Unemployment rejects ($\tau=-4.17$): very persistent, yet mean-reverting. A non-rejection is equally consistent with $\rho=0.99$. ::: --- ## 📝 Quiz #1: Critical Values {.quiz-question} A DF regression with a constant (no trend) gives $\tau=-2.40$. At the 5% level you should: - [Not reject the unit root, since $-2.40>-2.86$]{.correct data-explanation="✅ With a constant the reference distribution is DF̄_τ, whose 5% point is −2.86. −2.40 lies to its right, so H0: ρ = 1 is not rejected. The N(0,1) value −1.645 would wrongly reject."} - Reject the unit root, since $-2.40<-1.645$ - Reject the unit root, since $|-2.40|>1.96$ - Not reject, since $\tau$ must be positive to reject --- ## 📝 Quiz #2: Why ADF Adds Lags {.quiz-question} In the augmented DF regression $x_t=\rho x_{t-1}+\sum_{j=1}^p\alpha_j\Delta x_{t-j}+\varepsilon_t$, what is the role of the lagged differences? - [They absorb the serial correlation in $\Delta x_t$, so that $\varepsilon_t$ is serially uncorrelated and the DF tables apply]{.correct data-explanation="✅ Serial correlation adds λ/ω² to the DF limit. The lagged differences pre-whiten the error, and T(ρ̂−1)/(1−Σα̂_j) recovers the DF distribution."} - They remove the deterministic trend - They make $x_t$ stationary under the null - They raise the power against $\rho=0.999$ --- ## 📝 Quiz #3: Pivotal {.quiz-question} Why can the DF distribution be tabulated once and for all, while the stationary limit $N(0,1-\rho^2)$ cannot? - [The DF limit contains no unknown parameter; the stationary one depends on $\rho$]{.correct data-explanation="✅ In T(ρ̂ − 1) the σ² cancels between numerator and denominator, leaving a functional of standard Brownian motion only. The stationary limit has variance 1 − ρ², which must be estimated and standardised."} - The DF limit is normal - The DF limit does not depend on the deterministic terms - The stationary limit has infinite variance --- ## 📋 Key Formulas ::: {style="font-size: 28px"} | | | |---|---| | DF coefficient | $T(\hat\rho-1)\xrightarrow{d}\int_0^1\mathcal B\,d\mathcal B\big/\int_0^1\mathcal B^2dr$ | | DF $t$-ratio | $\tau\xrightarrow{d}\int_0^1\mathcal B\,d\mathcal B\big/\big(\int_0^1\mathcal B^2dr\big)^{1/2}$ | | constant / trend | replace $\mathcal B$ by $\bar{\mathcal B}=\mathcal B-\int\mathcal B$, or $\tilde{\mathcal B}=\mathcal B-b_0-b_1r$ | | 5% points, $\tau$ | $-1.94$, $-2.86$, $-3.41$ | | correlated errors | limit gains $+\lambda/\omega^2$ in the numerator, $\lambda=\sum_{j\ge1}\Gamma_j$ | | Phillips–Perron | $Z=T(\hat\rho-1)-\hat\lambda\big/\big(T^{-2}\sum x_{t-1}^2\big)$ | | ADF | $T(\hat\rho-1)/(1-\sum_j\hat\alpha_j)$, $p$ by BIC | ::: --- ## 📋 Summary ::: {style="font-size: 30px"} - Under a unit root $\hat\rho$ converges at rate $T$ to a pivotal, left-skewed DF distribution - The test is left-tailed; the critical values come from DF tables, never from $N(0,1)$ - Always include a constant; include a trend for trending series - Serial correlation adds a nuisance term: remove it with Phillips–Perron or with ADF lags - A unit root and $\rho=0.999$ are indistinguishable in a sample and differ only in the long run ::: --- ## 📚 Further Reading ::: {style="font-size: 28px"} - Hamilton, J. D. (1994), *Time Series Analysis*, ch. 17 - Tsay, R. S. (2010), *Analysis of Financial Time Series*, §2.7 - Dickey, D. A. and Fuller, W. A. (1979), *JASA*; Said, S. E. and Dickey, D. A. (1984), *Biometrika* - Phillips, P. C. B. and Perron, P. (1988), *Biometrika* - MacKinnon, J. G. (1996), *Journal of Applied Econometrics* (critical values) - Schwert, G. W. (1989), *JBES* (size of the tests with MA errors) **Next lecture: Nonlinear Models and Threshold Autoregression.** A series can be very persistent without a unit root, and white noise can still be predictable. [⬅ Previous lecture: 09 · Trends, Unit Roots and Brownian Motion](../09-trends-unit-root-asymptotics/trends_unit_root_asymptotics_lecture9.html) :::
