{
  "deckName": "Exam MAS-II — Time Series (ARIMA)",
  "examCode": "Exam MAS-II",
  "cards": [
    {
      "front": "Define a **white noise** process $\\{\\epsilon_t\\}$ and state its mean, variance, and autocovariance.",
      "back": "White noise is a sequence of uncorrelated, mean-zero, constant-variance shocks: $E[\\epsilon_t]=0$, $\\operatorname{Var}(\\epsilon_t)=\\sigma^2$ for all $t$, and $\\operatorname{Cov}(\\epsilon_t,\\epsilon_s)=0$ for $t\\neq s$.\nEquivalently $\\gamma_0=\\sigma^2$ and $\\gamma_k=0$ for $k\\neq 0$, so $\\rho_0=1$ and $\\rho_k=0$ for $k\\neq 0$. It is the irreducible random input that drives AR/MA/ARMA models.",
      "tag": "Stationarity & white noise"
    },
    {
      "front": "State the conditions for **weak (covariance) stationarity** of a series $\\{Y_t\\}$.",
      "back": "A process is weakly stationary if its first two moments do not depend on time:\n1. Constant mean: $E[Y_t]=\\mu$ for all $t$.\n2. Constant variance: $\\operatorname{Var}(Y_t)=\\gamma_0$ for all $t$.\n3. Autocovariance depends only on the lag, not the position: $\\operatorname{Cov}(Y_t,Y_{t+k})=\\gamma_k$ for all $t$.\nWeak stationarity is what Box–Jenkins modeling requires; a deterministic trend or changing variance violates it.",
      "tag": "Stationarity & white noise"
    },
    {
      "front": "Define the **autocovariance** $\\gamma_k$ and **autocorrelation** $\\rho_k$ of a stationary series.",
      "back": "Autocovariance at lag $k$: $\\gamma_k=\\operatorname{Cov}(Y_t,Y_{t+k})=E[(Y_t-\\mu)(Y_{t+k}-\\mu)]$.\nAutocorrelation at lag $k$: $\\rho_k=\\frac{\\gamma_k}{\\gamma_0}$, the autocovariance normalized by the variance $\\gamma_0$.\nProperties: $\\rho_0=1$, $\\rho_k=\\rho_{-k}$ (symmetry), and $|\\rho_k|\\le 1$. The plot of $\\rho_k$ against $k$ is the **ACF**.",
      "tag": "Stationarity & white noise"
    },
    {
      "front": "What is the **backshift (lag) operator** $B$, and how is it used to write AR, MA, and differencing?",
      "back": "The backshift operator shifts a series one step back: $B Y_t = Y_{t-1}$, and in general $B^k Y_t = Y_{t-k}$.\nAR(p): $\\phi(B)Y_t=\\epsilon_t$ with $\\phi(B)=1-\\phi_1 B-\\cdots-\\phi_p B^p$.\nMA(q): $Y_t=\\theta(B)\\epsilon_t$ with $\\theta(B)=1+\\theta_1 B+\\cdots+\\theta_q B^q$.\nDifferencing: $\\nabla Y_t=(1-B)Y_t=Y_t-Y_{t-1}$.",
      "tag": "Stationarity & white noise"
    },
    {
      "front": "Write the **AR(1)** model and state its stationarity condition.",
      "back": "Mean-zero AR(1): $Y_t=\\phi Y_{t-1}+\\epsilon_t$, where $\\{\\epsilon_t\\}$ is white noise with variance $\\sigma^2$.\nStationarity requires $|\\phi|<1$ (the root of the characteristic equation $1-\\phi z=0$, namely $z=1/\\phi$, lies outside the unit circle). If $\\phi=1$ the process is a non-stationary random walk; if $|\\phi|>1$ it explodes.",
      "tag": "AR models"
    },
    {
      "front": "For a stationary mean-zero AR(1), give the **variance** $\\gamma_0$ and the **autocorrelation** $\\rho_k$.",
      "back": "Variance: $\\gamma_0=\\operatorname{Var}(Y_t)=\\dfrac{\\sigma^2}{1-\\phi^2}$.\nAutocorrelation: $\\rho_k=\\phi^{\\,k}$ for $k\\ge 0$, so the ACF decays geometrically. When $0<\\phi<1$ it decays monotonically; when $-1<\\phi<0$ it alternates in sign while shrinking in magnitude.",
      "tag": "AR models"
    },
    {
      "front": "An AR(1) has $\\phi=0.6$ and shock variance $\\sigma^2=4$. Find $\\gamma_0$ and the autocorrelations $\\rho_1,\\rho_2,\\rho_3$.",
      "back": "Variance: $\\gamma_0=\\dfrac{\\sigma^2}{1-\\phi^2}=\\dfrac{4}{1-0.36}=\\dfrac{4}{0.64}=6.25$.\nAutocorrelations $\\rho_k=\\phi^k$:\n$\\rho_1=0.6$, $\\rho_2=0.6^2=0.36$, $\\rho_3=0.6^3=0.216$.\nThe corresponding autocovariances are $\\gamma_k=\\rho_k\\gamma_0$: $\\gamma_1=0.6(6.25)=3.75$, $\\gamma_2=0.36(6.25)=2.25$.",
      "tag": "AR models"
    },
    {
      "front": "An AR(1) is $Y_t=-0.5\\,Y_{t-1}+\\epsilon_t$ with $\\sigma^2=3$. Compute $\\gamma_0$, $\\rho_1$, $\\rho_2$ and describe the ACF pattern.",
      "back": "Variance: $\\gamma_0=\\dfrac{3}{1-(-0.5)^2}=\\dfrac{3}{1-0.25}=\\dfrac{3}{0.75}=4$.\n$\\rho_1=\\phi=-0.5$, $\\rho_2=\\phi^2=0.25$, $\\rho_3=\\phi^3=-0.125$.\nBecause $\\phi<0$ the ACF **alternates in sign** ($-,+,-,\\dots$) while decaying geometrically in magnitude.",
      "tag": "AR models"
    },
    {
      "front": "Include a nonzero mean in AR(1): write $Y_t=c+\\phi Y_{t-1}+\\epsilon_t$ and give the process mean $\\mu$.",
      "back": "Taking expectations of a stationary AR(1): $\\mu=c+\\phi\\mu$, so $\\mu=\\dfrac{c}{1-\\phi}$ (valid for $|\\phi|<1$).\nThe intercept $c$ is **not** the mean; the mean depends on $\\phi$ too. The autocorrelation structure is unchanged from the mean-zero case: $\\rho_k=\\phi^k$.",
      "tag": "AR models"
    },
    {
      "front": "An AR(1) is $Y_t=10+0.4\\,Y_{t-1}+\\epsilon_t$ with $\\sigma^2=5$. Find the process mean and variance.",
      "back": "Mean: $\\mu=\\dfrac{c}{1-\\phi}=\\dfrac{10}{1-0.4}=\\dfrac{10}{0.6}\\approx 16.667$.\nVariance: $\\gamma_0=\\dfrac{\\sigma^2}{1-\\phi^2}=\\dfrac{5}{1-0.16}=\\dfrac{5}{0.84}\\approx 5.952$.\nNote the mean uses $1-\\phi$ while the variance uses $1-\\phi^2$ — different denominators.",
      "tag": "AR models"
    },
    {
      "front": "Write the **AR(2)** model and state the **Yule–Walker equations** for its autocorrelations.",
      "back": "AR(2): $Y_t=\\phi_1 Y_{t-1}+\\phi_2 Y_{t-2}+\\epsilon_t$.\nMultiplying by $Y_{t-k}$ and taking expectations gives the Yule–Walker recursion $\\rho_k=\\phi_1\\rho_{k-1}+\\phi_2\\rho_{k-2}$ for $k\\ge 1$, with $\\rho_0=1$ and $\\rho_{-1}=\\rho_1$.\nIn particular $\\rho_1=\\dfrac{\\phi_1}{1-\\phi_2}$ and $\\rho_2=\\phi_1\\rho_1+\\phi_2$.",
      "tag": "AR models"
    },
    {
      "front": "State the **stationarity conditions** for an AR(2) model in terms of $\\phi_1$ and $\\phi_2$.",
      "back": "An AR(2) $Y_t=\\phi_1 Y_{t-1}+\\phi_2 Y_{t-2}+\\epsilon_t$ is stationary iff the roots of $1-\\phi_1 z-\\phi_2 z^2=0$ lie outside the unit circle. Equivalently the three inequalities hold:\n$\\phi_1+\\phi_2<1$, $\\quad \\phi_2-\\phi_1<1$, $\\quad |\\phi_2|<1$ (i.e. $-1<\\phi_2<1$).\nTogether these define the stationarity triangle in the $(\\phi_1,\\phi_2)$ plane.",
      "tag": "AR models"
    },
    {
      "front": "An AR(2) has $\\phi_1=0.5$ and $\\phi_2=0.2$. Verify stationarity and find $\\rho_1$, $\\rho_2$, $\\rho_3$.",
      "back": "Stationarity: $\\phi_1+\\phi_2=0.7<1$, $\\phi_2-\\phi_1=-0.3<1$, $|\\phi_2|=0.2<1$ — all satisfied, so stationary.\n$\\rho_1=\\dfrac{\\phi_1}{1-\\phi_2}=\\dfrac{0.5}{1-0.2}=\\dfrac{0.5}{0.8}=0.625$.\n$\\rho_2=\\phi_1\\rho_1+\\phi_2=0.5(0.625)+0.2=0.3125+0.2=0.5125$.\n$\\rho_3=\\phi_1\\rho_2+\\phi_2\\rho_1=0.5(0.5125)+0.2(0.625)=0.25625+0.125=0.38125$.",
      "tag": "AR models"
    },
    {
      "front": "An AR(2) has $\\phi_1=1.0$ and $\\phi_2=-0.25$. Is it stationary, and what are $\\rho_1$ and $\\rho_2$?",
      "back": "Check: $\\phi_1+\\phi_2=0.75<1$, $\\phi_2-\\phi_1=-1.25<1$, $|\\phi_2|=0.25<1$ — stationary.\n$\\rho_1=\\dfrac{\\phi_1}{1-\\phi_2}=\\dfrac{1.0}{1-(-0.25)}=\\dfrac{1}{1.25}=0.8$.\n$\\rho_2=\\phi_1\\rho_1+\\phi_2=1.0(0.8)+(-0.25)=0.8-0.25=0.55$.",
      "tag": "AR models"
    },
    {
      "front": "Given sample autocorrelations $\\hat\\rho_1=0.7$ and $\\hat\\rho_2=0.5$ from an AR(2) fit, solve the **Yule–Walker equations** for $\\hat\\phi_1$ and $\\hat\\phi_2$.",
      "back": "The lag-1 and lag-2 Yule–Walker equations are:\n$\\rho_1=\\phi_1+\\phi_2\\rho_1$ and $\\rho_2=\\phi_1\\rho_1+\\phi_2$.\nClosed form: $\\phi_1=\\dfrac{\\rho_1(1-\\rho_2)}{1-\\rho_1^2}$, $\\phi_2=\\dfrac{\\rho_2-\\rho_1^2}{1-\\rho_1^2}$.\n$\\phi_1=\\dfrac{0.7(1-0.5)}{1-0.49}=\\dfrac{0.35}{0.51}\\approx 0.686$.\n$\\phi_2=\\dfrac{0.5-0.49}{0.51}=\\dfrac{0.01}{0.51}\\approx 0.020$.",
      "tag": "AR models"
    },
    {
      "front": "Write the **MA(1)** model and give its mean, variance, and autocorrelations.",
      "back": "MA(1): $Y_t=\\epsilon_t+\\theta\\,\\epsilon_{t-1}$ (some texts use $-\\theta$; check the sign convention).\nMean $E[Y_t]=0$. Variance $\\gamma_0=\\sigma^2(1+\\theta^2)$.\nAutocorrelations: $\\rho_1=\\dfrac{\\theta}{1+\\theta^2}$ and $\\rho_k=0$ for $k\\ge 2$. An MA(1) is **always stationary** (finite sum of white noise), so no stationarity restriction on $\\theta$.",
      "tag": "MA models"
    },
    {
      "front": "What does it mean that the ACF of an MA(q) process **cuts off**, and at which lag?",
      "back": "For MA(q), $Y_t=\\epsilon_t+\\theta_1\\epsilon_{t-1}+\\cdots+\\theta_q\\epsilon_{t-q}$, the autocorrelation is **nonzero only out to lag $q$**: $\\rho_k\\neq 0$ for $k\\le q$ but $\\rho_k=0$ for all $k>q$.\nThis abrupt drop to zero after lag $q$ is the ACF \"cutting off.\" It is the signature used to identify the MA order $q$ from a sample ACF plot.",
      "tag": "MA models"
    },
    {
      "front": "An MA(1) has $\\theta=0.8$ and $\\sigma^2=2$. Find $\\gamma_0$, $\\gamma_1$, $\\rho_1$, and $\\rho_2$.",
      "back": "Variance: $\\gamma_0=\\sigma^2(1+\\theta^2)=2(1+0.64)=2(1.64)=3.28$.\nLag-1 autocovariance: $\\gamma_1=\\theta\\sigma^2=0.8(2)=1.6$.\n$\\rho_1=\\dfrac{\\gamma_1}{\\gamma_0}=\\dfrac{1.6}{3.28}\\approx 0.488$, or equivalently $\\dfrac{\\theta}{1+\\theta^2}=\\dfrac{0.8}{1.64}\\approx 0.488$.\n$\\rho_2=0$ (and all higher lags), the MA(1) cutoff.",
      "tag": "MA models"
    },
    {
      "front": "An MA(1) has $\\theta=-0.5$ and $\\sigma^2=9$. Compute $\\gamma_0$, $\\rho_1$, and the maximum possible $|\\rho_1|$ for any MA(1).",
      "back": "Variance: $\\gamma_0=\\sigma^2(1+\\theta^2)=9(1+0.25)=9(1.25)=11.25$.\n$\\rho_1=\\dfrac{\\theta}{1+\\theta^2}=\\dfrac{-0.5}{1.25}=-0.4$, and $\\rho_k=0$ for $k\\ge 2$.\nThe function $\\dfrac{\\theta}{1+\\theta^2}$ is maximized at $\\theta=\\pm 1$, giving $|\\rho_1|\\le 0.5$ — an MA(1) can never have lag-1 autocorrelation beyond $\\pm 0.5$.",
      "tag": "MA models"
    },
    {
      "front": "What is the **invertibility** condition for an MA(1), and why does it matter?",
      "back": "An MA(1) $Y_t=\\epsilon_t+\\theta\\epsilon_{t-1}$ is invertible iff $|\\theta|<1$. Invertibility lets you rewrite the MA as a convergent infinite-AR representation $\\epsilon_t=\\sum_{j\\ge 0}(-\\theta)^j Y_{t-j}$, so the shocks are recoverable from past observations.\nIt matters because two MA(1)s with reciprocal parameters $\\theta$ and $1/\\theta$ produce the **same** ACF; imposing $|\\theta|<1$ makes the model identifiable (a unique parameter for a given autocorrelation).",
      "tag": "MA models"
    },
    {
      "front": "Write a general **MA(2)** model and give its variance and lag-1, lag-2 autocorrelations.",
      "back": "MA(2): $Y_t=\\epsilon_t+\\theta_1\\epsilon_{t-1}+\\theta_2\\epsilon_{t-2}$.\nVariance: $\\gamma_0=\\sigma^2(1+\\theta_1^2+\\theta_2^2)$.\nAutocovariances: $\\gamma_1=\\sigma^2(\\theta_1+\\theta_1\\theta_2)$, $\\gamma_2=\\sigma^2\\theta_2$, and $\\gamma_k=0$ for $k\\ge 3$.\nThus $\\rho_1=\\dfrac{\\theta_1(1+\\theta_2)}{1+\\theta_1^2+\\theta_2^2}$, $\\rho_2=\\dfrac{\\theta_2}{1+\\theta_1^2+\\theta_2^2}$, and the ACF cuts off after lag $2$.",
      "tag": "MA models"
    },
    {
      "front": "An MA(2) has $\\theta_1=0.5$, $\\theta_2=0.3$, $\\sigma^2=1$. Compute $\\gamma_0$, $\\rho_1$, $\\rho_2$, $\\rho_3$.",
      "back": "$\\gamma_0=\\sigma^2(1+\\theta_1^2+\\theta_2^2)=1(1+0.25+0.09)=1.34$.\n$\\gamma_1=\\sigma^2(\\theta_1+\\theta_1\\theta_2)=1(0.5+0.5\\cdot0.3)=0.5+0.15=0.65$, so $\\rho_1=\\dfrac{0.65}{1.34}\\approx 0.485$.\n$\\gamma_2=\\sigma^2\\theta_2=0.3$, so $\\rho_2=\\dfrac{0.3}{1.34}\\approx 0.224$.\n$\\rho_3=0$ (MA(2) cutoff after lag 2).",
      "tag": "MA models"
    },
    {
      "front": "Write the general **ARMA(p,q)** model and state when it is stationary and invertible.",
      "back": "ARMA(p,q): $Y_t=\\phi_1 Y_{t-1}+\\cdots+\\phi_p Y_{t-p}+\\epsilon_t+\\theta_1\\epsilon_{t-1}+\\cdots+\\theta_q\\epsilon_{t-q}$, or compactly $\\phi(B)Y_t=\\theta(B)\\epsilon_t$.\n**Stationarity** is governed entirely by the AR part: roots of $\\phi(z)=0$ outside the unit circle.\n**Invertibility** is governed by the MA part: roots of $\\theta(z)=0$ outside the unit circle. The two conditions are separate.",
      "tag": "ARMA & ARIMA"
    },
    {
      "front": "For an **ARMA(1,1)** process $Y_t=\\phi Y_{t-1}+\\epsilon_t+\\theta\\epsilon_{t-1}$, describe the shape of the ACF beyond lag 1.",
      "back": "The lag-1 autocorrelation reflects both the AR and MA terms, but for $k\\ge 2$ the ACF satisfies the pure-AR recursion $\\rho_k=\\phi\\,\\rho_{k-1}$. So after the first lag the ACF **tails off geometrically** like an AR(1) at rate $\\phi$ — it never cuts off.\nBecause both ACF and PACF tail off (neither cuts off), a mixed ARMA is identified by the failure of either to truncate cleanly.",
      "tag": "ARMA & ARIMA"
    },
    {
      "front": "Define an **ARIMA(p,d,q)** model and the role of the differencing order $d$.",
      "back": "ARIMA(p,d,q) applies an ARMA(p,q) to the $d$-th difference of the series: $\\phi(B)(1-B)^d Y_t=\\theta(B)\\epsilon_t$.\nThe integer $d$ is the number of times you difference $Y_t$ to remove a trend and reach stationarity. With $W_t=(1-B)^d Y_t$, $\\{W_t\\}$ is a stationary ARMA(p,q). Most economic/loss series need $d=1$ (remove a linear trend); $d=2$ removes a quadratic trend.",
      "tag": "ARMA & ARIMA"
    },
    {
      "front": "Why difference a series, and how does **first differencing** $\\nabla Y_t=Y_t-Y_{t-1}$ remove a linear trend?",
      "back": "Stationary modeling needs a constant mean; a deterministic linear trend $Y_t=a+bt+\\epsilon_t$ has a mean that grows with $t$, violating stationarity.\nFirst differencing gives $\\nabla Y_t=Y_t-Y_{t-1}=b+(\\epsilon_t-\\epsilon_{t-1})$, whose mean is the constant $b$ — the trend slope becomes a constant level, restoring stationarity. A quadratic trend requires differencing twice ($d=2$).",
      "tag": "ARMA & ARIMA"
    },
    {
      "front": "Show that a **random walk** is an ARIMA(0,1,0), and explain why it is non-stationary.",
      "back": "A random walk is $Y_t=Y_{t-1}+\\epsilon_t$, i.e. $(1-B)Y_t=\\epsilon_t$ — differencing once ($d=1$) yields white noise, so it is ARIMA(0,1,0).\nIt is non-stationary because its variance grows without bound: $\\operatorname{Var}(Y_t)=t\\sigma^2$ (starting from $Y_0=0$), which depends on $t$. It is the boundary case $\\phi=1$ of an AR(1), a unit root.",
      "tag": "ARMA & ARIMA"
    },
    {
      "front": "A random walk with drift is $Y_t=Y_{t-1}+\\delta+\\epsilon_t$ with $\\delta=2$ and $\\sigma^2=9$, starting at $Y_0=50$. Find $E[Y_3]$ and $\\operatorname{Var}(Y_3)$.",
      "back": "Unrolling: $Y_t=Y_0+\\delta t+\\sum_{j=1}^{t}\\epsilon_j$.\nMean: $E[Y_3]=Y_0+\\delta\\cdot 3=50+2(3)=56$.\nVariance: $\\operatorname{Var}(Y_3)=3\\sigma^2=3(9)=27$ (the three independent shocks accumulate; the drift is deterministic).\nThe growing variance confirms non-stationarity — differencing gives the stationary $\\nabla Y_t=2+\\epsilon_t$.",
      "tag": "ARMA & ARIMA"
    },
    {
      "front": "Define the **partial autocorrelation function (PACF)** $\\phi_{kk}$ and what it measures.",
      "back": "The lag-$k$ partial autocorrelation $\\phi_{kk}$ is the correlation between $Y_t$ and $Y_{t-k}$ **after removing** the linear effect of the intervening values $Y_{t-1},\\dots,Y_{t-k+1}$.\nIt equals the last coefficient $\\phi_{kk}$ in the best linear AR($k$) fit. $\\phi_{11}=\\rho_1$. For an AR(p), $\\phi_{kk}=0$ for $k>p$ — the PACF cuts off at lag $p$.",
      "tag": "ACF/PACF identification"
    },
    {
      "front": "State the **Box–Jenkins identification rules** for distinguishing AR(p), MA(q), and ARMA from the ACF and PACF.",
      "back": "**AR(p):** ACF **tails off** (decays geometrically/sinusoidally); PACF **cuts off** after lag $p$.\n**MA(q):** ACF **cuts off** after lag $q$; PACF **tails off**.\n**ARMA(p,q):** **both** ACF and PACF tail off (neither truncates).\nWhite noise: both ACF and PACF are ~0 at all lags. Mnemonic: the function that *cuts off* names the model and its order.",
      "tag": "ACF/PACF identification"
    },
    {
      "front": "A sample ACF decays geometrically (lags ~$0.6, 0.36, 0.22,\\dots$) and the sample PACF shows a single spike at lag 1 ($\\hat\\phi_{11}\\approx0.6$) then ~0. Identify the model.",
      "back": "ACF tailing off + PACF cutting off after lag 1 is the textbook signature of an **AR(1)**.\nThe geometric ACF $0.6,0.36,0.22$ matches $\\rho_k=\\phi^k$ with $\\phi\\approx 0.6$, and the lone PACF spike at lag 1 confirms $p=1$. Estimated model: $Y_t\\approx 0.6\\,Y_{t-1}+\\epsilon_t$.",
      "tag": "ACF/PACF identification"
    },
    {
      "front": "A sample ACF has a significant spike only at lag 1 ($\\hat\\rho_1\\approx0.45$) and is ~0 thereafter, while the PACF tails off (decaying, alternating). Identify the model.",
      "back": "ACF cutting off after lag 1 + PACF tailing off is the signature of an **MA(1)**.\nThe single ACF spike at lag 1 sets $q=1$. Solving $\\rho_1=\\dfrac{\\theta}{1+\\theta^2}=0.45$ recovers the parameter. Estimated model: $Y_t=\\epsilon_t+\\theta\\epsilon_{t-1}$.",
      "tag": "ACF/PACF identification"
    },
    {
      "front": "A series' ACF decays slowly and nearly linearly (e.g. $0.98, 0.95, 0.92,\\dots$), barely dying out. What does this indicate and what is the remedy?",
      "back": "A very slowly decaying ACF that stays near 1 for many lags signals **non-stationarity** (a trend or near-unit root) — the series is not yet stationary, so AR/MA orders cannot be read off reliably.\nThe remedy is to **difference** the series ($d=1$, or $d=2$ if a single difference still decays slowly) and re-examine the ACF/PACF of the differenced series.",
      "tag": "ACF/PACF identification"
    },
    {
      "front": "Give the approximate **standard error** used to judge whether a sample autocorrelation $\\hat\\rho_k$ is significant, and the resulting bound.",
      "back": "Under the white-noise null, $\\operatorname{SE}(\\hat\\rho_k)\\approx \\dfrac{1}{\\sqrt{n}}$ for a series of length $n$.\nThe usual $95\\%$ significance band is $\\pm \\dfrac{1.96}{\\sqrt{n}}$ (often drawn as $\\pm\\dfrac{2}{\\sqrt{n}}$). A sample ACF/PACF value outside this band is treated as significantly nonzero; values inside are deemed indistinguishable from zero.",
      "tag": "ACF/PACF identification"
    },
    {
      "front": "With $n=100$ observations, a sample autocorrelation is $\\hat\\rho_3=0.15$. Is it statistically significant at the $5\\%$ level?",
      "back": "The $95\\%$ band is $\\pm\\dfrac{1.96}{\\sqrt{n}}=\\pm\\dfrac{1.96}{\\sqrt{100}}=\\pm\\dfrac{1.96}{10}=\\pm 0.196$.\nSince $|\\hat\\rho_3|=0.15<0.196$, it falls **inside** the band, so it is **not** significant — consistent with white noise at lag 3.\n(A value such as $0.25$ would have been significant.)",
      "tag": "ACF/PACF identification"
    },
    {
      "front": "Give the **one-step-ahead forecast** for a stationary AR(1) and its forecast error variance.",
      "back": "For $Y_t=c+\\phi Y_{t-1}+\\epsilon_t$, the minimum-MSE forecast made at time $t$ is $\\hat Y_{t+1}=c+\\phi Y_t$ (the future shock $\\epsilon_{t+1}$ has expectation 0).\nThe one-step forecast error is $Y_{t+1}-\\hat Y_{t+1}=\\epsilon_{t+1}$, so its variance is $\\sigma^2$ — the one-step error variance for an AR(1) equals the white-noise variance.",
      "tag": "Forecasting & diagnostics"
    },
    {
      "front": "An AR(1) is $Y_t=4+0.7\\,Y_{t-1}+\\epsilon_t$ with $\\sigma^2=2$, and the latest observation is $Y_{50}=20$. Give the 1-step and 2-step forecasts and their error variances.",
      "back": "1-step: $\\hat Y_{51}=4+0.7(20)=4+14=18$. Error variance $=\\sigma^2=2$.\n2-step: $\\hat Y_{52}=4+0.7\\hat Y_{51}=4+0.7(18)=4+12.6=16.6$. The 2-step error is $\\epsilon_{52}+\\phi\\epsilon_{51}$, so its variance $=\\sigma^2(1+\\phi^2)=2(1+0.49)=2(1.49)=2.98$.\nForecasts converge toward the mean $\\mu=\\dfrac{4}{1-0.7}\\approx 13.33$ as the horizon grows.",
      "tag": "Forecasting & diagnostics"
    },
    {
      "front": "Give the general **$h$-step forecast error variance** for a stationary AR(1).",
      "back": "Writing $Y_{t+h}-\\hat Y_{t+h}=\\sum_{j=0}^{h-1}\\phi^{j}\\epsilon_{t+h-j}$, the forecast error variance is\n$\\operatorname{Var}(e_h)=\\sigma^2\\sum_{j=0}^{h-1}\\phi^{2j}=\\sigma^2\\dfrac{1-\\phi^{2h}}{1-\\phi^2}$.\nAs $h\\to\\infty$ this rises to the unconditional variance $\\dfrac{\\sigma^2}{1-\\phi^2}$, so long-horizon forecasts carry the full process uncertainty.",
      "tag": "Forecasting & diagnostics"
    },
    {
      "front": "An MA(1) is $Y_t=\\epsilon_t+0.6\\epsilon_{t-1}$ with $\\sigma^2=4$, and the most recent residual is $\\hat\\epsilon_t=1.5$. Give the 1-step and 2-step forecasts and their error variances.",
      "back": "1-step: $\\hat Y_{t+1}=E[\\epsilon_{t+1}+0.6\\epsilon_t]=0.6\\hat\\epsilon_t=0.6(1.5)=0.9$. Error $=\\epsilon_{t+1}$, variance $=\\sigma^2=4$.\n2-step: $\\hat Y_{t+2}=E[\\epsilon_{t+2}+0.6\\epsilon_{t+1}]=0$ (both shocks are future). Error $=\\epsilon_{t+2}+0.6\\epsilon_{t+1}$, variance $=\\sigma^2(1+0.36)=4(1.36)=5.44$.\nBeyond the MA order ($h>q=1$) the forecast is just the mean (0 here).",
      "tag": "Forecasting & diagnostics"
    },
    {
      "front": "State the **Ljung–Box** statistic and how it is used to check residuals.",
      "back": "$Q=n(n+2)\\displaystyle\\sum_{k=1}^{m}\\dfrac{\\hat\\rho_k^2}{n-k}$, where $\\hat\\rho_k$ are the residual autocorrelations, $n$ is the sample size, and $m$ is the number of lags tested.\nUnder the null of no residual autocorrelation (an adequate model), $Q\\sim\\chi^2$ with $m-(p+q)$ degrees of freedom. A large $Q$ (small p-value) rejects adequacy — the residuals still carry structure, so the model needs more AR/MA terms.",
      "tag": "Forecasting & diagnostics"
    },
    {
      "front": "From a fitted ARMA(1,1) on $n=120$ residuals, the residual autocorrelations at lags 1–4 are $0.05, -0.08, 0.03, 0.06$. Compute the Ljung–Box $Q$ for $m=4$ and state the degrees of freedom.",
      "back": "$Q=n(n+2)\\sum_{k=1}^{4}\\dfrac{\\hat\\rho_k^2}{n-k}=120(122)\\Big[\\dfrac{0.05^2}{119}+\\dfrac{(-0.08)^2}{118}+\\dfrac{0.03^2}{117}+\\dfrac{0.06^2}{116}\\Big]$.\nTerms: $\\dfrac{0.0025}{119}=2.101\\times10^{-5}$, $\\dfrac{0.0064}{118}=5.424\\times10^{-5}$, $\\dfrac{0.0009}{117}=7.692\\times10^{-6}$, $\\dfrac{0.0036}{116}=3.103\\times10^{-5}$; sum $=1.140\\times10^{-4}$.\n$Q=14640(1.140\\times10^{-4})\\approx 1.67$. df $=m-(p+q)=4-2=2$. Since $\\chi^2_{0.95,2}=5.99>1.67$, do not reject — residuals look like white noise.",
      "tag": "Forecasting & diagnostics"
    },
    {
      "front": "Define the **AIC** for ARMA order selection and how to use it.",
      "back": "AIC $=-2\\ln(\\hat L)+2k$, where $\\hat L$ is the maximized likelihood and $k$ is the number of estimated parameters (e.g. $p+q$ plus intercept and $\\sigma^2$). It trades goodness of fit against complexity.\nAmong candidate models, choose the one with the **lowest** AIC. The BIC variant uses penalty $k\\ln n$ instead of $2k$, penalizing parameters more heavily and thus favoring more parsimonious models in large samples.",
      "tag": "Forecasting & diagnostics"
    },
    {
      "front": "Two candidate models are fitted: ARMA(1,0) with maximized log-likelihood $\\ln\\hat L=-145.0$ ($k=2$) and ARMA(2,1) with $\\ln\\hat L=-142.5$ ($k=4$). Which does AIC prefer?",
      "back": "AIC $=-2\\ln\\hat L+2k$.\nARMA(1,0): $-2(-145.0)+2(2)=290.0+4=294.0$.\nARMA(2,1): $-2(-142.5)+2(4)=285.0+8=293.0$.\nThe ARMA(2,1) has the lower AIC ($293.0<294.0$), so AIC prefers it — the $2.5$ gain in log-likelihood outweighs the extra-parameter penalty.",
      "tag": "Forecasting & diagnostics"
    },
    {
      "front": "Outline the three stages of the **Box–Jenkins** modeling cycle.",
      "back": "1. **Identification:** make the series stationary (difference if needed, choose $d$), then read the sample ACF/PACF to propose tentative $p$ and $q$.\n2. **Estimation:** fit the candidate ARIMA(p,d,q) by maximum likelihood (or least squares), estimating $\\phi$, $\\theta$, and $\\sigma^2$; compare models with AIC/BIC.\n3. **Diagnostic checking:** verify residuals are white noise (residual ACF inside bands, Ljung–Box not rejected). If checks fail, return to step 1 with a revised order; if they pass, forecast.",
      "tag": "Forecasting & diagnostics"
    }
  ]
}