{
  "deckName": "Exam SRM — Time Series",
  "examCode": "Exam SRM",
  "cards": [
    {
      "front": "Name the four components into which a time series $\\{Y_t\\}$ is classically decomposed.",
      "back": "**Trend** $(T_t)$ — the long-run upward or downward movement in the level.\n**Seasonal** $(S_t)$ — a pattern that repeats over a fixed, known period (e.g. 12 months, 4 quarters).\n**Cyclical** $(C_t)$ — longer wave-like swings of *variable* length not tied to a fixed calendar period (e.g. the business cycle).\n**Irregular / random** $(I_t)$ — the leftover unpredictable noise.\nAn additive model writes $Y_t=T_t+S_t+C_t+I_t$; a multiplicative model writes $Y_t=T_t\\cdot S_t\\cdot C_t\\cdot I_t$.",
      "tag": "Trend & seasonality"
    },
    {
      "front": "Define a **white noise** process $\\{\\epsilon_t\\}$ and state its three defining properties.",
      "back": "White noise is a sequence of uncorrelated random shocks with:\n1. constant mean $E[\\epsilon_t]=0$,\n2. constant variance $\\text{Var}(\\epsilon_t)=\\sigma^2$,\n3. zero autocorrelation: $\\text{Cov}(\\epsilon_t,\\epsilon_s)=0$ for $t\\neq s$.\nWhite noise is the simplest stationary series and is the building block (the error term) of every other model; a fitted model is adequate when its residuals look like white noise.",
      "tag": "Stationarity & white noise"
    },
    {
      "front": "State the conditions for **weak (covariance) stationarity** of a time series $\\{Y_t\\}$.",
      "back": "A series is weakly stationary if its first two moments do not change over time:\n1. **constant mean**: $E[Y_t]=\\mu$ for all $t$,\n2. **constant variance**: $\\text{Var}(Y_t)=\\sigma_Y^2$ for all $t$,\n3. **autocovariance depends only on the lag** $k$, not on $t$: $\\text{Cov}(Y_t,Y_{t+k})=\\gamma_k$.\nWhite noise and a stationary AR(1) are stationary; a random walk and any series with a trend or seasonal pattern are **not**.",
      "tag": "Stationarity & white noise"
    },
    {
      "front": "Why is **stationarity** required before fitting most time-series models, and how is non-stationarity usually removed?",
      "back": "Estimation of autocorrelations and model parameters assumes the statistical properties are stable over time; a moving mean or variance makes those estimates meaningless and forecasts unreliable.\nNon-stationarity from a **trend** is removed by **differencing**: $\\nabla Y_t = Y_t - Y_{t-1}$. Non-stationarity from changing **variance** is often tamed by a **log or power transform**. Seasonality is removed by **seasonal differencing** $Y_t - Y_{t-s}$.\nDifference repeatedly only as needed — over-differencing introduces spurious autocorrelation.",
      "tag": "Stationarity & white noise"
    },
    {
      "front": "Define the **autocorrelation function (ACF)** $\\rho_k$ at lag $k$ for a stationary series.",
      "back": "The lag-$k$ autocorrelation is the autocovariance scaled by the variance:\n$\\rho_k=\\frac{\\text{Cov}(Y_t,Y_{t+k})}{\\text{Var}(Y_t)}=\\frac{\\gamma_k}{\\gamma_0}$.\nProperties: $\\rho_0=1$, $-1\\le\\rho_k\\le 1$, and $\\rho_{-k}=\\rho_k$. The sample ACF $\\hat\\rho_k$ is plotted in a **correlogram**; for white noise all $\\hat\\rho_k$ for $k\\ge 1$ should be near $0$ (inside the $\\pm 2/\\sqrt{n}$ bands).",
      "tag": "Stationarity & white noise"
    },
    {
      "front": "Compute the lag-1 sample autocorrelation $\\hat\\rho_1$ for the series $4,\\,6,\\,5,\\,7,\\,8$ (mean $\\bar Y=6$).",
      "back": "Deviations $d_t=Y_t-\\bar Y$: $-2,\\,0,\\,-1,\\,1,\\,2$.\n**Numerator** $\\sum_{t=2}^{5} d_t d_{t-1}=(0)(-2)+(-1)(0)+(1)(-1)+(2)(1)=0+0-1+2=1$.\n**Denominator** $\\sum_{t=1}^{5} d_t^2=4+0+1+1+4=10$.\n$\\hat\\rho_1=\\frac{1}{10}=0.10$.\nThe small positive value indicates almost no lag-1 dependence in this short series.",
      "tag": "Stationarity & white noise"
    },
    {
      "front": "How is a **linear trend** model fitted to a time series, and what does each coefficient mean?",
      "back": "Regress the level on time: $Y_t=\\beta_0+\\beta_1 t+\\epsilon_t$, estimated by ordinary least squares with $t=1,2,\\dots,n$.\n$\\hat\\beta_0$ is the fitted level at $t=0$; $\\hat\\beta_1$ is the average change in $Y$ per period.\nThe forecast for a future period $t^{*}$ is $\\hat Y_{t^{*}}=\\hat\\beta_0+\\hat\\beta_1 t^{*}$. This **deterministic-trend** model assumes the trend continues unchanged, in contrast to a stochastic trend (random walk) removed by differencing.",
      "tag": "Trend & seasonality"
    },
    {
      "front": "A linear-trend fit on $n=10$ periods gives $\\hat\\beta_0=20.0$ and $\\hat\\beta_1=1.5$. Forecast $Y_{12}$ and state the implied change per period.",
      "back": "The fitted trend is $\\hat Y_t=20.0+1.5\\,t$.\nForecast for $t=12$: $\\hat Y_{12}=20.0+1.5(12)=20.0+18.0=38.0$.\nThe series rises by $\\hat\\beta_1=1.5$ units each period on average. Two periods beyond the last observation ($t=10$, fitted $35.0$) the forecast adds $2(1.5)=3.0$, giving $38.0$.",
      "tag": "Trend & seasonality"
    },
    {
      "front": "How are **seasonal indices** used to adjust a trend forecast in an additive (or multiplicative) model?",
      "back": "After estimating the trend, compute the average departure of each season from the trend.\n**Additive:** seasonal index $S_q$ is an amount added; the forecast is $\\hat Y=\\text{trend}+S_q$, with the indices summing to $0$ across the season.\n**Multiplicative:** seasonal index is a ratio averaging $1$ (summing to the number of seasons); the forecast is $\\hat Y=\\text{trend}\\times S_q$.\nTo **deseasonalize** an observation, subtract (additive) or divide by (multiplicative) its seasonal index.",
      "tag": "Trend & seasonality"
    },
    {
      "front": "Quarterly sales have trend forecast $200$ for next quarter and a multiplicative seasonal index of $1.15$ for that quarter. Give the seasonally adjusted forecast, and deseasonalize an observed value of $253$.",
      "back": "**Seasonal forecast** (multiplicative): $\\hat Y=\\text{trend}\\times S_q=200\\times 1.15=230$.\n**Deseasonalize** an actual of $253$: divide by the index, $\\frac{253}{1.15}=220$. The $220$ is the trend-level value with the seasonal effect stripped out, comparable across quarters.",
      "tag": "Trend & seasonality"
    },
    {
      "front": "Compute the four additive seasonal indices given average quarterly deviations from trend of $+12,\\,-8,\\,-10,\\,+4$.",
      "back": "Additive indices must sum to $0$. The raw deviations sum to $12-8-10+4=-2$, so subtract the mean $\\frac{-2}{4}=-0.5$ from each (equivalently add $0.5$).\nAdjusted indices: $12+0.5=12.5$, $-8+0.5=-7.5$, $-10+0.5=-9.5$, $4+0.5=4.5$.\nCheck: $12.5-7.5-9.5+4.5=0$. A Q1 trend forecast of $150$ becomes $150+12.5=162.5$.",
      "tag": "Trend & seasonality"
    },
    {
      "front": "Define the **random walk** $Y_t=Y_{t-1}+\\epsilon_t$ and explain why it is non-stationary.",
      "back": "Each value equals the previous value plus an independent white-noise shock $\\epsilon_t$ (mean $0$, variance $\\sigma^2$). Writing it from the start, $Y_t=Y_0+\\sum_{i=1}^{t}\\epsilon_i$, so it is the running sum of shocks.\nIt is **non-stationary** because its variance grows without bound: $\\text{Var}(Y_t)=t\\sigma^2$ depends on $t$. The mean is constant ($E[Y_t]=Y_0$) but the variance is not, so weak stationarity fails. It is the AR(1) boundary case $\\phi_1=1$ (a unit root).",
      "tag": "Random walk"
    },
    {
      "front": "For a random walk $Y_t=Y_{t-1}+\\epsilon_t$ with $\\sigma^2=9$ and $Y_0=50$, find $\\text{Var}(Y_6)$, the standard deviation of $Y_6$, and the one-step forecast of $Y_7$ given $Y_6$.",
      "back": "Since $Y_t=Y_0+\\sum_{i=1}^{t}\\epsilon_i$ with independent shocks, $\\text{Var}(Y_t)=t\\sigma^2$.\n$\\text{Var}(Y_6)=6(9)=54$, so $\\text{SD}(Y_6)=\\sqrt{54}\\approx 7.35$.\nThe optimal one-step forecast of a random walk is the **last observed value** (since $E[\\epsilon_{7}]=0$): $\\hat Y_7=Y_6$. The random walk is a \"no-change\" forecaster — tomorrow's best guess is today's value.",
      "tag": "Random walk"
    },
    {
      "front": "How does **first differencing** turn a random walk into a stationary series?",
      "back": "Define the first difference $\\nabla Y_t=Y_t-Y_{t-1}$. For the random walk $Y_t=Y_{t-1}+\\epsilon_t$ this gives $\\nabla Y_t=\\epsilon_t$, which is exactly **white noise** — constant mean $0$, constant variance $\\sigma^2$, zero autocorrelation — hence stationary.\nThis is why a series with a stochastic (unit-root) trend is made stationary by differencing rather than by subtracting a fitted line; the latter only works for a deterministic trend.",
      "tag": "Random walk"
    },
    {
      "front": "A **random walk with drift** is $Y_t=\\delta+Y_{t-1}+\\epsilon_t$. Give its mean function and explain the role of $\\delta$.",
      "back": "Iterating from $Y_0$, $Y_t=Y_0+\\delta t+\\sum_{i=1}^{t}\\epsilon_i$. So:\n$E[Y_t]=Y_0+\\delta t$ — a linear trend with slope $\\delta$ (the **drift**),\n$\\text{Var}(Y_t)=t\\sigma^2$ — still growing, so still non-stationary.\nFirst differencing gives $\\nabla Y_t=\\delta+\\epsilon_t$, white noise plus a constant mean $\\delta$. The drift adds a steady $\\delta$ per period on top of the pure random-walk behavior.",
      "tag": "Random walk"
    },
    {
      "front": "For a random walk with drift $\\delta=2$, $Y_0=10$, and $\\sigma^2=4$, find $E[Y_5]$ and $\\text{Var}(Y_5)$.",
      "back": "$E[Y_t]=Y_0+\\delta t=10+2(5)=20$.\n$\\text{Var}(Y_t)=t\\sigma^2=5(4)=20$, so $\\text{SD}(Y_5)=\\sqrt{20}\\approx 4.47$.\nThe expected level drifts up by $2$ per step while the uncertainty band widens as $\\sqrt{t}$.",
      "tag": "Random walk"
    },
    {
      "front": "Write the **AR(1)** model in mean-adjusted and intercept forms, and state the stationarity condition.",
      "back": "Intercept form: $Y_t=\\beta_0+\\phi_1 Y_{t-1}+\\epsilon_t$, with $\\epsilon_t$ white noise (mean $0$, variance $\\sigma^2$).\nMean-adjusted form: $Y_t-\\mu=\\phi_1(Y_{t-1}-\\mu)+\\epsilon_t$, where $\\mu=\\frac{\\beta_0}{1-\\phi_1}$.\n**Stationarity** requires $|\\phi_1|<1$. At $\\phi_1=1$ the process is a random walk (non-stationary); $|\\phi_1|>1$ explodes.",
      "tag": "Autoregressive models"
    },
    {
      "front": "For a stationary AR(1) $Y_t=\\beta_0+\\phi_1 Y_{t-1}+\\epsilon_t$, state the formulas for the mean, variance, and lag-$k$ autocorrelation.",
      "back": "**Mean:** $E[Y_t]=\\mu=\\frac{\\beta_0}{1-\\phi_1}$.\n**Variance:** $\\text{Var}(Y_t)=\\frac{\\sigma^2}{1-\\phi_1^{2}}$.\n**Autocorrelation:** $\\rho_k=\\phi_1^{k}$ for $k\\ge 0$.\nSo the ACF decays geometrically; it stays positive when $\\phi_1>0$ and alternates sign when $\\phi_1<0$. The lag-1 autocorrelation $\\rho_1=\\phi_1$ identifies the parameter directly.",
      "tag": "Autoregressive models"
    },
    {
      "front": "An AR(1) has $\\beta_0=6$, $\\phi_1=0.4$, and $\\sigma^2=5$. Find the process mean $\\mu$ and variance.",
      "back": "**Mean:** $\\mu=\\frac{\\beta_0}{1-\\phi_1}=\\frac{6}{1-0.4}=\\frac{6}{0.6}=10$.\n**Variance:** $\\text{Var}(Y_t)=\\frac{\\sigma^2}{1-\\phi_1^{2}}=\\frac{5}{1-0.16}=\\frac{5}{0.84}\\approx 5.952$.\nBecause $|\\phi_1|=0.4<1$ the process is stationary, so these constant moments are well defined.",
      "tag": "Autoregressive models"
    },
    {
      "front": "For an AR(1) with $\\phi_1=0.7$, compute the autocorrelations $\\rho_1,\\rho_2,\\rho_3$ and describe the ACF shape.",
      "back": "Using $\\rho_k=\\phi_1^{k}$:\n$\\rho_1=0.7$, $\\rho_2=0.7^{2}=0.49$, $\\rho_3=0.7^{3}=0.343$.\nThe ACF decays **geometrically toward zero**, all positive because $\\phi_1>0$. A correlogram showing this smooth exponential decay (with the partial ACF cutting off after lag 1) is the signature of an AR(1) process.",
      "tag": "Autoregressive models"
    },
    {
      "front": "An AR(1) with $\\phi_1=-0.6$, $\\beta_0=8$, and $\\sigma^2=10$: find $\\mu$, $\\rho_1$, $\\rho_2$, and describe the ACF.",
      "back": "$\\mu=\\frac{\\beta_0}{1-\\phi_1}=\\frac{8}{1-(-0.6)}=\\frac{8}{1.6}=5$.\n$\\rho_1=\\phi_1=-0.6$, $\\rho_2=\\phi_1^{2}=0.36$.\nWith $\\phi_1<0$ the autocorrelations **alternate in sign** while shrinking in magnitude ($-0.6,\\;0.36,\\;-0.216,\\dots$), giving an oscillating, damped correlogram. Variance $=\\frac{10}{1-0.36}=\\frac{10}{0.64}=15.625$.",
      "tag": "Autoregressive models"
    },
    {
      "front": "An AR(1) has $\\mu=20$, $\\phi_1=0.5$. The last observation is $Y_t=26$. Give the one-step and two-step forecasts $\\hat Y_{t+1}$ and $\\hat Y_{t+2}$.",
      "back": "Forecasts use the mean-adjusted recursion $\\hat Y_{t+h}-\\mu=\\phi_1^{h}(Y_t-\\mu)$.\n**One step:** $\\hat Y_{t+1}=\\mu+\\phi_1(Y_t-\\mu)=20+0.5(26-20)=20+3=23$.\n**Two step:** $\\hat Y_{t+2}=\\mu+\\phi_1^{2}(Y_t-\\mu)=20+0.25(6)=20+1.5=21.5$.\nAs the horizon grows the forecast decays geometrically back to the long-run mean $\\mu=20$.",
      "tag": "Autoregressive models"
    },
    {
      "front": "Show why the AR(1) variance is $\\frac{\\sigma^2}{1-\\phi_1^{2}}$ starting from $\\text{Var}(Y_t)=\\phi_1^{2}\\text{Var}(Y_{t-1})+\\sigma^2$.",
      "back": "Take variances of $Y_t-\\mu=\\phi_1(Y_{t-1}-\\mu)+\\epsilon_t$. Since $\\epsilon_t$ is independent of $Y_{t-1}$:\n$\\text{Var}(Y_t)=\\phi_1^{2}\\,\\text{Var}(Y_{t-1})+\\sigma^2$.\nStationarity means $\\text{Var}(Y_t)=\\text{Var}(Y_{t-1})=\\gamma_0$, so $\\gamma_0=\\phi_1^{2}\\gamma_0+\\sigma^2$.\nSolve: $\\gamma_0(1-\\phi_1^{2})=\\sigma^2\\Rightarrow \\gamma_0=\\frac{\\sigma^2}{1-\\phi_1^{2}}$, which requires $|\\phi_1|<1$ for a positive, finite variance.",
      "tag": "Autoregressive models"
    },
    {
      "front": "Given the lag-1 autocorrelation $\\hat\\rho_1=0.8$ for a series modeled as AR(1), estimate $\\phi_1$ and the lag-3 autocorrelation.",
      "back": "For an AR(1), $\\rho_1=\\phi_1$, so $\\hat\\phi_1=\\hat\\rho_1=0.8$.\nThen $\\rho_k=\\phi_1^{k}$ gives $\\hat\\rho_3=0.8^{3}=0.512$.\nThe single lag-1 autocorrelation pins down the whole geometric ACF of an AR(1).",
      "tag": "Autoregressive models"
    },
    {
      "front": "How does a **simple moving average (SMA)** of order $k$ smooth a series, and how is it used to forecast?",
      "back": "The $k$-period moving average at time $t$ is the mean of the most recent $k$ observations:\n$M_t=\\frac{Y_t+Y_{t-1}+\\dots+Y_{t-k+1}}{k}$.\nIt damps the irregular component; a larger $k$ smooths more but lags more. The naive **forecast** of the next value is the current moving average: $\\hat Y_{t+1}=M_t$. SMA weights the $k$ terms equally and ignores everything older.",
      "tag": "Smoothing"
    },
    {
      "front": "Given the last four observations $Y_{t-3},\\dots,Y_t = 22,\\,26,\\,24,\\,28$, compute the 3-period and 4-period moving-average forecasts of $Y_{t+1}$.",
      "back": "**3-period** (most recent three: $26,24,28$): $M_t=\\frac{26+24+28}{3}=\\frac{78}{3}=26.0$.\n**4-period** (all four: $22,26,24,28$): $M_t=\\frac{22+26+24+28}{4}=\\frac{100}{4}=25.0$.\nThe forecasts are $\\hat Y_{t+1}=26.0$ (3-period) and $25.0$ (4-period); the shorter window reacts more to the recent uptick.",
      "tag": "Smoothing"
    },
    {
      "front": "State the **single (simple) exponential smoothing** updating formula and explain the role of $\\alpha$.",
      "back": "$\\hat Y_{t+1}=\\alpha Y_t+(1-\\alpha)\\hat Y_t$, with smoothing constant $0<\\alpha<1$.\nEach new forecast blends the most recent actual $Y_t$ and the previous forecast $\\hat Y_t$. Expanding shows geometrically declining weights on past data: $\\alpha,\\;\\alpha(1-\\alpha),\\;\\alpha(1-\\alpha)^2,\\dots$.\nLarge $\\alpha$ → responsive, tracks recent changes; small $\\alpha$ → heavy smoothing, slow to react. It suits a series with **no trend or seasonality** (a locally constant level).",
      "tag": "Smoothing"
    },
    {
      "front": "Single exponential smoothing with $\\alpha=0.3$: the prior forecast is $\\hat Y_t=50$ and the actual is $Y_t=60$. Find $\\hat Y_{t+1}$.",
      "back": "$\\hat Y_{t+1}=\\alpha Y_t+(1-\\alpha)\\hat Y_t=0.3(60)+0.7(50)=18+35=53$.\nEquivalently, error-correction form: $\\hat Y_{t+1}=\\hat Y_t+\\alpha(Y_t-\\hat Y_t)=50+0.3(60-50)=50+3=53$. The forecast moves $30\\%$ of the way toward the latest observation.",
      "tag": "Smoothing"
    },
    {
      "front": "Run single exponential smoothing ($\\alpha=0.5$) on $Y_1,\\dots,Y_4 = 10,\\,14,\\,12,\\,16$, initializing $\\hat Y_2=Y_1=10$. Forecast $\\hat Y_5$.",
      "back": "Update $\\hat Y_{t+1}=0.5Y_t+0.5\\hat Y_t$:\n$\\hat Y_3=0.5(14)+0.5(10)=12.0$.\n$\\hat Y_4=0.5(12)+0.5(12.0)=12.0$.\n$\\hat Y_5=0.5(16)+0.5(12.0)=8+6=14.0$.\nThe one-step-ahead forecast of $Y_5$ is $\\mathbf{14.0}$.",
      "tag": "Smoothing"
    },
    {
      "front": "Compare a large versus a small smoothing constant $\\alpha$ in exponential smoothing.",
      "back": "$\\alpha$ controls how fast old data is forgotten (weights decay as $(1-\\alpha)^{j}$):\n**Large $\\alpha$ (near 1):** puts most weight on the latest actual — fast, responsive, but noisy; tracks real level shifts quickly.\n**Small $\\alpha$ (near 0):** weights are spread over a long history — heavy smoothing, stable, but slow to adapt to genuine changes.\n$\\alpha$ is often chosen to minimize in-sample one-step squared forecast error (SSE/MSE).",
      "tag": "Smoothing"
    },
    {
      "front": "Why does single exponential smoothing fail on a trending series, and what does **double exponential (Holt) smoothing** add?",
      "back": "Single smoothing forecasts a flat level, so on an upward trend it **lags behind systematically** (consistently under-forecasts).\n**Holt / double exponential smoothing** adds a second equation for the trend: it tracks a smoothed **level** $L_t=\\alpha Y_t+(1-\\alpha)(L_{t-1}+b_{t-1})$ and a smoothed **slope** $b_t=\\beta(L_t-L_{t-1})+(1-\\beta)b_{t-1}$. The $h$-step forecast is $\\hat Y_{t+h}=L_t+h\\,b_t$, projecting the current slope forward. (Holt-Winters extends this with a third seasonal equation.)",
      "tag": "Smoothing"
    },
    {
      "front": "Holt's method gives a current level $L_t=200$ and trend $b_t=5$. Forecast 1 and 3 periods ahead.",
      "back": "The Holt $h$-step forecast is $\\hat Y_{t+h}=L_t+h\\,b_t$.\n**One step** ($h=1$): $\\hat Y_{t+1}=200+1(5)=205$.\n**Three steps** ($h=3$): $\\hat Y_{t+3}=200+3(5)=215$.\nUnlike single smoothing's flat line, Holt extrapolates the slope $b_t=5$ per period into the future.",
      "tag": "Smoothing"
    },
    {
      "front": "Define **mean squared error (MSE)** and **root mean squared error (RMSE)** as forecast-accuracy measures.",
      "back": "Over $n$ forecasts with errors $e_t=Y_t-\\hat Y_t$:\n$\\text{MSE}=\\frac{1}{n}\\sum_{t=1}^{n} e_t^{2}=\\frac{1}{n}\\sum (Y_t-\\hat Y_t)^2$,\n$\\text{RMSE}=\\sqrt{\\text{MSE}}$.\nSquaring penalizes **large** errors heavily, so MSE/RMSE are sensitive to outliers. RMSE is in the original units of $Y$, making it easier to interpret than MSE.",
      "tag": "Forecast accuracy"
    },
    {
      "front": "Define **mean absolute error (MAE)** and **mean absolute percentage error (MAPE)**, and note when each is preferred.",
      "back": "$\\text{MAE}=\\frac{1}{n}\\sum_{t=1}^{n}|Y_t-\\hat Y_t|$ — average size of the errors in the data's units; less sensitive to outliers than MSE.\n$\\text{MAPE}=\\frac{1}{n}\\sum_{t=1}^{n}\\left|\\frac{Y_t-\\hat Y_t}{Y_t}\\right|\\times 100\\%$ — average error as a percent of the actual, which is **unit-free** and good for comparing across series of different scale. MAPE is undefined/unstable when any $Y_t$ is near $0$.",
      "tag": "Forecast accuracy"
    },
    {
      "front": "Forecast errors over four periods are $e_t = 3,\\,-2,\\,4,\\,-1$. Compute the MSE, RMSE, and MAE.",
      "back": "**MSE:** $\\frac{1}{4}(3^2+(-2)^2+4^2+(-1)^2)=\\frac{1}{4}(9+4+16+1)=\\frac{30}{4}=7.5$.\n**RMSE:** $\\sqrt{7.5}\\approx 2.74$.\n**MAE:** $\\frac{1}{4}(|3|+|-2|+|4|+|-1|)=\\frac{1}{4}(3+2+4+1)=\\frac{10}{4}=2.5$.\nMSE exceeds MAE$^2$ here because the large $|4|$ error is amplified by squaring.",
      "tag": "Forecast accuracy"
    },
    {
      "front": "Actuals $Y_t$ and forecasts $\\hat Y_t$ over three periods: $(100,95),\\,(120,126),\\,(80,76)$. Compute the MAPE.",
      "back": "Absolute percentage errors $\\left|\\frac{Y_t-\\hat Y_t}{Y_t}\\right|$:\nPeriod 1: $\\left|\\frac{100-95}{100}\\right|=0.05$.\nPeriod 2: $\\left|\\frac{120-126}{120}\\right|=\\frac{6}{120}=0.05$.\nPeriod 3: $\\left|\\frac{80-76}{80}\\right|=\\frac{4}{80}=0.05$.\n$\\text{MAPE}=\\frac{0.05+0.05+0.05}{3}\\times 100\\%=5.0\\%$.",
      "tag": "Forecast accuracy"
    },
    {
      "front": "Two models are compared on a hold-out set: Model A has $\\text{MAE}=4.2,\\ \\text{MAPE}=6.1\\%$; Model B has $\\text{MAE}=3.8,\\ \\text{MAPE}=7.5\\%$. Which is better?",
      "back": "It depends on the criterion — the measures disagree.\nModel B has the lower **absolute** error ($\\text{MAE}=3.8<4.2$), so it makes smaller errors in raw units.\nModel A has the lower **percentage** error ($\\text{MAPE}=6.1\\%<7.5\\%$), so it is more accurate relative to the size of the actuals.\nIf the costliest errors are the large-magnitude ones, prefer B; if proportional accuracy across scales matters, prefer A. Pick the metric that matches the decision being supported, and ideally also check MSE/RMSE.",
      "tag": "Forecast accuracy"
    },
    {
      "front": "Compute the MSE of a 3-period moving-average forecaster on the series $Y_4,Y_5,Y_6=30,\\,33,\\,31$, given moving-average forecasts $\\hat Y_4,\\hat Y_5,\\hat Y_6 = 28,\\,31,\\,34$.",
      "back": "Errors $e_t=Y_t-\\hat Y_t$:\n$e_4=30-28=2$, $e_5=33-31=2$, $e_6=31-34=-3$.\n$\\text{MSE}=\\frac{1}{3}(2^2+2^2+(-3)^2)=\\frac{1}{3}(4+4+9)=\\frac{17}{3}\\approx 5.67$.\nRMSE $=\\sqrt{5.67}\\approx 2.38$; the model would be compared against alternatives on the same hold-out periods.",
      "tag": "Forecast accuracy"
    },
    {
      "front": "Distinguish a **deterministic trend** from a **stochastic trend**, and the correct way to make each stationary.",
      "back": "**Deterministic trend:** $Y_t=\\beta_0+\\beta_1 t+\\epsilon_t$ — a fixed line plus stationary noise. Made stationary by **detrending** (subtracting the fitted line / regressing on $t$). Shocks have only temporary effects.\n**Stochastic trend:** a unit-root process like the random walk $Y_t=Y_{t-1}+\\epsilon_t$. Made stationary by **differencing** $\\nabla Y_t$. Shocks are **permanent** — they accumulate into the level.\nApplying the wrong remedy (differencing a deterministic trend, or detrending a random walk) leaves the residuals non-white.",
      "tag": "Random walk"
    },
    {
      "front": "For a stationary AR(1) with $\\phi_1=0.6$ and $\\sigma^2=8$, find the lag-1 autocovariance $\\gamma_1$.",
      "back": "First the variance: $\\gamma_0=\\frac{\\sigma^2}{1-\\phi_1^{2}}=\\frac{8}{1-0.36}=\\frac{8}{0.64}=12.5$.\nThen $\\gamma_1=\\rho_1\\gamma_0=\\phi_1\\gamma_0=0.6(12.5)=7.5$.\n(Equivalently $\\gamma_k=\\phi_1^{k}\\gamma_0$, so $\\gamma_2=0.36(12.5)=4.5$, etc.)",
      "tag": "Autoregressive models"
    },
    {
      "front": "A series has a clear upward straight-line trend with no seasonality. Which forecasting method is appropriate, and why not single exponential smoothing or a plain moving average?",
      "back": "Use a **linear-trend regression** $\\hat Y_t=\\hat\\beta_0+\\hat\\beta_1 t$ or **double (Holt) exponential smoothing**, both of which project the slope forward.\nA **simple moving average** and **single exponential smoothing** assume a locally flat level, so on a trend they **lag and systematically under-forecast** (always behind the rising actuals). Matching the method to the data's structure — trend, seasonality, or neither — is the key model-selection step.",
      "tag": "Smoothing"
    },
    {
      "front": "Single exponential smoothing with $\\alpha=0.2$, prior forecast $\\hat Y_t=40$, actual $Y_t=50$. Find the next forecast, then the forecast two steps ahead $\\hat Y_{t+2}$ (no new data after $t$).",
      "back": "**One step:** $\\hat Y_{t+1}=0.2(50)+0.8(40)=10+32=42$.\nWith no observation at $t+1$, single exponential smoothing produces a **flat** multi-step forecast: $\\hat Y_{t+2}=\\hat Y_{t+1}=42$. All horizons share the same value because the method has no trend term — a limitation that motivates Holt's method.",
      "tag": "Smoothing"
    },
    {
      "front": "Given sample autocovariances $\\gamma_0=20$, $\\gamma_1=12$, $\\gamma_2=6$, compute the autocorrelations $\\rho_1,\\rho_2$ and comment on whether an AR(1) is plausible.",
      "back": "$\\rho_1=\\frac{\\gamma_1}{\\gamma_0}=\\frac{12}{20}=0.6$ and $\\rho_2=\\frac{\\gamma_2}{\\gamma_0}=\\frac{6}{20}=0.3$.\nFor an AR(1) we would expect $\\rho_2=\\rho_1^{2}=0.6^{2}=0.36$. The observed $0.3$ is reasonably close to $0.36$, so an AR(1) with $\\phi_1\\approx 0.6$ is plausible; a large discrepancy would point to a higher-order model.",
      "tag": "Autoregressive models"
    }
  ]
}