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Exam SRM · Time-Series Model Selection · Tutorial 535 of 926

Uncertainty in Forecast Model Comparisons

Compare competing forecasts with paired loss differences, account for their dependence, and interpret statistical evidence alongside the size of the observed score difference.

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What You'll Learn

  • Define a loss differential and choose its sign so the favored model is clear.
  • Explain why competing models’ forecast losses are paired and can be dependent over time.
  • Derive the variance of an average loss differential from its autocovariances.
  • Calculate and interpret a horizon-specific Diebold–Mariano statistic.
  • Distinguish statistical evidence of a loss difference from practical importance.
  • Identify common errors in pairing forecasts and estimating uncertainty.

Why Forecast Comparisons Need Uncertainty

Two forecasting models can be evaluated on the same sequence of future observations. Their average scores may differ, but that difference is itself uncertain: it depends on which observations happened to occur during evaluation. A small score advantage does not establish that one model will generally forecast better.

The comparison should be paired. At each forecast origin, score both models against the same target, using the same forecast horizon and the same loss function. This pairing recognizes that both models face the same realized outcome and often respond to the same underlying time-series shocks.

Definition: For forecast origin \(t\), let \(e_{A,t}\) and \(e_{B,t}\) be the forecast errors from models A and B at the same horizon \(h\), both evaluated against the same observation \(Y_{t+h}\). If \(L(e)\) is a specified loss function, define the loss differential as \(d_t=L(e_{A,t})-L(e_{B,t})\). A positive \(d_t\) means model A has greater loss at origin \(t\), so the differential favors model B. A negative value favors model A.

For squared-error loss, \(L(e)=e^2\); for absolute-error loss, \(L(e)=|e|\). The comparison concerns expected loss under the chosen scoring rule, not whether the models’ forecasts are identical. A conclusion based on squared-error loss need not match one based on absolute-error loss, because the two rules weight forecast errors differently.

Worked Example: Constructing Paired Loss Differentials

A hypothetical forecasting team compares two models at a one-month horizon. For four forecast origins, the observed forecast errors are \(e_{A,t}=(2,-1,3,0)\) and \(e_{B,t}=(1,-2,1,1)\), in the same units as the outcome. Use squared-error loss to compute each paired differential and the mean differential. These four values are treated as fixed observations; no parametric error distribution is assumed, and no interest-rate or mortality assumption is relevant.

Method. Use the same origin and outcome for both models, and calculate \(d_t=e_{A,t}^2-e_{B,t}^2\). This sign convention makes a positive average favor model B.

At origin 1, the losses are \(2^2=4\) and \(1^2=1\), so \(d_1=4-1=3\).

At origin 2, the losses are \((-1)^2=1\) and \((-2)^2=4\), so \(d_2=1-4=-3\).

At origin 3, the losses are \(3^2=9\) and \(1^2=1\), so \(d_3=9-1=8\).

At origin 4, the losses are \(0^2=0\) and \(1^2=1\), so \(d_4=0-1=-1\).

$$ \bar d=\frac{3+(-3)+8+(-1)}{4}=\frac{7}{4}=1.75. $$

The mean squared loss is 3.50 for model A and 1.75 for model B, so the observed difference favors B by 1.75 squared units per forecast. The differential average agrees: \(\bar d=3.50-1.75=1.75\). This is a descriptive result from only four origins, not evidence that the population expected losses differ. Treating the two models’ scores as unrelated samples would discard the origin-by-origin pairing and would not be the paired comparison defined here.

Why Forecast Errors and Scores Are Dependent

At a given origin, errors from two models are generally related because both forecasts are made before the same outcome is observed. One model may be more responsive to a shared shock than the other, but the outcome itself is not separately generated for each model. Across origins, dependence can arise as well. In particular, successive \(h\)-step forecasts can cover overlapping periods: the forecast made at origin \(t\) targets \(Y_{t+h}\), while the next targets \(Y_{t+h+1}\). Their forecast errors may therefore share future shocks.

Even if the model-specific errors at one origin have variances \(\operatorname{Var}(L_{A,t})\) and \(\operatorname{Var}(L_{B,t})\), the variance of their difference also depends on their covariance. For any two random variables with finite variances,

$$ \operatorname{Var}(L_{A,t}-L_{B,t}) =\operatorname{Var}(L_{A,t})+\operatorname{Var}(L_{B,t}) -2\operatorname{Cov}(L_{A,t},L_{B,t}). $$

The same principle applies over time: the uncertainty in the average differential depends on the serial autocovariances of the differentials. Thus, the number of evaluated forecasts alone is not enough to calculate a valid standard error.

Key point: A Diebold–Mariano comparison is a test of equal expected loss for two forecasts under a specified loss function and horizon. It uses the paired series of loss differentials and accounts for their serial dependence. It does not require the two models’ forecast errors to be independent.

Deriving the Uncertainty of the Mean Differential

Let \(d_1,\ldots,d_T\) be the observed loss differentials from \(T\) evaluation origins. Suppose the differential process is stationary with finite second moments, and write \(\gamma_k=\operatorname{Cov}(d_t,d_{t-k})\), with \(\gamma_0=\operatorname{Var}(d_t)\). The sample mean is \(\bar d=T^{-1}\sum_{t=1}^T d_t\).

To see why autocovariances enter its variance, expand the variance of the sum. There are \(T\) terms with lag zero. For each lag \(k>0\), there are \(T-k\) pairs in each direction, and stationarity gives the same covariance \(\gamma_k\) for each pair.

$$ \begin{aligned} \operatorname{Var}(\bar d) &=\frac{1}{T^2}\sum_{t=1}^{T}\sum_{s=1}^{T}\operatorname{Cov}(d_t,d_s)\\ &=\frac{1}{T^2}\left(T\gamma_0+ 2\sum_{k=1}^{T-1}(T-k)\gamma_k\right)\\ &=\frac{1}{T}\left(\gamma_0+ 2\sum_{k=1}^{T-1}\left(1-\frac{k}{T}\right)\gamma_k\right). \end{aligned} $$

This identity explains why the independent-observation standard error, based only on \(\gamma_0\), can be wrong: it omits the terms involving \(\gamma_1,\gamma_2,\ldots\). Under standard weak-dependence conditions, the large-sample variance is approximated using the long-run variance \(\Omega=\gamma_0+2\sum_{k=1}^{\infty}\gamma_k\), so that \(\operatorname{Var}(\bar d)\) is approximately \(\Omega/T\).

In practice, the long-run variance is estimated from the observed differentials using a heteroskedasticity-and-autocorrelation-consistent (HAC) estimator. The estimator combines the sample variance with weighted estimates of autocovariances. A bandwidth and weighting rule must be chosen; for overlapping \(h\)-step forecast errors, dependence through at least the first \(h-1\) lags is a standard concern. More persistent dependence may call for additional lags. The appropriate choice depends on the forecast setup and sample size.

The Diebold–Mariano Comparison at a Specified Horizon

The null hypothesis is that the two forecasts have equal expected loss at the chosen horizon: \(H_0:E[d_t]=0\). The alternative may be two-sided, \(E[d_t]\ne0\), or one-sided if a direction was specified in advance. With a consistent estimate \(\widehat{\Omega}\) of the long-run variance and suitable stationarity, weak-dependence, and finite-moment conditions, the Diebold–Mariano statistic has an approximate standard normal reference distribution under the null.

$$ DM=\frac{\bar d}{\sqrt{\widehat{\Omega}/T}}. $$

A positive statistic points toward lower expected loss for model B under the sign convention above; a negative statistic points toward model A. The sign indicates direction, but the magnitude relative to its estimated standard error indicates how much statistical evidence the data provide. The normal approximation can be unreliable in short samples, especially when multi-step forecasts overlap. A test result should therefore be interpreted with attention to sample size and dependence, not as a mechanical verdict.

Worked Example: Interpreting a Diebold–Mariano Statistic

For a hypothetical one-step forecast comparison, suppose there are \(T=25\) paired origins. The estimated mean loss differential is \(\bar d=0.40\), where a positive value favors model B. A HAC calculation gives a long-run variance estimate \(\widehat{\Omega}=17/3\). Assume the differentials meet the conditions for the large-sample normal approximation. The reference distribution is \(N(0,1)\), with mean 0 and variance 1; no interest-rate or mortality assumption is involved.

Step 1: Calculate the standard error. The estimated variance of the mean is \(\widehat{\Omega}/T\):

$$ \frac{17/3}{25}=\frac{17}{75}=0.2267,\qquad SE(\bar d)=\sqrt{\frac{17}{75}}=0.4761. $$

Step 2: Calculate the test statistic.

$$ DM=\frac{0.40}{0.4761}=0.84. $$

Step 3: Interpret the evidence. For a two-sided test, the approximate \(p\)-value is \(2[1-\Phi(0.84)]\approx0.4005\), where \(\Phi\) is the standard normal cumulative distribution function. At the 5% level, this does not reject equal expected loss. The positive sample mean still favors B descriptively, but the data do not provide strong evidence of a persistent expected-loss difference.

Step 4: Put the estimate in context. A normal-approximation 95% interval for the mean differential is \(0.40\pm1.96(0.4761)\), or approximately \([-0.533,1.333]\). It includes zero and values favoring either model. The method avoids treating the observed average advantage as conclusive simply because it is positive.

Horizon, Practical Importance, and Common Traps

Forecast accuracy is horizon-specific. A comparison at horizon \(h=1\) uses forecasts of \(Y_{t+1}\); one at \(h=3\) uses forecasts of \(Y_{t+3}\). Each model pair must be evaluated on the same target dates at that horizon, with the same loss function. An average across horizons can conceal different performance patterns and is not the same test as a comparison at one specified horizon.

A statistically detectable difference may be too small to matter for a decision, while a potentially meaningful difference may be estimated too imprecisely to establish. Report the estimated average loss difference and its uncertainty, and relate the size of the difference to the units and consequences of the forecasting task. A failure to reject equal expected loss is not proof that the models are identical.

The standard test compares a specified pair of forecasting procedures on the evaluation differentials. It does not automatically turn a small difference into a robust finding, nor does it remove every uncertainty created by the evaluation design. In particular, the calculated uncertainty is only as credible as the pairing, horizon alignment, and dependence estimate.

Common trap: If the two loss series each have variance 4 and covariance 3 at a given origin, the variance of their difference is \(4+4-2(3)=2\). Assuming independence gives the incorrect value \(4+4=8\). The correction is to retain the covariance term; the forecast scores are paired, not independent.

Another frequent error is reversing the sign interpretation. With \(d_t=L_{A,t}-L_{B,t}\), a positive mean favors B, not A. State the differential definition before reporting a test statistic, and keep that convention through the calculation and conclusion.

Worked Example: A Three-Step Comparison With Overlapping Forecasts

A hypothetical analyst compares two models using three-step-ahead forecasts at 12 consecutive monthly origins. Both forecasts at each origin are evaluated against the same realized target three months later, using squared-error loss. The mean differential is \(\bar d=-0.60\) squared units, and a dependence-adjusted long-run variance estimate is \(\widehat{\Omega}=9\). Assume the approximate reference distribution is \(N(0,1)\), mean 0 and variance 1, and that the long-run variance estimate is appropriate. No interest-rate or mortality assumption is involved.

Step 1: Confirm the direction. Since \(d_t=L_{A,t}-L_{B,t}\), the negative mean means model A had lower average squared loss in this evaluation.

Step 2: Calculate uncertainty.

$$ SE(\bar d)=\sqrt{\frac{\widehat{\Omega}}{T}} =\sqrt{\frac{9}{12}}=\sqrt{0.75}=0.8660. $$

Step 3: Calculate the statistic and interval.

$$ DM=\frac{-0.60}{0.8660}=-0.69, \qquad -0.60\pm1.96(0.8660)=[-2.297,1.097]. $$

The two-sided normal-approximation \(p\)-value is about \(0.49\), so the result does not provide strong evidence of unequal expected loss. Although the point estimate favors A, the interval includes zero and positive values, which favor B. Because the origins are consecutive and the horizon is three steps, forecast errors may overlap in their future shocks; using a dependence-adjusted variance rather than an independence formula is important. This method avoids reading the negative point estimate as a decisive model ranking.

Check Your Understanding

For each problem, define the differential sign carefully and account for the horizon and dependence described.

  1. At one origin, model A has absolute error 3 and model B has absolute error 5. Under \(d=L_A-L_B\) and absolute-error loss, calculate \(d\) and state which model has lower loss at this origin. (Answer: \(d=-2\); model A has lower loss.)
  2. At a specified horizon, the paired squared-loss differentials are \(2,-1,3\). Calculate their mean and state its direction. (Answer: \(\bar d=4/3\); the observed mean favors model B.)
  3. A one-step comparison has \(\bar d=0.30\), \(T=36\), and \(\widehat{\Omega}=4\). Calculate the standard error and the Diebold–Mariano statistic. (Answer: \(SE=\sqrt{4/36}=1/3\); \(DM=0.90\).)
  4. For a three-step comparison using forecasts from consecutive origins, explain why an independence-based standard error may be inappropriate. (Answer: adjacent three-step forecast errors can share future shocks because their target periods overlap, inducing serial dependence in the loss differentials; the variance estimate should account for that dependence.)
  5. A comparison gives a positive mean differential, but its confidence interval includes zero and values below zero. Under the convention \(d=L_A-L_B\), what can be concluded? (Answer: the point estimate favors model B, but the interval does not establish a difference in expected loss; it includes values that favor model A as well.)
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