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Exam SRM · Time-Series Forecasting Cases · Tutorial 900 of 926

Worked Case: Claims Experience with Random-Walk Behavior

Learn to distinguish a random walk from a stationary claims process, compare a random walk with drift to a differenced ARIMA model, and interpret residual checks and multi-horizon forecast errors.

Advanced 12 min read

What You'll Learn

  • Assess level and first-difference behavior when considering random-walk claims models
  • State the innovation assumptions needed to use conditional-mean forecasts
  • Calculate fixed-origin forecasts from a random walk and a differenced ARIMA model
  • Use residual autocorrelations and a Ljung–Box statistic as diagnostic evidence
  • Compare forecast errors by horizon without confusing fixed-origin and rolling forecasts

Why Random-Walk Behavior Matters for Claims

A claims series may show a persistent upward or downward movement because each new observation builds on the previous level. In that situation, treating the levels as independent observations around a fixed mean can produce misleading forecasts. A random-walk model instead describes changes in the series; a differenced ARIMA model allows those changes to have their own dependence pattern.

The analysis is not just a matter of noticing that a plotted series rises. A trend could reflect persistent stochastic movement, a deterministic trend, a structural change, or another feature of the data. A useful case analysis considers the dated observations, examines first differences, compares plausible models, checks residuals, and evaluates forecasts at the horizon relevant to the decision. The hypothetical claims amounts below are measured in thousands of dollars and are not actual company or industry data.

Definition: A series \(Y_t\) is weakly stationary if its mean is constant over time, its variance is finite and constant over time, and its covariance between \(Y_t\) and \(Y_{t-k}\) depends only on the lag \(k\), not on the date \(t\). A random walk with drift is \(Y_t=Y_{t-1}+c+\varepsilon_t\), where \(c\) is a constant drift and the innovations \(\varepsilon_t\) have mean zero and constant variance. If the innovations are independent over time, the first differences \(\Delta Y_t=Y_t-Y_{t-1}\) are stationary around \(c\), but the levels are not stationary.

For a random walk, shocks do not fade from the level: a shock in one period remains in all later levels. Consequently, the variance of the level increases with time. This is different from a stationary series, in which the impact of a shock does not accumulate without bound. A plot and sample autocorrelations can suggest persistence, but they do not by themselves prove that a series has a unit root. The dated observations, plausible process mechanisms, and forecast performance all matter.

Worked Case: Claims Levels and First Differences

Consider hypothetical quarterly incurred claims totals, in thousands of dollars, from 2022 Q1 through 2025 Q4:

\(120, 126, 123, 131, 135, 132, 140, 145, 142, 151, 154, 152, 160, 165, 162, 171.\)

The levels generally rise, but the consecutive changes alternate between increases and decreases. The first differences, in the same units, are:

\(6,-3,8,4,-3,8,5,-3,9,3,-2,8,5,-3,9.\)

There are 15 differences. Their sum telescopes to the last level minus the first level: \(171-120=51\). Thus the sample mean difference is \(51/15=3.4\) thousand dollars per quarter. This average increase is evidence about the observed period, not proof that future claims will rise at exactly that rate.

Model assumptions for the calculations below: For the random-walk candidate, suppose the fitted model is \(Y_t=Y_{t-1}+c+\varepsilon_t\), with \(c=3.4\) and independent, identically distributed innovations \(\varepsilon_t\sim N(0,16)\). For the differenced ARIMA candidate, suppose \(\Delta Y_t=c+\phi(\Delta Y_{t-1}-c)+\varepsilon_t\), with \(c=3.4\), \(\phi=0.60\), and independent, identically distributed normal innovations with mean zero and variance 16. Treat these fitted parameters as fixed for the plug-in forecasts. Independence of future innovations from the observed past is an assumption of these forecast calculations; zero correlation alone would not generally be enough to establish conditional-mean forecasts.

Both candidates have first differences fluctuating around a positive average. Their forecasts differ because the ARIMA candidate allows a recent change to influence the next change, with that influence decaying geometrically. In contrast, the random walk with drift forecasts the same expected increase, \(c\), in every future quarter.

Worked Example: Fixed-Origin Forecasts at Horizons One Through Four

Worked Example: Comparing Fixed-Origin Claims Forecasts

At the end of 2025 Q4, the latest claims total is \(Y_T=171\), and the latest observed difference is \(\Delta Y_T=171-162=9\). Calculate forecasts for 2026 Q1 through Q4 from the two models above, using this one fixed forecast origin. Then compare them with subsequently observed hypothetical claims totals of \(175,181,186,194\), respectively.

Method: random walk with drift. Its conditional-mean forecast at horizon \(h\) is \(Y_T+hc\). Therefore the four forecasts are:

$$ \begin{aligned} \hat{Y}_{T+1|T}&=171+1(3.4)=174.4,\\ \hat{Y}_{T+2|T}&=171+2(3.4)=177.8,\\ \hat{Y}_{T+3|T}&=171+3(3.4)=181.2,\\ \hat{Y}_{T+4|T}&=171+4(3.4)=184.6. \end{aligned} $$

Method: ARIMA(1,1,0) with drift. The forecasted change at horizon \(h\) is \(c+\phi^h(\Delta Y_T-c)\). For horizons one through four, the forecasted changes are:

$$ \begin{aligned} \widehat{\Delta Y}_{T+1|T}&=3.4+0.60(9-3.4)=6.76,\\ \widehat{\Delta Y}_{T+2|T}&=3.4+0.60^2(9-3.4)=5.416,\\ \widehat{\Delta Y}_{T+3|T}&=3.4+0.60^3(9-3.4)=4.6096,\\ \widehat{\Delta Y}_{T+4|T}&=3.4+0.60^4(9-3.4)=4.12576. \end{aligned} $$

The level forecasts add the forecasted changes cumulatively to \(171\):

$$ \begin{aligned} \hat{Y}_{T+1|T}&=171+6.76=177.76,\\ \hat{Y}_{T+2|T}&=171+6.76+5.416=183.176,\\ \hat{Y}_{T+3|T}&=171+6.76+5.416+4.6096=187.7856,\\ \hat{Y}_{T+4|T}&=171+6.76+5.416+4.6096+4.12576=191.91136. \end{aligned} $$

At the four horizons, the actual-minus-forecast errors for the random walk are \(0.6, 3.2, 4.8, 9.4\). Its squared errors are \(0.36, 10.24, 23.04, 88.36\), so the mean squared error is \((0.36+10.24+23.04+88.36)/4=30.50\). The ARIMA errors are \(-2.76, -2.176, -1.7856, 2.08864\); their squared errors are \(7.6176, 4.734976, 3.18836736,\) and approximately \(4.362416\). The mean squared error is approximately \(19.90336/4=4.97584\). On this short, hypothetical four-quarter evaluation, the ARIMA forecasts have the smaller mean squared error.

Check: The ARIMA change forecasts approach \(3.4\), the drift, as the horizon increases; its level forecasts therefore continue to rise, but the additional increase per quarter gradually approaches the random walk’s constant \(3.4\). These are four forecasts from one origin, not four successive one-step-ahead forecasts. The comparison is illustrative: four forecast errors are too few to establish that one model will generally forecast better.

Why the Forecast Formulas Hold

The conditional-mean forecast is valid when future innovations have conditional expectation zero given the information available at the forecast origin. Independent, identically distributed innovations with mean zero satisfy this condition. Martingale-difference innovations with respect to that information set also satisfy it. Merely assuming that innovations are uncorrelated does not, in general, imply zero conditional expectation.

For the random walk with drift, expand the model forward \(h\) periods:

$$ Y_{T+h}=Y_T+hc+\sum_{j=1}^{h}\varepsilon_{T+j}. $$

Given the data through time \(T\), \(Y_T\) and \(c\) are known for this calculation. If each future innovation has conditional mean zero, the conditional expectation of the sum is zero. Therefore:

$$ E[Y_{T+h}\mid\mathcal{F}_T]=Y_T+hc. $$

Here \(\mathcal{F}_T\) represents the information available through time \(T\). Under the additional assumption of independent normal innovations with variance \(\sigma^2\), the forecast error is the sum of \(h\) independent innovations, with variance \(h\sigma^2\). Thus, in the example, the four forecast-error variances are \(16,32,48,64\) thousand-dollars-squared. The corresponding standard errors are \(4,\ 4\sqrt{2},\ 4\sqrt{3},\ 8\). These calculations do not include uncertainty from estimating the model parameters.

For the ARIMA candidate, the differenced series follows an AR(1) model around \(c\). Its conditional forecast for the next difference is \(c+\phi(\Delta Y_T-c)\). Repeated substitution gives \(c+\phi^h(\Delta Y_T-c)\) at horizon \(h\). The level forecast must sum all forecasted changes because \(Y_{T+h}=Y_T+\sum_{j=1}^h\Delta Y_{T+j}\). This cumulative step is essential: a forecast for a difference is not a forecast for the level.

Residual Checks and Candidate Comparison

After fitting a candidate, inspect its residuals, which are the observed values minus fitted values on the scale modeled. For a random walk, the relevant residuals are innovations in the level equation; for a differenced ARIMA model, they are innovations in the differenced equation. A useful residual series should have no systematic pattern left that the model could have captured. Residual plots can reveal changing spread, outliers, or runs; residual autocorrelations can reveal remaining serial dependence.

The Ljung–Box statistic combines residual autocorrelations over a specified number of lags. At \(m\) lags, it is:

$$ Q=m(m+2)\sum_{k=1}^{m}\frac{r_k^2}{n-k}, $$

where \(n\) is the number of residuals and \(r_k\) is the sample residual autocorrelation at lag \(k\). Under the approximate null distribution, \(Q\) is compared with a chi-square distribution. When parameters are estimated, a common degrees-of-freedom adjustment subtracts the number of fitted AR and MA terms; the exact testing convention should be stated.

Worked Example: Interpreting Residual Autocorrelations

Suppose the differenced ARIMA fit has \(n=20\) usable residuals, with residual autocorrelations \(r_1=0.10\) and \(r_2=-0.05\). Test for autocorrelation through lag 2 using the Ljung–Box statistic. The model has one fitted AR term and no fitted MA terms.

Set \(m=2\). The statistic is:

$$ \begin{aligned} Q&=2(2+2)\left(\frac{0.10^2}{20-1}+\frac{(-0.05)^2}{20-2}\right)\\ &=8\left(\frac{0.01}{19}+\frac{0.0025}{18}\right)\\ &\approx8(0.00052632+0.00013889)\\ &\approx0.0053217. \end{aligned} $$

The arithmetic above needs the outer factor: the Ljung–Box multiplier is \(m(m+2)=2(4)=8\), so \(Q\approx8(0.00052632+0.00013889)=0.0053217\). This value is very small. With the common adjustment, the reference degrees of freedom are \(2-1=1\); the 5% upper critical value is approximately \(3.841\). Since \(0.0053217<3.841\), these data do not provide evidence against the null of no residual autocorrelation through lag 2.

Interpretation and check: Failing to reject this null is not proof that the model is correct. This short-lag check does not establish normality, constant variance, or good forecasts. Also, the calculation uses the stated \(n\) and autocorrelations; it does not claim these diagnostic values came from the claims series in the worked case.

A model should not be selected from one diagnostic alone. Compare plausible candidates using parsimony, fit criteria when appropriate, residual behavior, and forecast evaluation designed to match the decision. If the task is to forecast four quarters ahead from a single cutoff, evaluate four-horizon fixed-origin forecasts. If the task concerns repeated next-quarter forecasting, use successive origins and update the information set with each newly observed value. Those are different evaluation designs and produce different forecast errors.

Worked Example: A One-Step Forecast With an Updated Origin

Worked Example: Updating Before the Next Forecast

Use the fitted random walk with drift \(c=3.4\), with the same innovation assumptions, for the first two held-out claims values in the preceding case. The first actual value is 175. After observing it, forecast the next quarter. Calculate the two one-step forecast errors.

At the end of 2025 Q4, the one-step forecast for 2026 Q1 is \(171+3.4=174.4\). The actual value is 175, so the first actual-minus-forecast error is \(175-174.4=0.6\).

Now update the origin: the latest observed value is \(175\). The one-step forecast for 2026 Q2 is \(175+3.4=178.4\). The actual value is 181, so the second error is \(181-178.4=2.6\).

Check: The second forecast uses the newly observed 175, not the original 171. The one-step errors are \(0.6\) and \(2.6\), and their mean squared error is \((0.6^2+2.6^2)/2=(0.36+6.76)/2=3.56\). This differs from the fixed-origin two-quarter forecast comparison, which does not update after seeing the first held-out value.

Common Mistakes in Random-Walk Case Analysis

  • Calling a rising series stationary because its differences have a stable average. Stationarity of the differences does not make the levels stationary; a random walk’s level variance grows over time.
  • Using uncorrelated innovations as if they guaranteed conditional mean zero. For conditional-mean forecasts, state a sufficient assumption such as independence from the past or the martingale-difference property.
  • Forecasting differences but reporting them as levels. For a differenced model, add the forecasted changes cumulatively to the latest observed level.
  • Confusing fixed-origin multi-step forecasts with rolling one-step forecasts. A fixed-origin forecast holds the information set fixed; a rolling evaluation updates it after each new observation.
  • Treating a residual test as a certificate of model validity. A Ljung–Box result addresses residual autocorrelation at selected lags; it does not check every model assumption or guarantee good future performance.
Key takeaway: A random-walk interpretation is credible only when the dated series, differenced behavior, model assumptions, residual diagnostics, and horizon-specific forecast evidence fit together. The random walk’s simple forecast is a valuable benchmark; a more elaborate differenced ARIMA candidate is useful only if its added structure is supported and improves the forecasting task being evaluated. This case-study approach is aligned with the SOA Exam SRM (Statistics for Risk Modeling) syllabus.

Check Your Understanding

Work through each item using the stated model assumptions. Claims amounts are hypothetical.

  1. A random walk with drift has latest level \(Y_T=240\) and drift \(c=2.5\), both in thousands of dollars. What is its fixed-origin forecast three periods ahead? (Answer: \(247.5\) thousand dollars.)
  2. A random walk has independent, mean-zero innovations with variance \(9\). What is the forecast-error variance and standard error at horizon 4, ignoring parameter uncertainty? (Answer: variance \(36\), standard error \(6\), in the corresponding level units.)
  3. A differenced ARIMA model has \(c=2\), \(\phi=0.5\), and latest difference \(\Delta Y_T=6\). What is the forecasted difference at horizon 2? (Answer: \(2+0.5^2(6-2)=3\).)
  4. For the model in question 3, the latest level is 100. Calculate its level forecast at horizon 2 by summing forecasted differences. (Answer: the differences are \(4\) and \(3\), so the level forecast is \(100+4+3=107\).)
  5. A fixed-origin forecast is 52 for the next period, whose actual value is 54. After observing 54, a random walk with drift 1.5 is used to forecast the following period. The next actual value is 56. Find both one-step errors and their mean squared error. (Answer: errors \(2\) and \(0.5\); mean squared error \((4+0.25)/2=2.125\).)
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