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Deriving Random-Walk Forecasts and Their Uncertainty

Learn to derive random-walk forecasts and quantify how their uncertainty changes across forecast horizons.

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What You'll Learn

  • Define a random walk with and without drift using its innovation assumptions.
  • Derive the conditional point forecast from the model’s recursive form.
  • Derive forecast-error variance and standard deviation at horizon h.
  • Distinguish the innovation variance from the variance of the observed series.
  • Explain how estimating drift adds a separate source of forecast uncertainty.
  • Identify horizon, drift, and variance-versus-standard-deviation traps.

Why Random-Walk Forecasts Need a Horizon

A random walk carries its current level forward, then adds a new random innovation at each time step. Its forecast therefore depends on the forecast origin, the horizon, and whether the model includes drift. The forecast level may remain constant when there is no drift, even though the forecast becomes less certain at every longer horizon.

This distinction matters in a hypothetical risk index: a forecast for next period and one for several periods ahead can have the same point estimate under a no-drift random walk, but they do not have the same forecast-error variance. The treatment here is aligned with the SOA Exam SRM (Statistics for Risk Modeling) syllabus.

Definition: A random walk without drift satisfies \(y_t=y_{t-1}+\varepsilon_t\). A random walk with drift satisfies \(y_t=c+y_{t-1}+\varepsilon_t\), where \(c\) is the drift per time period. In either model, assume the innovations \(\varepsilon_t\) are independent and identically distributed with mean \(0\) and variance \(\sigma_\varepsilon^2\), and are independent of the observed history through time \(T\). Normality is not required for the forecast means or variances derived below.

The forecast origin is \(T\), the observed level at that origin is \(y_T\), and \(h\) is a positive integer counting future time steps. Write \(\hat{y}_{T+h|T}\) for the forecast of \(y_{T+h}\) using information through time \(T\). The forecast error is the future value minus that forecast. The drift \(c\) and innovation variance \(\sigma_\varepsilon^2\) are treated as known unless stated otherwise.

Deriving the Forecast and Its Error Variance

Repeatedly substitute the random-walk equation forward from the forecast origin. Under the model with drift, the first future value is \(y_{T+1}=y_T+c+\varepsilon_{T+1}\). Each subsequent step adds another \(c\) and another innovation. After \(h\) steps, this gives:

$$ y_{T+h}=y_T+hc+\sum_{j=1}^{h}\varepsilon_{T+j}. $$

The no-drift model is the special case \(c=0\). Since \(y_T\) is known at the origin and each future innovation has conditional expectation zero, taking the conditional expectation given the observed history through \(T\) gives the point forecast.

$$ \hat{y}_{T+h|T}=E[y_{T+h}\mid\mathcal{F}_T]=y_T+hc. $$

Here \(\mathcal{F}_T\) denotes the information observed through time \(T\). Subtract this conditional mean from the future value to obtain the forecast error. The terms \(y_T\) and \(hc\) cancel, leaving only the future innovations:

$$ e_{T+h|T}=y_{T+h}-\hat{y}_{T+h|T} =\sum_{j=1}^{h}\varepsilon_{T+j}. $$

The future innovations are independent, each with variance \(\sigma_\varepsilon^2\). Variances of independent terms add, so the forecast-error variance and standard deviation are:

$$ \operatorname{Var}(e_{T+h|T}\mid\mathcal{F}_T) =\sum_{j=1}^{h}\operatorname{Var}(\varepsilon_{T+j}) =h\sigma_\varepsilon^2, \qquad \operatorname{SD}(e_{T+h|T}\mid\mathcal{F}_T) =\sqrt{h}\,\sigma_\varepsilon. $$

This derivation explains why the variance grows linearly with the horizon: every additional step contributes one more innovation variance. The standard deviation grows as \(\sqrt{h}\), not as \(h\). Neither result says that the observed level \(y_{T+h}\) has variance \(h\sigma_\varepsilon^2\) unconditionally; the formula is for the forecast error conditional on the information at the origin.

Key result: With known drift and innovation variance, the random-walk forecast is \(y_T+hc\), and its forecast-error variance is \(h\sigma_\varepsilon^2\). Without drift, set \(c=0\), so the forecast is \(y_T\) at every horizon. The uncertainty still increases with \(h\).

Worked Forecasts

Worked Example: A No-Drift Forecast Several Steps Ahead

A hypothetical risk index has current value \(y_T=82\). Model it as a random walk without drift, with independent normal innovations \(\varepsilon_t\sim N(0,9)\) in index points per period. Find the point forecast and forecast-error variance four periods ahead. State the forecast-error standard deviation as well.

There is no interest-rate or mortality component in this index forecast. The innovation distribution has mean \(0\) and variance \(\sigma_\varepsilon^2=9\), so \(\sigma_\varepsilon=3\). The model has \(c=0\), the origin level is \(82\), and the horizon is \(h=4\).

Point forecast. Set \(c=0\) in the forecast equation:

$$ \hat{y}_{T+4|T}=y_T+4c=82+4(0)=82. $$

Forecast-error variance. Four future innovations contribute to the error:

$$ \operatorname{Var}(e_{T+4|T}\mid\mathcal{F}_T) =4(9)=36. $$

Forecast-error standard deviation.

$$ \sqrt{36}=6. $$

The four-period forecast is \(82\), with forecast-error variance \(36\) and standard deviation \(6\) index points. The check is that the variance is four times the one-period innovation variance, not \(4^2\) times that variance. This method avoids treating a constant point forecast as evidence of constant uncertainty.

Worked Example: A Random Walk With Known Drift

A hypothetical portfolio-risk measure is \(124\) at the forecast origin. A random walk with known drift \(c=2.5\) measure-units per month is specified, with independent normal innovations \(\varepsilon_t\sim N(0,16)\) measure-units per month. Find the forecast and forecast-error variance four months ahead.

No interest-rate or mortality assumption is involved. The model parameters are \(y_T=124\), \(c=2.5\), \(\sigma_\varepsilon^2=16\), and \(h=4\). The innovation standard deviation is \(\sqrt{16}=4\).

Point forecast. The drift contributes \(2.5\) units in each of four steps, for a total expected increase of \(4(2.5)=10\):

$$ \hat{y}_{T+4|T}=124+4(2.5)=124+10=134. $$

Forecast-error variance. Drift changes the conditional mean but does not add random variation when it is known. The four future innovations contribute:

$$ \operatorname{Var}(e_{T+4|T}\mid\mathcal{F}_T) =4(16)=64. $$

The standard deviation is \(\sqrt{64}=8\) units. As a check, the forecast is \(10\) units above the origin value, while the forecast error has variance \(64\); the deterministic drift is not part of that error variance. This method avoids adding \(c\) to the innovation variance.

Worked Example: Forecasting When Drift Is Estimated

A hypothetical index is \(60\) at the forecast origin. Its four most recent one-period changes are \(1, 3, 2,\) and \(4\). Assume the changes follow an independent normal model with unknown mean drift \(c\) and unknown innovation variance \(\sigma_\varepsilon^2\). Estimate the drift by the sample mean and innovation variance by the sample variance. Find the three-period point forecast and an estimate of forecast-error variance that includes drift-estimation uncertainty.

There is no interest-rate or mortality assumption. The observed increments are treated as a sample from \(N(c,\sigma_\varepsilon^2)\). First calculate the estimated drift:

$$ \hat{c}=\frac{1+3+2+4}{4}=\frac{10}{4}=2.5. $$

The deviations from \(2.5\) are \(-1.5, 0.5, -0.5,\) and \(1.5\). Their squared deviations are \(2.25, 0.25, 0.25,\) and \(2.25\), which sum to \(5\). Thus the unbiased sample variance estimate is:

$$ s_\varepsilon^2=\frac{5}{4-1}=\frac{5}{3}\approx1.6667. $$

Point forecast. Use the estimated drift over three periods:

$$ \hat{y}_{T+3|T}=60+3(2.5)=67.5. $$

Variance from future innovations. Using the estimated innovation variance, this component is:

$$ 3s_\varepsilon^2=3\left(\frac{5}{3}\right)=5. $$

Additional variance from estimating drift. The estimated drift is the mean of \(n=4\) independent increments, so its variance is \(\sigma_\varepsilon^2/4\). Its contribution to a forecast \(h=3\) steps ahead is \(h^2\sigma_\varepsilon^2/4\), estimated as:

$$ 3^2\left(\frac{s_\varepsilon^2}{4}\right) =9\left(\frac{5/3}{4}\right) =9\left(\frac{5}{12}\right) =\frac{45}{12}=3.75. $$

Adding the two components gives an estimated repeated-sample forecast-error variance of \(5+3.75=8.75\), and standard deviation \(\sqrt{8.75}\approx2.9580\). The check is that the additional component is positive: uncertain drift matters more at longer horizons because a drift error is accumulated \(h\) times. This calculation accounts for drift-estimation uncertainty across repeated historical samples; conditional on the observed fitted drift as fixed, the future-innovation component alone is \(5\).

How Uncertainty Changes With Horizon

For a model with known parameters, the forecast-error variance at horizon \(h\) is \(h\sigma_\varepsilon^2\). Moving from \(h\) to \(h+1\) adds exactly one more innovation variance, \(\sigma_\varepsilon^2\). The point forecast under drift moves by \(c\) per added step. In the no-drift case, the point forecast stays at \(y_T\), but the forecast-error variance continues to increase.

When drift is estimated from \(n\) independent historical increments, the distinction between the two uncertainty calculations matters. The variance conditional on the fitted drift treats it as fixed and counts future innovations only. Across repeated estimation samples, the estimated drift also varies. If the drift estimate is the sample mean, its variance is \(\sigma_\varepsilon^2/n\). The estimation contribution to an \(h\)-step forecast is therefore \(h^2\sigma_\varepsilon^2/n\), because the drift error is multiplied by \(h\). Under independent historical and future innovations, the repeated-sample forecast-error variance is:

$$ h\sigma_\varepsilon^2+h^2\frac{\sigma_\varepsilon^2}{n}. $$

This expression assumes a sample-mean drift estimate from \(n\) independent increments and describes uncertainty across repeated samples and future paths. It should not be confused with the conditional forecast-error variance when the fitted drift is held fixed.

When a normal forecast distribution applies: If the innovations are independent \(N(0,\sigma_\varepsilon^2)\) and the drift and variance are known, the forecast error is normally distributed with mean \(0\) and variance \(h\sigma_\varepsilon^2\), because it is a sum of \(h\) independent normal innovations. Without normality, the mean and variance results still hold under the stated innovation assumptions, but normality of the forecast error does not follow.

Common Exam Traps

A frequent error is to use the horizon incorrectly. For example, with innovation variance \(9\) and horizon \(4\), writing \(4^2(9)=144\) treats the four independent innovations as if they were the same error repeated four times. The correct calculation is:

$$ \operatorname{Var}\left(\sum_{j=1}^{4}\varepsilon_{T+j}\right) =\sum_{j=1}^{4}\operatorname{Var}(\varepsilon_{T+j}) =4(9)=36. $$

Another common error is to report \(h\sigma_\varepsilon^2\) as a standard deviation. It is a variance; the standard deviation is \(\sqrt{h\sigma_\varepsilon^2}\). Also check the time unit: if the model is monthly, a horizon of one year is \(h=12\), not \(h=1\). Use the model’s period consistently for the drift and innovation variance.

Finally, drift and innovation variance play different roles. Known drift moves the point forecast by \(hc\) but does not change the forecast-error variance. An estimated drift can add parameter-estimation uncertainty, as shown above. Do not insert the drift itself into the innovation variance.

Check Your Understanding

For each problem, identify the horizon and distinguish the point forecast from forecast-error uncertainty.

  1. A no-drift random walk has \(y_T=35\) and independent innovations with mean \(0\) and variance \(4\). Find the forecast and forecast-error variance six periods ahead. (Answer: forecast \(35\); variance \(24\).)
  2. A random walk has \(y_T=18\), known drift \(c=-0.75\) per quarter, and innovation variance \(2.25\). Find the forecast and forecast-error standard deviation eight quarters ahead. (Answer: forecast \(12\); standard deviation \(\sqrt{18}\approx4.2426\).)
  3. A no-drift model has innovation standard deviation \(5\). Find the forecast-error variance and standard deviation at horizon \(9\). (Answer: variance \(225\); standard deviation \(15\).)
  4. A random walk with known drift \(1.2\) has current value \(70\) and innovation variance \(3\). Find the point forecast and forecast-error variance five steps ahead. (Answer: forecast \(76\); variance \(15\).)
  5. A drift estimate is the mean of \(n=5\) independent increments, with innovation variance \(4\). For a horizon of \(h=3\), calculate the repeated-sample forecast-error variance including drift-estimation uncertainty. (Answer: \(3(4)+3^2(4/5)=12+7.2=19.2\).)

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