Conditional variance and the variance budget

What you will learn

Derive and audit total variance through conditional residual cancellation, then distinguish average information gains from the spread of one selected group.

Prerequisites: Conditional forecasts and the tower · Variance and standard deviation

Define conditional spread on an exact population

Assume one of six distinct tickets is drawn uniformly. Their scores, in ticket order, are −2, 2, 4, 4, 4 and 8. The first two tickets carry report A and the last four carry report B. Repeated scores belong to separate tickets, each with probability 1/6. These are stipulated numerical measurements, not observed returns or estimated parameters. All moments exist because the population is finite. We make no independence, normality or causal assumption. The report is simply a function of the selected ticket. Population variance uses probability weights; there is no sample-size correction.

Let X be the score and S the report. The overall mean μ is 20/6=10/3. A report changes which tickets are compatible with the observation. Within A the mean mA is zero and the variance vA is [(-2)²+2²]/2=4. Within B the mean mB is 5 and the variance vB is [(-1)²+(-1)²+(-1)²+3²]/4=3. Thus Var(X given S) is itself a random variable: it equals 4 with probability 1/3 and 3 with probability 2/3. It is different from the variance of the conditional mean, whose possible values are centers rather than spreads.

Build a squared-deviation ledger before using a formula

For each ticket write X−μ=(X−mS)+(mS−μ). The first part measures deviation from the reported group center; the second measures displacement of that center from the population center. Squaring gives (X−μ)²=(X−mS)²+(mS−μ)²+2(X−mS)(mS−μ). This equality holds ticket by ticket. The last quantity is signed, so a diagram with only two positive square areas would be incomplete at the individual-ticket level. A negative cross term is allowed; a negative variance is not.

Why may the cross term disappear after averaging? Inside a fixed report s, the center displacement mS−μ is constant. Multiplying it by the sum of within-group residuals gives zero, because Σ(X−ms)=ΣX−ns ms=0. Therefore the cross contributions cancel separately inside each report. This uses the conditional-mean definition, not independence between residual and report. For example, ticket 1 has residual −2 and center displacement −10/3, giving cross term 40/3; ticket 2 contributes −40/3. Neither term is zero, but their equal-probability contributions cancel. The four B residuals −1,−1,−1,3 cancel for the same reason.

Reconcile within variation with movement of centers

The expected conditional variance is W=(1/3)4+(2/3)3=10/3. Each report must receive its probability weight; (4+3)/2=3.5 incorrectly treats the two reports as equally likely. The between component is B=Var(mS)=(1/3)(0−10/3)²+(2/3)(5−10/3)²=100/27+50/27=50/9. Both W and B have squared-score units. The law of total variance is Var(X)=W+B=10/3+50/9=80/9. Standard deviations cannot be added this way, because taking a square root does not distribute across a sum.

Independently audit the result using raw second moments. The squared scores sum to 4+4+16+16+16+64=120, so E[X²]=20. Subtract μ²=100/9 to obtain 80/9 again. A third check centers each ticket directly: deviations are −16/3,−4/3,2/3,2/3,2/3,14/3. Their squared numerators sum to 256+16+4+4+4+196=480. Dividing by 9 and then by six yields 80/9. Agreement among grouped, second-moment and direct-deviation calculations checks different arithmetic steps rather than repeating the same formula.

Change the report while preserving the score distribution

Suppose a richer report separates the three score-4 tickets from the score-8 ticket while retaining A. The new reports have masses 2/6,3/6,1/6, means 0,4,8 and conditional variances 4,0,0. Their expected within variation is now 4/3. The between component becomes 68/9, so the total stays 80/9. Information has moved two squared-score units from unresolved variation into variation of forecast centers. It has not changed the draw mechanism, any score, or the unconditional variance.

At the other extreme, suppress the report completely. The one conditional center is μ, W becomes 80/9 and B becomes zero. Revealing the exact ticket makes W zero and B equal 80/9. Under nested refinements, average within variation cannot increase, but a particular newly observed group can have a larger conditional variance than the population. For a concrete counterexample in this same population, report whether the ticket is one of the two extreme tickets −2 and 8. That group has mean 3 and variance 25, greater than 80/9. Its probability is only 1/3. The identity constrains a weighted average; it does not bound every individual conditional spread.

Stress-test the missing cross term in the workshop

The workshop lets you move individual physical tickets between two reports, preserving every ticket probability and the entire score law. Duplicate score-4 tickets have separate controls. Read each group count and center, then construct the averaged residual square, averaged center-displacement square and signed cross term. Changing group membership invalidates all submitted entries. Empty groups are absent reports with zero probability; the model never divides by their counts. Putting every ticket in one group is therefore a legitimate boundary case.

Next activate shifted centers, which replaces each conditional center ms by cs=ms+1. The ticket identity still holds, but cross cancellation fails. For the original partition, residual-square average is 13/3, center-displacement-square average is 59/9, and cross average is −2. Their sum remains 80/9. Calling the first two quantities expected conditional variance and variance of conditional expectation would be wrong in this shifted mode: the chosen centers are biased and their average is no longer μ. Use this controlled failure to explain why checking only the sum of two positive terms is insufficient. These are exact population accounting identities, not evidence that an estimated forecasting system is accurate on future data.

Hands-on: Reallocate a fixed variance budget

Move physical tickets between reports, compare conditional and shifted centers, and reconcile all three squared-deviation terms.

Common mistakes

• Averaging group variances without probability weights.
• Discarding a cross term ticket by ticket instead of proving groupwise cancellation.
• Adding standard deviations instead of squared deviations.
• Claiming every realized conditional variance is smaller than the unconditional variance.
• Calling residual squares around shifted centers conditional variances.

Keep these ideas

• Conditional variance is a report-dependent random variable.
• Total variance equals expected within variance plus variance of conditional means.
• Zero conditional residual means justify cross cancellation without independence.
• Nested refinement reallocates variation without changing the score law.
• A particular high-spread report can exceed the population variance.

Further reading

• Data 140: Variance by Conditioning, section 22.3.1 ↗

  Read deviation decomposition, cross cancellation and the total-variance formula. Original six-ticket derivation and regrouping activity; no video reviewed. Checked 2026-10-24.

• Data 140: Least Squares Predictor, sections 22.2.3–22.2.4 ↗

  Read conditional variance definition and its expected residual-square interpretation. Original population, shifted-center audit and diagnostics; no source exercises or video reproduced. Checked 2026-10-24.