Conditional forecasts and the tower

What you will learn

Construct the law of E[X given S], derive iterated expectation and verify a nested tower under information coarsening while rejecting forecasts with only a correct global mean.

Prerequisites: Transforming distributions and losing information · Combining conditional branches · Expected value as a weighted average

Specify the observation before the forecast

Assume one ticket is selected uniformly from eight physical tickets. Their numerical scores are −4, 0, 4, 2, 5, 5, 8 and 8. A reporting system reveals only a band: A contains the first two tickets; B contains the ticket scoring 4; C contains the three tickets scoring 2, 5 and 5; D contains the two tickets scoring 8. Scores are stipulated measurements, not profits, and the eight tickets are a complete exact population rather than an empirical sample. No independence, normal distribution, fitted model or repeated-trial approximation is assumed. Let X be the selected score and S the reported band.

A forecast based only on S must give the same answer to every ticket in a band. The observer cannot distinguish the two A tickets, so assigning −4 to one and 0 to the other uses hidden information. Conditional expectation specifies a particular legal forecast: average X within each observed band using its conditional ticket weights. The band weights are 2/8, 1/8, 3/8 and 2/8. The corresponding forecasts are −2, 4, 4 and 8. A forecast may be a value that no ticket actually carries; −2 is a mean, not a newly created physical outcome.

A table of conditional means is itself a random variable

For a fixed report s, write m(s)=E[X given S=s]. Each m(s) is a number. Before the ticket is selected, however, the report S is random, and M=m(S) is a random variable. Its value is the forecast that will be issued once the report arrives. Do not confuse X, the eventual score, with M, the report-dependent mean, or with E[X], the single unconditional average. Those three objects answer different questions even when some numerical values happen to coincide.

Build the law of M by collecting report probabilities at equal forecast values. A produces −2 with mass 2/8. Both B and C produce 4, so their disjoint masses combine to (1+3)/8=4/8. D produces 8 with mass 2/8. Thus M has three possible values even though S has four possible reports. This is an information compression: observing a forecast of 4 does not reveal whether B or C occurred. It also shows why assigning probability 1/4 to each distinct forecast value would be wrong. There are three values, four bands and eight tickets; only the stipulated ticket mechanism determines their probabilities.

Recover the original mean by weighting forecasts

Directly summing the eight scores gives −4+0+4+2+5+5+8+8=28, hence E[X]=28/8=3.5. Averaging M from its law gives [−2×2+4×4+8×2]/8=(−4+16+16)/8=3.5. The equality E[E[X given S]]=E[X] is iterated expectation. The outer expectation averages over which report occurs, while the inner expectation averages over scores compatible with that report. An unweighted average of the four band forecasts is (−2+4+4+8)/4=3.5 here too, but that agreement is accidental: the unequal band weights cannot generally be discarded.

For an algebraic audit, let band j contain total ticket weight wj and weighted score sum tj, with total weight W. Its conditional mean is tj/wj. The contribution to the outer mean is (wj/W)(tj/wj)=tj/W. Adding over a disjoint, exhaustive partition counts every score contribution exactly once, yielding Σtj/W=E[X]. This cancellation explains the identity without requiring independent variables. Every displayed band has positive mass. An impossible report would have denominator zero, so the ordinary conditional ratio would be undefined there; no fabricated zero forecast should be used to conceal that modeling problem.

Merge reports and verify the nested tower

Now replace S with a coarser report G that merges A with B into L and C with D into H. The observer has less information, because G is determined by S but S cannot be reconstructed from G. L has three tickets and score total −4+0+4=0, so E[X given L]=0. H has five tickets and total 2+5+5+8+8=28, so E[X given H]=28/5=5.6. The new forecast law has value 0 with probability 3/8 and value 5.6 with probability 5/8. Its overall mean remains (3/8)0+(5/8)(28/5)=3.5.

The same coarse forecasts can be obtained without returning to the raw scores. Inside L, the original bands have conditional weights 2/3 and 1/3, giving (2/3)(−2)+(1/3)4=0. Inside H, they have weights 3/5 and 2/5, giving (3/5)4+(2/5)8=28/5. This is the finite nested tower: E[E[X given S] given G]=E[X given G] when G is a coarsening of S. Averaging the two fine forecasts equally within H would give 6, an incorrect result. The nesting assumption matters: arbitrary overlapping reports do not license this same reduction. The workshop only merges adjacent whole bands, so every selected coarse partition satisfies it.

A global mean check cannot certify conditional correctness

Consider a forecast that always announces 3.5. It has the correct overall average, yet its A residual total is (−4−3.5)+(0−3.5)=−11. A proper conditional-mean forecast has weighted residual sum zero within every reported group. Under the fine forecast −2 in A, residuals are −2 and +2, which cancel. Under the coarse forecast 0 in L, the three residuals −4,0,+4 cancel. Cancellation is an aggregate property, not a promise that any one prediction equals the eventual score.

Use the workshop to split and merge boundaries, enter a forecast for each visible group and inspect both its mass and signed residual total. Your entries create a candidate forecast law, so equal entries combine into a single value. Audit each group before trusting the outer mean. With the default two groups, guessing 3.5 everywhere passes the global average but fails both local residual tests. Returning to all four bands exposes another subtlety: B and C can share forecast 4 while retaining different conditional score laws. A mean does not identify a full distribution or establish independence. These exact model identities validate the stated arithmetic, not the accuracy of an unknown real-world reporting system.

Hands-on: Merge information and balance conditional residuals

Choose an information partition, construct its forecast law and test each group residual before checking the tower mean.

Common mistakes

• Weighting bands equally rather than weighting tickets, even if an accidental equality hides the mistake.
• Counting B and C as distinct forecast values when both issue 4.
• Using a hidden ticket score as a forecast when only a merged band is observed.
• Inferring conditional correctness from one global mean or zero overall residual.
• Applying the nested tower to nonnested reports without proving the required information relation.

Keep these ideas

• A fixed report gives a number; the report-dependent conditional mean is a random variable.
• Collect equal conditional means to build its law.
• Average with report probabilities to recover E[X].
• Whole-band merging permits a nested tower with renormalized within-group weights.
• Check residual balance inside every observable group, not only across the full population.

Further reading

• Data 140 textbook: Expectation by Conditioning, sections 9.2.1–9.2.2 ↗

  Read conditional expectation as a random variable and the finite iterated-expectation proof. Our eight-ticket score model, merging derivation and diagnostics are original. No video review claimed. Checked 2026-10-24.

• Probability for Data Science: Conditional Expectation As a Projection, review and within-strip deviations ↗

  Read the conditional-mean definition and zero within-strip residual derivation. Scope excludes later optimal-prediction results. Original finite partition activity; no source exercise or illustration reproduced. Checked 2026-10-24.