Value of Imperfect Information

Lecture

Dr. James Doss-Gollin

Wednesday, February 25, 2026

Draft: This content is under development.

Recap: Monday’s Umbrella Problem

	Bring umbrella	Don’t bring
Rain (\(p = 0.3\))	\(-5\)	\(-50\)
No rain (\(1 - p = 0.7\))	\(-5\)	\(0\)

• Without information: bring the umbrella, expected cost \(= \$5.00\)
• With perfect information: expected cost \(= 0.3 \times 5 + 0.7 \times 0 = \$1.50\)
• EVPI \(= 5.00 - 1.50 = \$3.50\)

Target: 1:00. “Let’s make sure everyone has the key result from Monday.” Have the payoff table visible. Give 15 seconds for students to recall the EVPI formula.

Decision Trees

A decision tree represents a decision problem as a sequence of nodes.

• Squares are decisions (you choose)
• Circles are chance events (nature chooses)
• The order of nodes matches the order of \(\max\) and \(\sum\) in the formula

Target: 1:03. “A decision tree is just a picture of the formula. Let me show you what I mean.”

Tree: No Information

You choose before nature reveals the state. The decision node comes first — this computes \(\max_a \sum_s U(a,s) \, p(s)\).

digraph {
    rankdir=LR
    node [fontname="Helvetica", fontsize=11]
    edge [fontname="Helvetica", fontsize=10]

    D [shape=box, label="Decision", style=filled, fillcolor="#cde0f5"]

    NB [shape=circle, label="Nature", width=0.7, style=filled, fillcolor="#e0e0e0"]
    ND [shape=circle, label="Nature", width=0.7, style=filled, fillcolor="#e0e0e0"]

    p1 [shape=plaintext, label="-$5"]
    p2 [shape=plaintext, label="-$5"]
    p3 [shape=plaintext, label="-$50"]
    p4 [shape=plaintext, label="$0"]

    D -> NB [label="  Bring umbrella", penwidth=2.5]
    D -> ND [label="  Don't bring"]

    NB -> p1 [label="  Rain (0.3)"]
    NB -> p2 [label="  No rain (0.7)"]

    ND -> p3 [label="  Rain (0.3)"]
    ND -> p4 [label="  No rain (0.7)"]
}

Figure 1: Decision tree without information. You choose first, then nature reveals the state.

Target: 1:06. “Decision first, then nature. You have to commit before you know.” “The formula says the same thing: \(\max\) is on the outside, \(\sum\) is on the inside.”

Solving the No-Information Tree

Work from the leaves back to the root:

“Bring” branch: \[E[\text{cost}] = 0.3 \times 5 + 0.7 \times 5 = 5\]

“Don’t bring” branch: \[E[\text{cost}] = 0.3 \times 50 + 0.7 \times 0 = 15\]

Best action: bring the umbrella. Expected cost \(= \$5.00\).

Target: 1:08. Walk through each branch on the board, pointing back to the tree. “The decision node picks the branch with the lower expected cost.”

Tree: Perfect Information

Nature reveals the state before you decide. The chance node comes first — this computes \(\sum_s \max_a U(a,s) \, p(s)\).

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    rankdir=LR
    node [fontname="Helvetica", fontsize=11]
    edge [fontname="Helvetica", fontsize=10]

    N [shape=circle, label="Nature", width=1.0, style=filled, fillcolor="#e0e0e0"]

    DR [shape=box, label="Decision", style=filled, fillcolor="#cde0f5"]
    DD [shape=box, label="Decision", style=filled, fillcolor="#cde0f5"]

    p1 [shape=plaintext, label="-$5"]
    p2 [shape=plaintext, label="-$50"]
    p3 [shape=plaintext, label="-$5"]
    p4 [shape=plaintext, label="$0"]

    N -> DR [label="  Rain (0.3)"]
    N -> DD [label="  No rain (0.7)"]

    DR -> p1 [label="  Bring", penwidth=2.5]
    DR -> p2 [label="  Don't", style=dashed, color="#999999"]

    DD -> p3 [label="  Bring", style=dashed, color="#999999"]
    DD -> p4 [label="  Don't", penwidth=2.5]
}

Figure 2: Decision tree with perfect information. Nature reveals the state, then you choose.

Target: 1:10. “Now nature goes first. If you know it’s raining, bring the umbrella. If you know it’s dry, leave it.”

Solving the Perfect-Information Tree

At each decision node, pick the cheapest action:

• Rain (prob 0.3): bring umbrella, cost \(= 5\) (vs. 50 without)
• No rain (prob 0.7): leave umbrella, cost \(= 0\) (vs. 5 with)

Average over nature:

\[E[\text{cost}] = 0.3 \times 5 + 0.7 \times 0 = \$1.50\]

Target: 1:12. “With perfect information, you never pay the loss AND you skip the umbrella hassle on dry days.”

EVPI From the Trees

The EVPI is the gap between the two trees:

\[\text{EVPI} = 5.00 - 1.50 = \$3.50\]

The only difference is the order of the square and the circle — which is the order of \(\max\) and \(\sum\) in the formula.

Scaling note: imagine drawing a tree for the dike problem. The decision node fans into infinitely many heights, each into branches for sea level rise, storm surge, discount rate… The integral \(\int \max_a U(a,s) \, p(s) \, ds\) is the compact version of an infinite tree.

Target: 1:15. Key point: “The only difference between these two trees is the order of the square and the circle. The tree is just a picture of the math.”

Scaling: “For the umbrella, the tree and the formula say the same thing. For hard problems, you need the formula.”

An Imperfect Forecast

No real forecast is perfect. A forecast’s accuracy is characterized by two conditional probabilities:

• Hit rate: \(P(\text{f. rain} \mid \text{rain}) = 0.90\)
• Correct rejection rate: \(P(\text{f. dry} \mid \text{dry}) = 0.80\)

These describe the forecast system, independent of the base rate.

To draw the imperfect-information tree, we need \(P(\text{f. rain})\) and \(P(\text{rain} \mid \text{f. rain})\) — those require Bayes’ rule.

Target: 1:18. “On Monday the oracle was always right. Now the oracle is your weather app — and sometimes it’s wrong.”

“We know the hit rate and the correct rejection rate. But to build the tree, we need different numbers. Bayes’ rule bridges the gap.”

Building the Tree: Signal Probabilities

First, the complementary likelihoods:

\[P(\text{f. dry} \mid \text{rain}) = 1 - 0.90 = 0.10, \qquad P(\text{f. rain} \mid \text{dry}) = 1 - 0.80 = 0.20\]

Then the law of total probability gives the signal probabilities:

\[ P(\text{f. rain}) = \underbrace{0.90 \times 0.30}_{0.27} + \underbrace{0.20 \times 0.70}_{0.14} = 0.41 \]

\[ P(\text{f. dry}) = 1 - 0.41 = 0.59 \]

These are the first branches of the imperfect-information tree.

Target: 1:22. Walk through each step on the board. “We start with two numbers (hit rate, correct rejection) and the base rate. Total probability gives us how often the forecast says rain vs. dry.”

The Incomplete Tree

We can draw the first two layers, but the nature nodes need posterior probabilities we haven’t computed yet.

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    node [fontname="Helvetica", fontsize=11]
    edge [fontname="Helvetica", fontsize=10]

    F [shape=circle, label="Signal", width=1.0, style=filled, fillcolor="#e0e0e0"]

    DR [shape=box, label="Decision", style=filled, fillcolor="#cde0f5"]
    DD [shape=box, label="Decision", style=filled, fillcolor="#cde0f5"]

    NRB [shape=circle, label="Nature", width=0.7, style=filled, fillcolor="#ffe0e0"]
    NRD [shape=circle, label="Nature", width=0.7, style=filled, fillcolor="#ffe0e0"]
    NDB [shape=circle, label="Nature", width=0.7, style=filled, fillcolor="#ffe0e0"]
    NDD [shape=circle, label="Nature", width=0.7, style=filled, fillcolor="#ffe0e0"]

    F -> DR [label="  Forecast rain (0.41)"]
    F -> DD [label="  Forecast dry (0.59)"]

    DR -> NRB [label="  Bring"]
    DR -> NRD [label="  Don't"]

    DD -> NDB [label="  Bring"]
    DD -> NDD [label="  Don't"]

    NRB -> QRB1 [label="  Rain (???)"]
    NRB -> QRB2 [label="  Dry (???)"]
    NRD -> QRD1 [label="  Rain (???)"]
    NRD -> QRD2 [label="  Dry (???)"]
    NDB -> QDB1 [label="  Rain (???)"]
    NDB -> QDB2 [label="  Dry (???)"]
    NDD -> QDD1 [label="  Rain (???)"]
    NDD -> QDD2 [label="  Dry (???)"]

    QRB1 [shape=plaintext, label="-$5"]
    QRB2 [shape=plaintext, label="-$5"]
    QRD1 [shape=plaintext, label="-$50"]
    QRD2 [shape=plaintext, label="$0"]
    QDB1 [shape=plaintext, label="-$5"]
    QDB2 [shape=plaintext, label="-$5"]
    QDD1 [shape=plaintext, label="-$50"]
    QDD2 [shape=plaintext, label="$0"]
}

Figure 3: Incomplete imperfect-information tree. The signal and decision layers are known, but the posterior probabilities at the nature nodes require Bayes’ rule.

Target: 1:24. “We know the signal probabilities and the payoffs. But to fill in the nature nodes, we need the posterior probabilities: what’s the chance of rain given the forecast? That’s Bayes’ rule.”

Point to the red-highlighted nature nodes.

Building the Tree: Posterior Probabilities

Apply Bayes’ rule to fill in the nature nodes:

\[ P(\text{rain} \mid \text{f. rain}) = \frac{P(\text{f. rain} \mid \text{rain}) \cdot p}{P(\text{f. rain})} = \frac{0.90 \times 0.30}{0.41} = \frac{0.27}{0.41} \approx 0.66 \]

\[ P(\text{rain} \mid \text{f. dry}) = \frac{P(\text{f. dry} \mid \text{rain}) \cdot p}{P(\text{f. dry})} = \frac{0.10 \times 0.30}{0.59} = \frac{0.03}{0.59} \approx 0.05 \]

The forecast shifts the probability from the base rate of 0.30 to 0.66 (forecast rain) or 0.05 (forecast dry).

Target: 1:28. “Bayes’ rule does the rest. 0.27 / 0.41 = 0.66.” “If the forecast says rain, the chance of rain jumps from 30% to 66%. If it says dry, the chance drops to 5%.”

Key normalization: the denominator P(f. rain) = 0.41 ensures the posteriors are proper probabilities.

The Complete Tree

Now we can fill in every number. Three layers: Signal (chance) \(\to\) Decision (you) \(\to\) Nature (chance).

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    edge [fontname="Helvetica", fontsize=10]

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    DR [shape=box, label="Decision", style=filled, fillcolor="#cde0f5"]
    DD [shape=box, label="Decision", style=filled, fillcolor="#cde0f5"]

    NRB [shape=circle, label="Nature", width=0.7, style=filled, fillcolor="#e0e0e0"]
    NRD [shape=circle, label="Nature", width=0.7, style=filled, fillcolor="#e0e0e0"]
    NDB [shape=circle, label="Nature", width=0.7, style=filled, fillcolor="#e0e0e0"]
    NDD [shape=circle, label="Nature", width=0.7, style=filled, fillcolor="#e0e0e0"]

    p1 [shape=plaintext, label="-$5"]
    p2 [shape=plaintext, label="-$5"]
    p3 [shape=plaintext, label="-$50"]
    p4 [shape=plaintext, label="$0"]
    p5 [shape=plaintext, label="-$5"]
    p6 [shape=plaintext, label="-$5"]
    p7 [shape=plaintext, label="-$50"]
    p8 [shape=plaintext, label="$0"]

    F -> DR [label="  Forecast rain (0.41)"]
    F -> DD [label="  Forecast dry (0.59)"]

    DR -> NRB [label="  Bring", penwidth=2.5]
    DR -> NRD [label="  Don't", style=dashed, color="#999999"]

    DD -> NDB [label="  Bring", style=dashed, color="#999999"]
    DD -> NDD [label="  Don't", penwidth=2.5]

    NRB -> p1 [label="  Rain (0.66)"]
    NRB -> p2 [label="  Dry (0.34)"]

    NRD -> p3 [label="  Rain (0.66)"]
    NRD -> p4 [label="  Dry (0.34)"]

    NDB -> p5 [label="  Rain (0.05)"]
    NDB -> p6 [label="  Dry (0.95)"]

    NDD -> p7 [label="  Rain (0.05)"]
    NDD -> p8 [label="  Dry (0.95)"]
}

Figure 4: Decision tree with an imperfect forecast. Posterior probabilities from Bayes’ rule replace the prior at the nature nodes.

Target: 1:30. “Compare the three trees. Without info: Decision → Nature. Perfect info: Nature → Decision. Imperfect info: Signal → Decision → Nature.”

“The signal doesn’t tell you the state directly — it shifts your probability via Bayes’ rule. That’s the whole story of EVII in one picture.”

Note: the bold/dashed branches already show the optimal actions — we’ll verify that next.

Solving the Tree: Optimal Actions

At each decision node, apply Monday’s cost-loss rule: bring the umbrella when \(P(\text{rain}) \times L > C\).

• Forecast rain (\(P(\text{rain}) = 0.66\)): \(0.66 \times 50 = 33 > 5\) \(\Rightarrow\) bring umbrella
• Forecast dry (\(P(\text{rain}) = 0.05\)): \(0.05 \times 50 = 2.50 < 5\) \(\Rightarrow\) leave umbrella

The bold branches in the tree confirm the optimal action at each decision node.

Target: 1:33. “The cost-loss rule from Monday works at every decision node — just use the posterior probability instead of the prior.”

“When the forecast says rain, expected loss without protection is $33 — much more than $5, so bring it. When the forecast says dry, expected loss is only $2.50 — not worth the $5 hassle.”

Solving the Tree: Expected Cost

Roll back to the signal node. The expected cost along each optimal branch:

• Forecast rain (prob 0.41): bring umbrella \(\to\) cost \(= \$5.00\)
• Forecast dry (prob 0.59): leave umbrella \(\to\) expected cost \(= 0.05 \times 50 + 0.95 \times 0 = \$2.50\)

Average over the signal:

\[ E[\text{cost with forecast}] = 0.41 \times 5.00 + 0.59 \times 2.50 = 2.05 + 1.475 = \$3.55 \]

Target: 1:36. “For each signal, we’ve already solved the optimal action. Now we average across the two signals, weighting by how often each signal occurs.”

Note for precision: with exact (unrounded) posteriors, this equals exactly $3.55 because \(0.41 \times 5 + 0.03 \times 50 = 2.05 + 1.50 = 3.55\).

EVII

Scenario	Expected cost
No information	\(\$5.00\)
Imperfect forecast	\(\$3.55\)
Perfect information	\(\$1.50\)

\[ \text{EVII} = 5.00 - 3.55 = \$1.45 \]

\[ \text{EVPI} = 5.00 - 1.50 = \$3.50 \]

The forecast captures \(1.45 / 3.50 \approx 41\%\) of the perfect-information value.

Target: 1:39. “The forecast isn’t perfect, but it’s still worth $1.45 — about 41% of what a perfect oracle would give.”

“The dangerous branch is forecast dry + actually rains: you get soaked. But that only happens about 3% of the time (0.10 × 0.30).”

Ask: “What would make the forecast more or less valuable?” Key answers: better hit/rejection rates increase EVII; if the base rate is far from \(C/L\), the decision is already obvious and the forecast adds little.

The General Formula

The umbrella illustrates a general three-step pattern:

1. Enumerate signals: each possible signal \(z\) has probability \(p(z)\)
2. Update and optimize: for each signal, compute the posterior \(p(s \mid z)\) via Bayes’ rule, then choose the best action
3. Average over signals

\[ \text{EVII} = \underbrace{\int \left[ \max_a \int U(a, s) \, p(s \mid z) \, ds \right] p(z) \, dz}_{\text{get signal, update, optimize, average}} \;-\; \underbrace{\max_a \int U(a, s) \, p(s) \, ds}_{\text{optimize with prior alone}} \]

This reduces to EVPI when the signal perfectly reveals \(s\), and to zero when the signal is pure noise (\(p(s \mid z) = p(s)\)).

Target: 1:42. Walk through the three steps verbally before showing the formula. “EVPI is the ceiling — you proved that with the umbrella numbers.”

On scaling: “The decision tree for EVII has three layers: signal, decision, nature. For the umbrella, that’s manageable — 2 signals × 2 actions × 2 states = 8 leaves. For the dike problem with a continuous forecast, you’d need to integrate over all possible signals. The formula handles this naturally. The tree would be infinite.”

Blackwell’s Theorem

A key result (Blackwell, 1953): \(0 \leq \text{EVII} \leq \text{EVPI}\).

An imperfect forecast can never hurt in expectation.

The relationship between forecast quality and value is nonlinear. Forecasts are most valuable near the decision boundary (\(C/L \approx p\)). When the decision is already obvious, even a much better forecast adds little value.

Target: 1:45. “Forecasts are most valuable to decision-makers who are on the fence.” Draw a number line with \(C/L\) and \(p\) marked. When they’re close, a nudge from the forecast flips the decision. When they’re far apart, no forecast can change the outcome. “Spending money to improve a forecast is only worthwhile if the forecast actually changes decisions.”

Summary

1. Decision trees visualize the order of \(\max\) and \(\sum\) in expected value calculations. The EVPI is the gap between “nature first” and “decision first” trees.
2. EVII extends this to noisy signals: Signal \(\to\) Decision \(\to\) Nature. Bayes’ rule computes posteriors; Blackwell’s theorem guarantees \(0 \leq \text{EVII} \leq \text{EVPI}\).
3. For continuous or high-dimensional problems, trees become intractable — the integral formulation is the compact representation.

Target: 1:48. “Three things to remember: trees picture the formula, EVII adds a signal layer, and integrals replace trees when the problem gets big.”

References

Blackwell, D. (1953). Equivalent comparisons of experiments. The Annals of Mathematical Statistics, 24(2), 265–272. https://doi.org/10.1214/aoms/1177729032