Question
A delivery coordinator initially receives a broad bulletin saying that a route is amber. On an amber day, she believes a delivery will arrive on time with probability 1/2. A colleague proposes adding one of several more detailed messages and insists, “Whichever detail we get, the probability of an on-time delivery will go up.”
Assume the detailed messages form a partition of the amber days: every amber day produces exactly one message, and the messages considered all have positive conditional probability. Can the colleague be right while remaining probabilistically coherent? Prove your answer. To make the point concrete, consider 20 equally likely amber-day scenarios: 8 produce message North (6 on time, 2 late), while 12 produce message South (4 on time, 8 late). Calculate the probabilities after North and South and reconcile them with the original 1/2.
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Hint 1
Use the law of total probability within the amber group, conditioning on which detailed message arrives.
Check your answer
Solution
Let A be the event that delivery is on time, C the observed coarse bulletin “amber”, and \(M_{1},\ldots,M_{r}\) the mutually exclusive, collectively exhaustive detailed messages that can occur given C. Set \(w_{j}=P(M_{j}\mid C)>0\). Conditional total probability gives the coherence identity
\(P(A\mid C)=\sum_{j=1}^{r}w_jP(A\mid C,M_j)\), with \(\sum_{j}w_{j}=1\).
If all detailed posteriors were strictly larger than the original \(p=P(A\mid C)\), their weighted average would also be strictly greater than p. That would contradict the identity. More generally, if every posterior were at least p and one were strictly greater, the average would exceed p; therefore a genuine upward revision on some positive-probability branches must be offset by downward revisions on others (unless no branch changes probability).
In the proposed 20-scenario refinement, North has 8 cases with 6 successes, so \(P(A\mid C,\text{North})=6/8=\)\(3/4\). South has 12 cases with 4 successes, so \(P(A\mid C,\text{South})=4/12=\)\(1/3\). Their conditional message probabilities are \(P(\text{North}\mid C)=8/20=2/5\) and \(P(\text{South}\mid C)=12/20=3/5\). The weighted posterior average is
\((2/5)(3/4)+(3/5)(1/3)=3/10+1/5=\)\(1/2\).
The unconditional-on-detail amber probability is also \((6+4)/20=10/20=1/2\), as required. More information can raise the probability after one particular message, but it cannot guarantee a rise after every possible positive-probability message.
Assumptions
The broad bulletin amber occurs with positive probability and is already observed.
Detailed messages partition amber cases and all mentioned branches have positive conditional probabilities.
The 20 amber scenarios in the example are equally likely.
Final Answer
No: a positive-probability refinement cannot make every posterior strictly greater than the coarse probability. In the example, \(P(\text{on time}\mid\text{North},\text{amber})=3/4\), \(P(\text{on time}\mid\text{South},\text{amber})=1/3\), and their \(8/20\) and \(12/20\) weighted average is \(1/2\).