The Central Idea
Suppose \(H\) is a hypothesis and evidence arrives in stages:
\[ E_1,E_2,\ldots,E_n. \]The first update produces
\[ P(H\mid E_1). \]That posterior then becomes the starting probability for the next update:
\[ P(H\mid E_1,E_2). \]Continuing this process gives
\[ \boxed{ P(H) \longrightarrow P(H\mid E_1) \longrightarrow P(H\mid E_1,E_2) \longrightarrow \cdots }. \]Posterior Becomes the Next Prior
After observing \(E_1\), the posterior is
\[ P(H\mid E_1). \]When \(E_2\) later arrives, this updated probability plays the role of the prior for the next stage.
Conceptually,
\[ \boxed{ \text{new prior} = \text{previous posterior} }. \]One-Step Bayes Update
For one piece of evidence \(E\),
\[ P(H\mid E) = \frac{P(E\mid H)P(H)} {P(E)}. \]In a two-hypothesis setting with \(H\) and \(H^c\),
\[ \boxed{ P(H\mid E) = \frac{P(H)P(E\mid H)} {P(H)P(E\mid H)+P(H^c)P(E\mid H^c)} }. \]Two Sequential Updates
Suppose \(E_1\) is observed first.
Then
\[ P(H\mid E_1) = \frac{P(E_1\mid H)P(H)} {P(E_1)}. \]Next, after observing \(E_2\),
\[ \boxed{ P(H\mid E_1,E_2) = \frac{ P(E_2\mid H,E_1)P(H\mid E_1) }{ P(E_2\mid E_1) } }. \]The Second Update Uses Conditional Evidence Probabilities
The likelihood at the second stage is generally
\[ P(E_2\mid H,E_1), \]not automatically
\[ P(E_2\mid H). \]The first piece of evidence may change how likely the second piece is.
Only additional independence assumptions allow this expression to simplify.
Conditional Independence of Evidence
If \(E_1\) and \(E_2\) are conditionally independent given \(H\), then
\[ P(E_2\mid H,E_1)=P(E_2\mid H). \]Similarly, if they are conditionally independent given \(H^c\),
\[ P(E_2\mid H^c,E_1)=P(E_2\mid H^c). \]Under these assumptions, sequential updating becomes much simpler.
Conditional Independence Is Not Ordinary Independence
Evidence variables can be dependent overall while becoming independent once the underlying hypothesis is known.
Thus,
\[ E_1\not\!\perp E_2 \]can coexist with
\[ E_1\perp E_2\mid H. \]The distinction matters in repeated Bayesian updates.
A First Sequential Example
Suppose
\[ P(H)=0.20. \]Evidence \(E_1\) has
\[ P(E_1\mid H)=0.80 \]and
\[ P(E_1\mid H^c)=0.20. \]Perform the First Update
Since
\[ P(H^c)=0.80, \]total probability gives
\[ P(E_1) = 0.20(0.80)+0.80(0.20). \]Therefore,
\[ P(E_1)=0.16+0.16=0.32. \]Bayes' theorem gives
\[ P(H\mid E_1) = \frac{0.20(0.80)}{0.32} = \boxed{0.50}. \]The First Posterior Becomes the Second Prior
Before observing \(E_2\), the updated starting probability is now
\[ \boxed{ P(H\mid E_1)=0.50 }. \]The complement is also updated:
\[ P(H^c\mid E_1)=0.50. \]Introduce the Second Piece of Evidence
Suppose \(E_2\) is conditionally independent of \(E_1\) given each hypothesis and
\[ P(E_2\mid H)=0.90, \] \[ P(E_2\mid H^c)=0.30. \]The second update begins from the new prior \(0.50\).
Perform the Second Update
The evidence probability relative to the updated state is
\[ P(E_2\mid E_1) = 0.50(0.90)+0.50(0.30). \]Thus,
\[ P(E_2\mid E_1)=0.45+0.15=0.60. \]Therefore,
\[ P(H\mid E_1,E_2) = \frac{0.50(0.90)}{0.60} = \boxed{0.75}. \]The Probability Path
The sequential update is
\[ \boxed{ 0.20 \longrightarrow 0.50 \longrightarrow 0.75 }. \]Each new piece of evidence changes the probability relative to the information already accumulated.
Batch Updating
Instead of updating one piece of evidence at a time, we can sometimes combine all evidence at once.
For two pieces,
\[ P(H\mid E_1,E_2) = \frac{ P(E_1,E_2\mid H)P(H) }{ P(E_1,E_2) }. \]This is a batch update.
Sequential and Batch Updating Should Agree
If the same joint probability model is used consistently, updating sequentially and updating all evidence at once produce the same final posterior.
Symbolically,
\[ \boxed{ P(H\mid E_1,E_2) } \]is the same target probability regardless of whether it is computed in one step or several mathematically consistent steps.
Verify the Example by Batch Updating
Under conditional independence,
\[ P(E_1,E_2\mid H) = P(E_1\mid H)P(E_2\mid H). \]Therefore,
\[ P(E_1,E_2\mid H) = 0.80(0.90) = 0.72. \]Similarly,
\[ P(E_1,E_2\mid H^c) = 0.20(0.30) = 0.06. \]Complete the Batch Calculation
Bayes' theorem gives
\[ P(H\mid E_1,E_2) = \frac{0.20(0.72)} {0.20(0.72)+0.80(0.06)}. \]Therefore,
\[ P(H\mid E_1,E_2) = \frac{0.144}{0.144+0.048} = \frac{0.144}{0.192}. \]Thus,
\[ \boxed{ P(H\mid E_1,E_2)=0.75 }. \]This matches the sequential calculation.
Odds Form of Sequential Updating
Sequential updating becomes especially compact in odds form.
If the prior odds are
\[ O_0, \]and successive evidence pieces have likelihood ratios
\[ LR_1,LR_2,\ldots,LR_n, \]then under the appropriate factorization assumptions,
\[ \boxed{ O_n = O_0 \prod_{i=1}^{n}LR_i }. \]First Odds Update
In the previous example,
\[ P(H)=0.20. \]The prior odds are
\[ \frac{0.20}{0.80} = \frac14. \]The first likelihood ratio is
\[ LR_1 = \frac{0.80}{0.20} = 4. \]Therefore,
\[ O_1=\frac14(4)=1. \]Odds \(1:1\) correspond to probability \(0.50\).
Second Odds Update
The second likelihood ratio is
\[ LR_2 = \frac{0.90}{0.30} = 3. \]Starting from odds
\[ 1:1, \]the second update gives
\[ O_2=1(3)=3. \]Odds \(3:1\) correspond to
\[ \boxed{ P(H\mid E_1,E_2)=\frac34=0.75 }. \]Why Odds Form Is Useful
In probability form, each update requires normalization.
In odds form, each new likelihood ratio simply multiplies the current odds.
This is particularly convenient when many pieces of evidence arrive in sequence.
Sequential Evidence Can Support or Oppose a Hypothesis
One piece of evidence may increase the posterior while another decreases it.
For example,
\[ LR_1>1 \]favors \(H\), while
\[ LR_2<1 \]favors \(H^c\).
The final odds reflect the product of all evidence contributions.
Example: Conflicting Evidence
Suppose prior odds are
\[ 1:1. \]The first piece of evidence has
\[ LR_1=5, \]while the second has
\[ LR_2=0.20. \]Then
\[ O_2 = 1(5)(0.20) = 1. \]The final odds return to
\[ \boxed{1:1}. \]Conflicting Evidence Can Cancel
In the previous example,
\[ 5(0.20)=1. \]The second evidence item exactly reverses the odds multiplier created by the first.
The final posterior therefore equals the original prior.
Uninformative Evidence Leaves the Posterior Unchanged
If a new observation has likelihood ratio
\[ LR=1, \]then
\[ O_{\text{new}} = O_{\text{old}}. \]The new evidence does not distinguish the competing hypotheses.
Repeated Strong Evidence Can Accumulate Quickly
Suppose prior odds are
\[ 1:9. \]Three conditionally independent evidence items each have likelihood ratio
\[ 3. \]Then
\[ O_3 = \frac19(3^3) = \frac{27}{9} = 3. \]The final odds are
\[ \boxed{3:1}. \]Convert the Accumulated Odds
Odds of
\[ 3:1 \]correspond to probability
\[ \frac{3}{3+1}. \]Therefore,
\[ \boxed{ P(H\mid E_1,E_2,E_3)=0.75 }. \]Repeated evidence can overcome a low starting probability when the likelihood ratios consistently favor the hypothesis.
Dependence Between Evidence Items
The simple product
\[ LR_1LR_2 \]can be invalid when the evidence items contain overlapping information.
If \(E_2\) depends on \(E_1\) even after conditioning on the hypothesis, the correct second-stage likelihood ratio is
\[ \boxed{ \frac{P(E_2\mid H,E_1)} {P(E_2\mid H^c,E_1)} }. \]Do Not Double Count Correlated Evidence
Suppose two reports are based on the same underlying source.
Treating them as independent may count essentially the same information twice.
This can produce a posterior that is too extreme.
Example: Repeated Measurement with Shared Error
Suppose a sensor has a persistent calibration error.
Two readings from that sensor may be strongly correlated even after the system state is fixed.
Therefore, the second reading may provide less new information than an independent second sensor would provide.
Independent Evidence Is New Information
When evidence items are conditionally independent given the underlying hypothesis, each contributes a distinct multiplicative likelihood factor.
This is the cleanest setting for repeated Bayesian updating.
Sequential Updating with More Than Two Hypotheses
Suppose
\[ H_1,H_2,\ldots,H_k \]form a complete set of competing hypotheses.
After observing \(E_1\),
\[ P(H_i\mid E_1) \]becomes the updated probability for each hypothesis.
Second Update with Several Hypotheses
After observing \(E_2\), update each hypothesis:
\[ \boxed{ P(H_i\mid E_1,E_2) = \frac{ P(E_2\mid H_i,E_1)P(H_i\mid E_1) }{ \sum_j P(E_2\mid H_j,E_1)P(H_j\mid E_1) } }. \]Each previous posterior acts as the next prior weight.
Example: Three Hypotheses
Suppose the initial priors are
\[ P(H_1)=0.50, \qquad P(H_2)=0.30, \qquad P(H_3)=0.20. \]Evidence \(E_1\) has likelihoods
\[ 0.20,\qquad0.40,\qquad0.80. \]First Three-Hypothesis Update
The unnormalized weights are
\[ 0.50(0.20)=0.10, \] \[ 0.30(0.40)=0.12, \] \[ 0.20(0.80)=0.16. \]Their sum is
\[ 0.38. \]Therefore,
\[ P(H_1\mid E_1)=\frac{0.10}{0.38}\approx0.2632, \] \[ P(H_2\mid E_1)=\frac{0.12}{0.38}\approx0.3158, \] \[ P(H_3\mid E_1)=\frac{0.16}{0.38}\approx0.4211. \]Second Evidence Item
Suppose new evidence \(E_2\) has conditional likelihoods
\[ P(E_2\mid H_1)=0.90, \] \[ P(E_2\mid H_2)=0.50, \] \[ P(E_2\mid H_3)=0.20. \]Assume the model justifies using these likelihoods at the second stage.
Use the First Posterior as the New Prior
Multiply the updated probabilities by the second-stage likelihoods:
\[ 0.2632(0.90)\approx0.2369, \] \[ 0.3158(0.50)\approx0.1579, \] \[ 0.4211(0.20)\approx0.0842. \]The new normalization total is approximately
\[ 0.4790. \]Normalize the Second Update
Therefore,
\[ P(H_1\mid E_1,E_2) \approx \frac{0.2369}{0.4790} \approx \boxed{0.4946}, \] \[ P(H_2\mid E_1,E_2) \approx \boxed{0.3296}, \] \[ P(H_3\mid E_1,E_2) \approx \boxed{0.1758}. \]The second evidence item reverses the ranking created by the first.
Sequential Updating Can Change Rankings Repeatedly
A hypothesis with the largest prior need not have the largest posterior.
A later evidence item can also reverse an earlier posterior ranking.
Bayesian updating is dynamic because each stage responds to the evidence accumulated so far.
Updating from Natural Frequencies
Sequential reasoning can also be represented with counts.
Begin with a hypothetical population, apply the first evidence filter, then treat the surviving cases as the population for the next stage.
Example: First Filter
Suppose \(10\%\) of \(10{,}000\) cases satisfy \(H\).
Then:
\[ 1{,}000 \]satisfy \(H\), and
\[ 9{,}000 \]satisfy \(H^c\).
Suppose \(E_1\) occurs in \(80\%\) of \(H\) cases and \(20\%\) of \(H^c\) cases.
Count the Cases after the First Filter
From \(H\):
\[ 1000(0.80)=800. \]From \(H^c\):
\[ 9000(0.20)=1800. \]So among the
\[ 2600 \]cases satisfying \(E_1\),
\[ \frac{800}{2600} \approx \boxed{0.3077} \]satisfy \(H\).
Apply a Second Filter
Suppose among the remaining \(H\) cases, \(E_2\) occurs with probability
\[ 0.90, \]while among the remaining \(H^c\) cases it occurs with probability
\[ 0.30. \]Then:
\[ 800(0.90)=720 \]\(H\)-cases survive, and
\[ 1800(0.30)=540 \]\(H^c\)-cases survive.
Find the Final Frequency-Based Posterior
After both filters,
\[ 720+540=1260 \]cases remain.
Therefore,
\[ P(H\mid E_1,E_2) = \frac{720}{1260} = \boxed{\frac47} \approx \boxed{0.5714}. \]Natural frequencies make the sequential elimination process visible.
Sequential Updating in Bayesian Networks
In a Bayesian network, new observations can be incorporated one at a time.
Each observed node changes the posterior distribution over unobserved variables.
A later observation updates the already-conditioned network further.
Example: Hidden State with Two Sensors
Suppose
\[ H\longrightarrow S_1, \qquad H\longrightarrow S_2. \]If \(S_1\) and \(S_2\) are conditionally independent given \(H\), then the joint sensor likelihood under \(H\) factors:
\[ P(S_1,S_2\mid H) = P(S_1\mid H)P(S_2\mid H). \]First Sensor Update
Observe \(S_1=s_1\).
The prior
\[ P(H) \]becomes
\[ P(H\mid S_1=s_1). \]Second Sensor Update
Then observe
\[ S_2=s_2. \]The updated posterior becomes
\[ P(H\mid S_1=s_1,S_2=s_2). \]This is sequential Bayesian-network inference.
Order of Evidence under a Consistent Model
Conditioning on the same final evidence set should produce the same final posterior regardless of the order in which the evidence is processed:
\[ \boxed{ P(H\mid E_1,E_2) = P(H\mid E_2,E_1) }. \]The intermediate posteriors may differ, but the final conditioning event is the same.
Why Order Can Appear to Matter in Practice
If different approximations, changing models, or inconsistent likelihoods are used at different stages, sequential calculations can produce different answers.
Such differences arise from the modeling procedure, not from the basic commutativity of conditioning on the same final evidence set.
Updating with Negative Evidence
Evidence can be informative because something happened or because something failed to happen.
If \(E\) is not observed, the relevant likelihood is
\[ P(E^c\mid H). \]Since
\[ P(E^c\mid H)=1-P(E\mid H), \]absence of expected evidence can lower the posterior probability of a hypothesis.
Example: Missing Expected Evidence
Suppose
\[ P(E\mid H)=0.90 \]and
\[ P(E\mid H^c)=0.40. \]If \(E\) is not observed, then
\[ P(E^c\mid H)=0.10 \]and
\[ P(E^c\mid H^c)=0.60. \]The likelihood ratio for the absence is
\[ \frac{0.10}{0.60} = \boxed{\frac16}. \]The missing evidence strongly lowers the odds for \(H\).
Evidence Quality Matters
Sequential updating does not automatically make a model more accurate.
If new evidence is noisy, biased, poorly measured, or modeled incorrectly, the posterior can move in the wrong direction.
Bayesian updating is only as reliable as its probability inputs.
Model Drift
In long-running applications, the data-generating process may change over time.
Then likelihoods estimated from older conditions may no longer describe new evidence accurately.
Sequential updating should therefore be paired with attention to whether the underlying model remains appropriate.
Sequential Updating Does Not Mean Infinite Certainty
Repeated evidence can push posterior probabilities toward \(0\) or \(1\), but extreme confidence is justified only when the evidence model supports it.
Correlated evidence, model uncertainty, and measurement error can limit how much confidence repeated observations should generate.
Common Error: Reusing the Original Prior
After the first update, the next stage should begin with
\[ P(H\mid E_1), \]not the original
\[ P(H). \]Ignoring the previous posterior discards information already learned.
Common Error: Multiplying Raw Probabilities Instead of Likelihood Ratios
In odds form, the update is
\[ O_{\text{new}} = O_{\text{old}}LR. \]The likelihood ratio is
\[ \frac{P(E\mid H)} {P(E\mid H^c)}, \]not simply \(P(E\mid H)\).
Common Error: Assuming Every Evidence Item Is Independent
Repeated observations can be correlated.
Multiplying independent-style likelihood ratios without justification can overcount information.
Common Error: Ignoring Evidence Dependence after Conditioning
Ordinary independence is not enough.
The important simplification for sequential Bayes updates is often conditional independence given each competing hypothesis.
Common Error: Forgetting to Normalize in Probability Form
After multiplying a current prior by a likelihood, the result is an unnormalized weight.
It must be divided by the total evidence probability to become a posterior probability.
Common Error: Treating Intermediate Probabilities as Final
A posterior after \(E_1\) is final only relative to \(E_1\).
Once \(E_2\) arrives, the relevant posterior becomes
\[ P(H\mid E_1,E_2). \]Common Error: Ignoring Evidence against a Hypothesis
Sequential updating must incorporate evidence whether it supports or opposes the hypothesis.
Selectively including only favorable evidence produces a distorted posterior.
Common Error: Assuming More Evidence Always Raises the Probability
New evidence may increase, decrease, or leave unchanged the posterior probability.
The direction depends on the likelihood comparison under the competing hypotheses.
A Reliable Probability-Form Procedure
- Start with the current prior probabilities.
- Identify the new evidence.
- Write the likelihood of that evidence under each hypothesis.
- Multiply each current prior by its corresponding likelihood.
- Add the resulting weights to form the normalization constant.
- Divide each weight by the total to obtain the posterior.
- Use that posterior as the prior for the next evidence item.
- Repeat as evidence accumulates.
- Check dependence assumptions before simplifying likelihoods.
A Reliable Odds-Form Procedure
- Convert the current probability to odds.
- Compute the likelihood ratio for the new evidence.
- Multiply the current odds by the likelihood ratio.
- Treat the result as the new odds.
- Repeat for later evidence items when the model justifies the factorization.
- Convert final odds back to probability if desired.
Example: Full Probability-Form Workflow
Suppose
\[ P(H)=0.30. \]Evidence \(E_1\) satisfies
\[ P(E_1\mid H)=0.70, \qquad P(E_1\mid H^c)=0.20. \]First,
\[ P(E_1) = 0.30(0.70)+0.70(0.20) = 0.21+0.14 = 0.35. \]Therefore,
\[ P(H\mid E_1) = \frac{0.21}{0.35} = \boxed{0.60}. \]Continue the Full Workflow
Suppose new evidence \(E_2\) satisfies
\[ P(E_2\mid H,E_1)=0.80 \]and
\[ P(E_2\mid H^c,E_1)=0.40. \]The updated complement probability is
\[ P(H^c\mid E_1)=0.40. \]Therefore,
\[ P(E_2\mid E_1) = 0.60(0.80)+0.40(0.40) = 0.64. \]Thus,
\[ P(H\mid E_1,E_2) = \frac{0.60(0.80)}{0.64} = \boxed{0.75}. \]Example: Full Odds-Form Workflow
Suppose the prior probability is
\[ 0.20. \]Prior odds are
\[ 1:4. \]Suppose three evidence items have likelihood ratios
\[ 2,\qquad3,\qquad0.50. \]Then final odds are
\[ \frac14(2)(3)(0.50) = \frac34. \]Therefore, final odds are
\[ 3:4. \]The final probability is
\[ \boxed{ \frac{3}{7}\approx0.4286 }. \]Practice 1: Posterior Becomes Prior
A hypothesis has prior probability \(0.25\). After evidence \(E_1\), its posterior becomes \(0.60\).
What probability should be used as the starting prior for the next update?
Practice 2: First Sequential Update
Suppose
\[ P(H)=0.40, \] \[ P(E_1\mid H)=0.75, \] \[ P(E_1\mid H^c)=0.25. \]Find
\[ P(H\mid E_1). \]Practice 3: Second Sequential Update
Continue Practice 2. Suppose
\[ P(E_2\mid H,E_1)=0.80 \]and
\[ P(E_2\mid H^c,E_1)=0.40. \]Find
\[ P(H\mid E_1,E_2). \]Practice 4: Odds Update
Prior odds are
\[ 1:3. \]Two conditionally independent evidence items have likelihood ratios
\[ 4 \]and
\[ 2. \]Find the final posterior odds.
Practice 5: Convert Final Odds
Convert the final odds from Practice 4 into a probability.
Practice 6: Conflicting Evidence
Prior odds are
\[ 2:1. \]The first likelihood ratio is
\[ 3 \]and the second is
\[ \frac16. \]Find the final posterior odds.
Practice 7: Uninformative Evidence
Current odds are
\[ 5:2. \]New evidence has likelihood ratio
\[ 1. \]Find the updated odds.
Practice 8: Dependence Warning
Two pieces of evidence are generated by the same faulty sensor.
Why might multiplying separate likelihood ratios as though the observations were conditionally independent be inappropriate?
Practice 9: Negative Evidence
Suppose
\[ P(E\mid H)=0.90 \]and
\[ P(E\mid H^c)=0.30. \]Find the likelihood ratio associated with observing \(E^c\).
Practice 10: Identify the Error
A student starts with
\[ P(H)=0.20. \]After the first update, the posterior is
\[ 0.55. \]For the second update, the student again uses \(0.20\) as the prior.
Explain the error.
Practice Answers
Practice 1
The posterior from the first stage becomes the prior for the next:
\[ \boxed{0.60}. \]Practice 2
Since
\[ P(H^c)=0.60, \]we have
\[ P(E_1) = 0.40(0.75)+0.60(0.25). \]Therefore,
\[ P(E_1)=0.30+0.15=0.45. \]Thus,
\[ P(H\mid E_1) = \frac{0.30}{0.45} = \boxed{\frac23} \approx \boxed{0.6667}. \]Practice 3
After Practice 2,
\[ P(H\mid E_1)=\frac23, \qquad P(H^c\mid E_1)=\frac13. \]Therefore,
\[ P(E_2\mid E_1) = \frac23(0.80)+\frac13(0.40). \]This equals
\[ \frac{8}{15}+\frac{2}{15} = \frac{10}{15} = \frac23. \]Hence,
\[ P(H\mid E_1,E_2) = \frac{\frac23(0.80)}{\frac23} = \boxed{0.80}. \]Practice 4
The prior odds value is
\[ \frac13. \]Therefore,
\[ \frac13(4)(2)=\frac83. \]The final posterior odds are
\[ \boxed{8:3}. \]Practice 5
Odds of
\[ 8:3 \]correspond to probability
\[ \frac{8}{8+3} = \boxed{\frac{8}{11}} \approx \boxed{0.7273}. \]Practice 6
The odds value is initially \(2\).
Update:
\[ 2(3)\left(\frac16\right)=1. \]Therefore, the final posterior odds are
\[ \boxed{1:1}. \]Practice 7
A likelihood ratio of \(1\) leaves the odds unchanged:
\[ \boxed{5:2}. \]Practice 8
The two readings may share the same source of error, so they may remain dependent even after the underlying state is fixed.
Treating them as conditionally independent could count the same information twice and exaggerate the posterior update.
Practice 9
First,
\[ P(E^c\mid H)=0.10 \]and
\[ P(E^c\mid H^c)=0.70. \]Therefore,
\[ LR_{E^c} = \frac{0.10}{0.70} = \boxed{\frac17} \approx \boxed{0.1429}. \]Practice 10
The first posterior already incorporates \(E_1\).
The second update must begin from
\[ \boxed{0.55}, \]not the original \(0.20\).
Reusing the original prior discards the information gained from the first evidence item.
What to Remember
Sequential Bayesian updating follows the pattern
\[ \boxed{ \text{prior} \longrightarrow \text{posterior} \longrightarrow \text{new prior} \longrightarrow \text{new posterior} }. \]In probability form,
\[ \boxed{ P(H\mid E_1,E_2) = \frac{ P(E_2\mid H,E_1)P(H\mid E_1) }{ P(E_2\mid E_1) } }. \]In odds form, under appropriate likelihood-ratio factorization,
\[ \boxed{ O_n = O_0 \prod_{i=1}^{n}LR_i }. \]- Each posterior becomes the prior for the next update.
- Sequential and batch updating agree when based on the same joint model.
- Evidence can increase, decrease, or leave unchanged a posterior probability.
- Likelihood ratios make repeated updates especially compact.
- Conditional independence must be justified before likelihood factors are multiplied separately.
- Correlated evidence can cause double counting if treated as independent.
- Negative evidence can be informative through complement likelihoods.
- For the same final evidence set, consistent Bayesian conditioning gives the same final posterior regardless of processing order.
- Repeated updating remains dependent on the quality and stability of the underlying probability model.
The central habit is: carry forward everything already learned, then measure only what the new evidence adds.
Where We Go Next
Sequential updating completes the progression from conditional probability through dependence, total probability, Bayes' theorem, likelihoods, testing, conditional expectation, dependency diagrams, and Bayesian networks.
Together, these tools form a general framework for reasoning under uncertainty when information arrives in stages and probabilities must be revised consistently.