T.8 · Conditioning, Bayesian inference and sampling error

For P(B)>0, P(A|B)=P(A∩B)/P(B): restrict the population to B before computing the fraction in A. Independence means P(A∩B)=P(A)P(B), equivalently P(A|B)=P(A) when the denominator is nonzero. Mutually exclusive events with positive probabilities cannot be independent, because their joint probability is zero. Sampling without replacement usually changes later probabilities; a fixed denominator for successive draws would describe a different experiment.

Build joint probabilities by multiplying a prior category proportion by the relevant conditional probability. If D is the condition and + a positive result, P(D∩+)=P(D)P(+|D). The total positive probability is P(D)P(+|D)+P(not D)P(+|not D), because these are disjoint and cover every positive result. Bayes divides the first joint probability by this total. Sensitivity is P(+|D), while specificity is P(−|not D); the false-positive rate is one minus specificity.

For a hypothetical test with prevalence 2%, sensitivity 90% and specificity 95%, use a population of 10,000 for clarity. Of 200 people with the condition, 180 test positive. Of 9,800 without it, 490 test positive. Therefore 180 of 670 positive results have the condition: posterior 18/67, about 26.9%. Reversing the conditional would give 90%, answering a different question. This is a mathematical model, not a recommendation about clinical decisions.

For independent identically distributed observations with finite variance σ², the sample mean has expectation μ and variance σ²/n; its standard error is σ/√n. This is spread of repeated sample means, not the spread of individual observations. Quadrupling n halves the standard error, rather than quartering it. Normal population data give an exactly normal mean; otherwise a central-limit approximation needs adequate conditions and sample size. Correlation invalidates the simple independent variance calculation.

Let P(A)=0.4, P(B)=0.5 and P(A∩B)=0.3. Then P(A|B)=0.6, so A and B are not independent, since 0.3≠0.4·0.5. If a variable has standard deviation 12, an independent sample of size 36 has mean standard error 2. Increasing the sample size to 144 gives 1; the individual-observation standard deviation stays 12.

Do not reverse P(+|D) into P(D|+). Include false positives in the posterior denominator. Standard deviation of individual observations and standard error of their mean have different sample-size behaviour.

In a stated model, 10% of devices have a defect. A test is positive for 80% of defective and 5% of nondefective devices. Find the probability of a defect given a positive result using 1000 expected devices.

Worked solution

Expected counts are 100 defective and 900 nondefective. Positive counts are $100(0.8)=80$ and $900(0.05)=45$, respectively. The conditioned group contains 125 positives, hence

Negative counts 20 and 855 check both row totals. The sensitivity 0.8 is $P(+\mid D)$, not the requested reversed condition. These are illustrative model rates, not measured performance claims.

Independent observations have variance 25. Find the standard error of their mean for n=100. If instead every distinct pair has covariance 1, find the mean's variance. Does quadrupling n necessarily halve the standard error in that second model?

Check after attempting

Independence gives $\operatorname{Var}(\overline X)=25/n$ and at 100 standard error $5/10=0.5$. With covariance one,

At 100 it is 1.24. The standard error tends to one as n grows, rather than to zero; quadrupling n need not halve it. Covariance terms may be dropped only under justified uncorrelatedness, with independence a sufficient condition.