4.2 Statistics and probability - AHL content

Bayes updates a probability after new evidence.

Bayes’ theorem reverses a conditional relationship by weighting the evidence with the prior probabilities of its possible causes.

Example

If a rare condition affects 1% and a test detects it 90% of the time but has a 5% false-positive rate, a positive result is not a 90% diagnosis.

Build the joint probabilities first, then divide the relevant cause-and-evidence cell by all positive evidence.

Sensitivity is P(positive|condition), not P(condition|positive).

For mutually exclusive and exhaustive causes $C_1,\ldots,C_n$ with $n\le3$, $P(C_i|E)=P(E|C_i)P(C_i)/\sum_j P(E|C_j)P(C_j)$. In the test example, $P(positive)=0.90(0.01)+0.05(0.99)=0.0585$, so $P(condition|positive)=0.009/0.0585\approx0.154$. Thus about 15.4%, not 90%, of positive tests correspond to the condition. Independence would instead mean the evidence leaves the prior unchanged.

A probability density describes area, not height as probability.

For a continuous variable, P(a≤X≤b) is the area under f(x) between a and b; the total area is one and a single point has probability zero.

Example

If f(x)=2x on 0≤x≤1, P(X≤.5)=∫₀^.5 2x dx=.25.

Check the support and integrate over the interval before interpreting the result.

A density can exceed 1; only area, not height alone, is a probability.

Moment toolkit: for discrete $X$, $E(X)=\sum xP(X=x)$ and $E(X^2)=\sum x^2P(X=x)$; for continuous $X$, replace sums by integrals over its support. In both cases $Var(X)=E(X^2)-[E(X)]^2$ and $SD(X)=\sqrt{Var(X)}$. A continuous mode maximizes $f(x)$, while a median $m$ satisfies $\int_{-\infty}^{m}f(x)\,dx=1/2$. Under $Y=aX+b$, $E(Y)=aE(X)+b$ and $Var(Y)=a^2Var(X)$. For piecewise densities, integrate every relevant piece and require non-negative density with total area one; density height itself may exceed one.