104

9

Probability and Likelihood

the notationupper P left brace upper A vertical bar upper H right braceP{A|H} (read as “the conditional probability ofupper AA on hypothesisupper HH” or

“the conditional probability of upper AA for a given event upper HH”) and

upper P left brace upper A vertical bar upper H right brace equals StartFraction upper P left brace upper A upper H right brace Over upper P left brace upper H right brace EndFraction periodP{A|H} = ^P{AH}

P{H} ^.

(9.15)

This result can be derived by noting that we are asking “to what extent isupper HH contained

inupper AA?” which means “to what extent areupper HH andupper AA likely to occur simultaneously?” In

set notation, this is upper P left brace upper A intersection upper H right brace equals upper P left brace upper H intersection upper A right braceP{A ∩H} = P{H ∩A}. Therefore, upper P left brace upper A vertical bar upper H right brace equals k upper P left brace upper A intersection upper H right braceP{A|H} = kP{A ∩H},

where kk is a constant. If upper A equals upper HA = H, then upper P left brace upper H vertical bar upper H right brace equals k upper P left brace upper H intersection upper H right brace equals k upper P left brace upper H right brace equals 1P{H|H} = kP{H ∩H} = kP{H} = 1;

hence, k equals 1 divided by upper P left brace upper H right bracek = 1/P{H} and we obtain

upper P left brace upper A vertical bar upper H right brace equals StartFraction upper P left brace upper A intersection upper H right brace Over upper P left brace upper H right brace EndFractionP{A|H} = ^{P{A ∩H}}

P{H}

(9.16)

(i.e., Eq. 9.15). If all sample points have equal probabilities, then

upper P left brace upper A vertical bar upper H right brace equals StartFraction upper N left brace upper A upper H right brace Over upper N left brace upper H right brace EndFraction commaP{A|H} = ^N{AH}

N{H} ^,

(9.17)

where upper N left brace upper A upper H right braceN{AH} is the number of sample points common to upper AA and upper HH.

From this comes a theorem, due to Bayes, of great importance and widely referred

to, which gives the probability that the event upper AA, which has occurred, is the result of

the cause upper E Subscript kEk:

upper P left brace upper E Subscript k Baseline vertical bar upper A right brace equals StartFraction upper P left brace upper A vertical bar upper E Subscript k Baseline right brace upper P left brace upper E Subscript k Baseline right brace Over sigma summation Underscript j equals 1 Overscript n Endscripts upper P left brace upper A vertical bar upper E Subscript j Baseline right brace upper P left brace upper E Subscript j Baseline right brace EndFraction normal f normal o normal r k equals 1 comma ellipsis comma n commaP{Ek|A} =

P{A|Ek}P{Ek}

En

j=1 ^{P{A|E j}P{E j}}

for k = 1, . . . , n ,

(9.18)

where the upper E Subscript jE j are mutually exclusive hypotheses.

Proof. Let the simple events upper E Subscript iEi be labelled such that

upper A equals upper E 1 union upper E 2 union midline horizontal ellipsis union upper E Subscript m Baseline comma 1 less than or equals m less than or equals n periodA = E1 ∪E2 ∪· · · ∪Em , 1 ≤m ≤n .

(9.19)

Then

upper P left brace upper A right brace equals sigma summation Underscript j equals 1 Overscript m Endscripts upper P left brace upper E Subscript j Baseline right brace periodP{A} =

m

E

j=1

P{E j} .

(9.20)

From the definition (9.15),

sigma summation Underscript j equals 1 Overscript n Endscripts upper P left brace upper A vertical bar upper E Subscript j Baseline right brace upper P left brace upper E Subscript j Baseline right brace equals sigma summation Underscript j equals 1 Overscript n Endscripts upper P left brace upper A intersection upper E Subscript j Baseline right brace comma

n

E

j=1

P{A|E j}P{E j} =

n

E

j=1

P{A ∩E j} ,

(9.21)

which can be equated to the right-hand side of (9.20)