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Mathematics
Physics
Bayes Theorem
Bayes Theorem
BAYES' THEOREM
Let
A
A
A
and
B
B
B
be events with
P
(
A
)
>
0
\mathbb{P}(A)>0
P
(
A
)
>
0
and
P
(
B
)
>
0
\mathbb{P}(B)>0
P
(
B
)
>
0
. Then
P
(
A
∣
B
)
⏟
posterior
=
P
(
B
∣
A
)
⏟
likelihood
P
(
A
)
⏟
prior
P
(
B
)
⏟
evidence
.
\begin{equation*}\underbrace{\mathbb{P}(A\mid B)}_{\text{posterior}} =\frac{ \underbrace{\mathbb{P}(B\mid A)}_{\text{likelihood}} \underbrace{\mathbb{P}(A)}_{\text{prior}}}{\underbrace{\mathbb{P}(B)}_{\text{evidence}}}.\end{equation*}
posterior
P
(
A
∣
B
)
=
evidence
P
(
B
)
likelihood
P
(
B
∣
A
)
prior
P
(
A
)
.
Proof
The proof is only the definition of conditional probability used twice. Since
P
(
A
∣
B
)
=
P
(
A
∩
B
)
P
(
B
)
and
P
(
B
∣
A
)
=
P
(
A
∩
B
)
P
(
A
)
,
\begin{equation*}\mathbb{P}(A\mid B)=\frac{\mathbb{P}(A\cap B)}{\mathbb{P}(B)}\quad \text{and} \quad\mathbb{P}(B\mid A)=\frac{\mathbb{P}(A\cap B)}{\mathbb{P}(A)},\end{equation*}
P
(
A
∣
B
)
=
P
(
B
)
P
(
A
∩
B
)
and
P
(
B
∣
A
)
=
P
(
A
)
P
(
A
∩
B
)
,
we get
P
(
A
∩
B
)
=
P
(
B
∣
A
)
P
(
A
)
.
\begin{equation*}\mathbb{P}(A\cap B)=\mathbb{P}(B\mid A)\mathbb{P}(A).\end{equation*}
P
(
A
∩
B
)
=
P
(
B
∣
A
)
P
(
A
)
.
Substitution into the first identity gives
P
(
A
∣
B
)
=
P
(
B
∣
A
)
P
(
A
)
P
(
B
)
.
□
\begin{equation*}\mathbb{P}(A\mid B)=\frac{\mathbb{P}(B\mid A)\mathbb{P}(A)}{\mathbb{P}(B)}.\qquad \Box\end{equation*}
P
(
A
∣
B
)
=
P
(
B
)
P
(
B
∣
A
)
P
(
A
)
.
□
References
[Durrett2019]
Durrett, Rick.
Probability: Theory and Examples
. Fifth edition. Cambridge University Press, 2019.
PDF
.