In probability theory, the Doob–Dynkin lemma, named after Joseph L. Doob and Eugene Dynkin (also known as the factorization lemma), characterizes the situation when one random variable is a function of another by the inclusion of the σ {\displaystyle \sigma } -algebras generated by the random variables. The usual statement of the lemma is formulated in terms of one random variable being measurable with respect to the σ {\displaystyle \sigma } -algebra generated by the other.
The lemma plays an important role in the conditional expectation in probability theory, where it allows replacement of the conditioning on a random variable by conditioning on the σ {\displaystyle \sigma } -algebra that is generated by the random variable.
Notations and introductory remarks
In the lemma below, B [ 0 , 1 ] {\displaystyle {\mathcal {B}}[0,1]} is the σ {\displaystyle \sigma } -algebra of Borel sets on [ 0 , 1 ] . {\displaystyle [0,1].} If T : X → Y , {\displaystyle T\colon X\to Y,} and ( Y , Y ) {\displaystyle (Y,{\mathcal {Y}})} is a measurable space, then
σ ( T ) = def { T − 1 ( S ) ∣ S ∈ Y } {\displaystyle \sigma (T)\ {\stackrel {\text{def}}{=}}\ \{T^{-1}(S)\mid S\in {\mathcal {Y}}\}}is the smallest σ {\displaystyle \sigma } -algebra on X {\displaystyle X} such that T {\displaystyle T} is σ ( T ) / Y {\displaystyle \sigma (T)/{\mathcal {Y}}} -measurable.
Statement of the lemma
Let T : Ω → Ω ′ {\displaystyle T\colon \Omega \rightarrow \Omega '} be a function, and ( Ω ′ , A ′ ) {\displaystyle (\Omega ',{\mathcal {A}}')} a measurable space. A function f : Ω → [ 0 , 1 ] {\displaystyle f\colon \Omega \rightarrow [0,1]} is σ ( T ) / B [ 0 , 1 ] {\displaystyle \sigma (T)/{\mathcal {B}}[0,1]} -measurable if and only if f = g ∘ T , {\displaystyle f=g\circ T,} for some A ′ / B [ 0 , 1 ] {\displaystyle {\mathcal {A}}'/{\mathcal {B}}[0,1]} -measurable g : Ω ′ → [ 0 , 1 ] . {\displaystyle g\colon \Omega '\to [0,1].} 1
Remark. The "if" part simply states that the composition of two measurable functions is measurable. The "only if" part is proven below.
| Proof. |
Let f {\displaystyle f} be σ ( T ) / B [ 0 , 1 ] {\displaystyle \sigma (T)/{\mathcal {B}}[0,1]} -measurable. First, note that, by the above descriptive definition of σ ( T ) {\displaystyle \sigma (T)} as the set of preimages of A ′ {\displaystyle {\mathcal {A}}'} -measurable sets under T {\displaystyle T} , we know that if A ∈ σ ( T ) {\displaystyle A\in \sigma (T)} , then there exists some A ′ ∈ A ′ {\displaystyle A'\in {\mathcal {A}}'} such that A = T − 1 ( A ′ ) {\displaystyle A=T^{-1}(A')} . Now, assume that f = 1 A {\displaystyle f=\mathbf {1} _{A}} is an indicator of some set A ∈ σ ( T ) {\displaystyle A\in \sigma (T)} . If we identify A ′ ∈ A ′ {\displaystyle A'\in {\mathcal {A}}'} such that A = T − 1 ( A ′ ) {\displaystyle A=T^{-1}(A')} , then the function g = 1 A ′ {\displaystyle g=\mathbf {1} _{A'}} suits the requirement, and since A ∈ σ ( T ) {\displaystyle A\in \sigma (T)} , such a set A ′ ∈ A ′ {\displaystyle A'\in {\mathcal {A}}'} always exists. By linearity, the claim extends to any simple measurable function f . {\displaystyle f.} Let f {\displaystyle f} be measurable but not necessarily simple. As explained in the article on simple functions, f {\displaystyle f} is a pointwise limit of a monotonically non-decreasing sequence f n ≥ 0 {\displaystyle f_{n}\geq 0} of simple functions. The previous step guarantees that f n = g n ∘ T , {\displaystyle f_{n}=g_{n}\circ T,} for some measurable g n . {\displaystyle g_{n}.} The supremum g ( x ) = sup n ≥ 1 g n ( x ) {\displaystyle \textstyle g(x)=\sup _{n\geq 1}g_{n}(x)} exists on the entire Ω ′ {\displaystyle \Omega '} and is measurable. (The article on measurable functions explains why supremum of a sequence of measurable functions is measurable). For every x ∈ Im T , {\displaystyle x\in \operatorname {Im} T,} the sequence g n ( x ) {\displaystyle g_{n}(x)} is non-decreasing, so g | Im T ( x ) = lim n → ∞ g n | Im T ( x ) {\displaystyle \textstyle g|_{\operatorname {Im} T}(x)=\lim _{n\to \infty }g_{n}|_{\operatorname {Im} T}(x)} which shows that f = g ∘ T . {\displaystyle f=g\circ T.} |
Remark. The lemma remains valid if the space ( [ 0 , 1 ] , B [ 0 , 1 ] ) {\displaystyle ([0,1],{\mathcal {B}}[0,1])} is replaced with ( S , B ( S ) ) , {\displaystyle (S,{\mathcal {B}}(S)),} where S ⊆ [ − ∞ , ∞ ] , {\displaystyle S\subseteq [-\infty ,\infty ],} S {\displaystyle S} is bijective with [ 0 , 1 ] , {\displaystyle [0,1],} and the bijection is measurable in both directions.
By definition, the measurability of f {\displaystyle f} means that f − 1 ( S ) ∈ σ ( T ) {\displaystyle f^{-1}(S)\in \sigma (T)} for every Borel set S ⊆ [ 0 , 1 ] . {\displaystyle S\subseteq [0,1].} Therefore σ ( f ) ⊆ σ ( T ) , {\displaystyle \sigma (f)\subseteq \sigma (T),} and the lemma may be restated as follows.
Lemma. Let T : Ω → Ω ′ , {\displaystyle T\colon \Omega \rightarrow \Omega ',} f : Ω → [ 0 , 1 ] , {\displaystyle f\colon \Omega \rightarrow [0,1],} and ( Ω ′ , A ′ ) {\displaystyle (\Omega ',{\mathcal {A}}')} is a measurable space. Then f = g ∘ T , {\displaystyle f=g\circ T,} for some A ′ / B [ 0 , 1 ] {\displaystyle {\mathcal {A}}'/{\mathcal {B}}[0,1]} -measurable g : Ω ′ → [ 0 , 1 ] , {\displaystyle g\colon \Omega '\to [0,1],} if and only if σ ( f ) ⊆ σ ( T ) {\displaystyle \sigma (f)\subseteq \sigma (T)} .
See also
- A. Bobrowski: Functional analysis for probability and stochastic processes: an introduction, Cambridge University Press (2005), ISBN 0-521-83166-0
- M. M. Rao, R. J. Swift : Probability Theory with Applications, Mathematics and Its Applications, vol. 582, Springer-Verlag (2006), ISBN 0-387-27730-7 doi:10.1007/0-387-27731-5
References
Kallenberg, Olav (1997). Foundations of Modern Probability. Springer. p. 7. ISBN 0-387-94957-7. 0-387-94957-7 ↩