Doob–Dynkin lemma

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 $\sigma$ -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 $\sigma$ -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 $\sigma$ -algebra that is generated by the random variable.

Notations and introductory remarks

In the lemma below, ${\mathcal {B}}[0,1]$ is the $\sigma$ -algebra of Borel sets on $[0,1].$ If $T\colon X\to Y,$ and $(Y,{\mathcal {Y}})$ is a measurable space, then

\sigma (T)\ {\stackrel {\text{def}}{=}}\ \{T^{-1}(S)\mid S\in {\mathcal {Y}}\}

is the smallest $\sigma$ -algebra on $X$ such that $T$ is $\sigma (T)/{\mathcal {Y}}$ -measurable.

Statement of the lemma

Let $T\colon \Omega \rightarrow \Omega '$ be a function, and $(\Omega ',{\mathcal {A}}')$ a measurable space. A function $f\colon \Omega \rightarrow [0,1]$ is $\sigma (T)/{\mathcal {B}}[0,1]$ -measurable if and only if $f=g\circ T,$ for some ${\mathcal {A}}'/{\mathcal {B}}[0,1]$ -measurable $g\colon \Omega '\to [0,1].$ [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 $f$ be $\sigma (T)/{\mathcal {B}}[0,1]$ -measurable.

Assume that $f=\mathbf {1} _{A}$ is an indicator of some set $A\in \sigma (T).$ If $A=T^{-1}(A'),$ then the function $g={\mathbf {1} }_{A'}$ suits the requirement. By linearity, the claim extends to any simple measurable function $f.$

Let $f$ be measurable but not necessarily simple. As explained in the article on simple functions, $f$ is a pointwise limit of a monotonically non-decreasing sequence $f_{n}\geq 0$ of simple functions. The previous step guarantees that $f_{n}=g_{n}\circ T,$ for some measurable $g_{n}.$ The supremum $\textstyle g(x)=\sup _{n\geq 1}g_{n}(x)$ exists on the entire $\Omega '$ and is measurable. (The article on measurable functions explains why supremum of a sequence of measurable functions is measurable). For every $x\in \operatorname {Im} T,$ the sequence $g_{n}(x)$ is non-decreasing, so $\textstyle g|_{\operatorname {Im} T}(x)=\lim _{n\to \infty }g_{n}|_{\operatorname {Im} T}(x)$ which shows that $f=g\circ T.$

Remark. The lemma remains valid if the space $([0,1],{\mathcal {B}}[0,1])$ is replaced with $(S,{\mathcal {B}}(S)),$ where $S\subseteq [-\infty ,\infty ],$ $S$ is bijective with $[0,1],$ and the bijection is measurable in both directions.

By definition, the measurability of $f$ means that $f^{-1}(S)\in \sigma (T)$ for every Borel set $S\subseteq [0,1].$ Therefore $\sigma (f)\subseteq \sigma (T),$ and the lemma may be restated as follows.

Lemma. Let $T\colon \Omega \rightarrow \Omega ',$ $f\colon \Omega \rightarrow [0,1],$ and $(\Omega ',{\mathcal {A}}')$ is a measurable space. Then $f=g\circ T,$ for some ${\mathcal {A}}'/{\mathcal {B}}[0,1]$ -measurable $g\colon \Omega '\to [0,1],$ if and only if $\sigma (f)\subseteq \sigma (T)$ .

References

Kallenberg, Olav (1997). Foundations of Modern Probability. Springer. p. 7. ISBN 0-387-94957-7.

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

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[1] Kallenberg, Olav (1997). Foundations of Modern Probability. Springer. p. 7. ISBN 0-387-94957-7.

Doob–Dynkin lemma

Notations and introductory remarks

Statement of the lemma

See also

References