Doob–Dynkin lemma

Template:Short description Template:Redirect 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, $ℬ[0,1]$ is the $\sigma$-algebra of Borel sets on $[0,1].$ If $T:X\to Y,$ and $(Y,𝒴)$ is a measurable space, then

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

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

Statement of the lemma

Let $T:\mathrm{\Omega }\to {\mathrm{\Omega }}^{\prime }$ be a function, and $({\mathrm{\Omega }}^{\prime },{𝒜}^{\prime })$ a measurable space. A function $f:\mathrm{\Omega }\to [0,1]$ is $\sigma (T)/ℬ[0,1]$-measurable if and only if $f=g\circ T,$ for some ${𝒜}^{\prime }/ℬ[0,1]$-measurable $g:{\mathrm{\Omega }}^{\prime }\to [0,1].$

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 $\sigma (T)/ℬ[0,1]$-measurable.

First, note that, by the above descriptive definition of $\sigma (T)$ as the set of preimages of ${𝒜}^{\prime }$-measurable sets under $T$, we know that if $A\in \sigma (T)$, then there exists some $A^{\prime }\in {𝒜}^{\prime }$ such that $A=T^{-1}(A^{\prime })$.

Now, assume that $f={\mathrm{𝟏}}_A$ is an indicator of some set $A\in \sigma (T)$. If we identify $A^{\prime }\in {𝒜}^{\prime }$ such that $A=T^{-1}(A^{\prime })$, then the function $g={\mathrm{𝟏}}_{A^{\prime }}$ suits the requirement, and since $A\in \sigma (T)$, such a set $A^{\prime }\in {𝒜}^{\prime }$ always exists. 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 $f_n\ge 0$ of simple functions. The previous step guarantees that $f_n=g_n\circ T,$ for some measurable $g_n.$ The supremum $g(x)={sup}_{n\ge 1}{g}_n(x)$ exists on the entire ${\mathrm{\Omega }}^{\prime }$ and is measurable. (The article on measurable functions explains why supremum of a sequence of measurable functions is measurable). For every $x\in \mathrm{Im}T,$ the sequence $g_n(x)$ is non-decreasing, so $g{|}_{\mathrm{Im}T}(x)=\underset{n\to \mathrm{\infty }}{lim}{g}_n{|}_{\mathrm{Im}T}(x)$ which shows that $f=g\circ T.$

Remark. The lemma remains valid if the space $([0,1],ℬ[0,1])$ is replaced with $(S,ℬ(S)),$ where $S\subseteq [-\mathrm{\infty },\mathrm{\infty }],$ $S$ 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 $T:\mathrm{\Omega }\to {\mathrm{\Omega }}^{\prime },$ $f:\mathrm{\Omega }\to [0,1],$ and $({\mathrm{\Omega }}^{\prime },{𝒜}^{\prime })$ is a measurable space. Then $f=g\circ T,$ for some ${𝒜}^{\prime }/ℬ[0,1]$-measurable $g:{\mathrm{\Omega }}^{\prime }\to [0,1],$ if and only if $\sigma (f)\subseteq \sigma (T)$.

See also

• Conditional expectation

References

• A. Bobrowski: Functional analysis for probability and stochastic processes: an introduction, Cambridge University Press (2005), Template:ISBN
• M. M. Rao, R. J. Swift : Probability Theory with Applications, Mathematics and Its Applications, vol. 582, Springer-Verlag (2006), Template:ISBN Template:Doi