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 [math]\displaystyle{ \sigma }[/math]-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 [math]\displaystyle{ \sigma }[/math]-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 [math]\displaystyle{ \sigma }[/math]-algebra that is generated by the random variable.
Notations and introductory remarks
In the lemma below, [math]\displaystyle{ \mathcal{B}[0,1] }[/math] is the [math]\displaystyle{ \sigma }[/math]-algebra of Borel sets on [math]\displaystyle{ [0,1]. }[/math] If [math]\displaystyle{ T\colon X\to Y, }[/math] and [math]\displaystyle{ (Y,{\mathcal Y}) }[/math] is a measurable space, then
- [math]\displaystyle{ \sigma(T)\ \stackrel{\text{def}}{=}\ \{T^{-1}(S)\mid S\in {\mathcal Y}\} }[/math]
is the smallest [math]\displaystyle{ \sigma }[/math]-algebra on [math]\displaystyle{ X }[/math] such that [math]\displaystyle{ T }[/math] is [math]\displaystyle{ \sigma(T) / {\mathcal Y} }[/math]-measurable.
Statement of the lemma
Let [math]\displaystyle{ T\colon \Omega\rightarrow\Omega' }[/math] be a function, and [math]\displaystyle{ (\Omega',\mathcal{A}') }[/math] a measurable space. A function [math]\displaystyle{ f\colon \Omega\rightarrow [0,1] }[/math] is [math]\displaystyle{ \sigma(T) / \mathcal{B}[0,1] }[/math]-measurable if and only if [math]\displaystyle{ f=g\circ T, }[/math] for some [math]\displaystyle{ \mathcal{A}' / \mathcal{B}[0,1] }[/math]-measurable [math]\displaystyle{ g\colon \Omega' \to [0,1]. }[/math][1]
Remark. The "if" part simply states that the composition of two measurable functions is measurable. The "only if" part is proven below.
| Proof. |
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Let [math]\displaystyle{ f }[/math] be [math]\displaystyle{ \sigma(T) / \mathcal{B}[0,1] }[/math]-measurable. Assume that [math]\displaystyle{ f = \mathbf{1}_A }[/math] is an indicator of some set [math]\displaystyle{ A \in \sigma(T). }[/math] If [math]\displaystyle{ A = T^{-1}(A'), }[/math] then the function [math]\displaystyle{ g={\mathbf 1}_{A'} }[/math] suits the requirement. By linearity, the claim extends to any simple measurable function [math]\displaystyle{ f. }[/math] Let [math]\displaystyle{ f }[/math] be measurable but not necessarily simple. As explained in the article on simple functions, [math]\displaystyle{ f }[/math] is a pointwise limit of a monotonically non-decreasing sequence [math]\displaystyle{ f_n \geq 0 }[/math] of simple functions. The previous step guarantees that [math]\displaystyle{ f_n = g_n \circ T, }[/math] for some measurable [math]\displaystyle{ g_n. }[/math] The supremum [math]\displaystyle{ \textstyle g(x)=\sup_{ n \geq 1} g_n(x) }[/math] exists on the entire [math]\displaystyle{ \Omega' }[/math] and is measurable. (The article on measurable functions explains why supremum of a sequence of measurable functions is measurable). For every [math]\displaystyle{ x \in \operatorname{Im}T, }[/math] the sequence [math]\displaystyle{ g_n(x) }[/math] is non-decreasing, so [math]\displaystyle{ \textstyle g|_{\operatorname{Im}T}(x) = \lim_{ n \to \infty} g_n|_{\operatorname{Im}T}(x) }[/math] which shows that [math]\displaystyle{ f = g \circ T. }[/math] |
Remark. The lemma remains valid if the space [math]\displaystyle{ ([0,1],\mathcal{B}[0,1]) }[/math] is replaced with [math]\displaystyle{ (S,\mathcal{B}(S)), }[/math] where [math]\displaystyle{ S \subseteq [-\infty,\infty], }[/math] [math]\displaystyle{ S }[/math] is bijective with [math]\displaystyle{ [0,1], }[/math] and the bijection is measurable in both directions.
By definition, the measurability of [math]\displaystyle{ f }[/math] means that [math]\displaystyle{ f^{-1}(S)\in \sigma(T) }[/math] for every Borel set [math]\displaystyle{ S \subseteq [0,1]. }[/math] Therefore [math]\displaystyle{ \sigma(f) \subseteq \sigma(T), }[/math] and the lemma may be restated as follows.
Lemma. Let [math]\displaystyle{ T\colon \Omega\rightarrow\Omega', }[/math] [math]\displaystyle{ f\colon \Omega\rightarrow [0,1], }[/math] and [math]\displaystyle{ (\Omega',\mathcal{A}') }[/math] is a measurable space. Then [math]\displaystyle{ f = g\circ T, }[/math] for some [math]\displaystyle{ \mathcal{A}' / \mathcal{B}[0,1] }[/math]-measurable [math]\displaystyle{ g\colon \Omega' \to [0,1], }[/math] if and only if [math]\displaystyle{ \sigma(f) \subseteq \sigma(T) }[/math].
See also
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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