The π-𝜆 theorem

Let ${\displaystyle D}$ be a 𝜆-system, and let  ${\displaystyle {\mathcal {I}}\subseteq D}$ be a π-system contained in ${\displaystyle D.}$ The π-𝜆 theorem^[1] states that the 𝜎-algebra ${\displaystyle \sigma ({\mathcal {I}})}$ generated by ${\displaystyle {\mathcal {I}}}$ is contained in ${\displaystyle D~:~}$ ${\displaystyle \sigma ({\mathcal {I}})\subseteq D.}$

The π-𝜆 theorem can be used to prove many elementary measure theoretic results. For instance, it is used in proving the uniqueness claim of the Carathéodory extension theorem for 𝜎-finite measures.^[2]

The π-𝜆 theorem is closely related to the monotone class theorem, which provides a similar relationship between monotone classes and algebras, and can be used to derive many of the same results. Since π-systems are simpler classes than algebras, it can be easier to identify the sets that are in them while, on the other hand, checking whether the property under consideration determines a 𝜆-system is often relatively easy. Despite the difference between the two theorems, the π-𝜆 theorem is sometimes referred to as the monotone class theorem.^[1]

Example

Let ${\displaystyle \mu _{1},\mu _{2}:F\to \mathbb {R} }$ be two measures on the 𝜎-algebra ${\displaystyle F,}$ and suppose that ${\displaystyle F=\sigma (I)}$ is generated by a π-system ${\displaystyle I.}$ If

1. ${\displaystyle \mu _{1}(A)=\mu _{2}(A)}$ for all ${\displaystyle A\in I,}$ and
2. ${\displaystyle \mu _{1}(\Omega )=\mu _{2}(\Omega )<\infty ,}$

then ${\displaystyle \mu _{1}=\mu _{2}.}$ This is the uniqueness statement of the Carathéodory extension theorem for finite measures. If this result does not seem very remarkable, consider the fact that it usually is very difficult or even impossible to fully describe every set in the 𝜎-algebra, and so the problem of equating measures would be completely hopeless without such a tool.

Idea of the proof^[2] Define the collection of sets

$${\displaystyle D=\left\{A\in \sigma (I)\colon \mu _{1}(A)=\mu _{2}(A)\right\}.}$$

By the first assumption, ${\displaystyle \mu _{1}}$ and ${\displaystyle \mu _{2}}$ agree on ${\displaystyle I}$ and thus ${\displaystyle I\subseteq D.}$ By the second assumption, ${\displaystyle \Omega \in D,}$ and it can further be shown that ${\displaystyle D}$ is a 𝜆-system. It follows from the π-𝜆 theorem that ${\displaystyle \sigma (I)\subseteq D\subseteq \sigma (I),}$ and so ${\displaystyle D=\sigma (I).}$ That is to say, the measures agree on ${\displaystyle \sigma (I).}$

Equality in distribution

The π-𝜆 theorem motivates the common definition of the probability distribution of a random variable ${\displaystyle X:(\Omega ,{\mathcal {F}},\operatorname {P} )\to \mathbb {R} }$ in terms of its cumulative distribution function. Recall that the cumulative distribution of a random variable is defined as

$${\displaystyle F_{X}(a)=\operatorname {P} [X\leq a],\qquad a\in \mathbb {R} ,}$$

whereas the seemingly more general law of the variable is the probability measure

$${\displaystyle {\mathcal {L}}_{X}(B)=\operatorname {P} \left[X^{-1}(B)\right]\quad {\text{ for all }}B\in {\mathcal {B}}(\mathbb {R} ),}$$

where ${\displaystyle {\mathcal {B}}(\mathbb {R} )}$ is the Borel 𝜎-algebra. The random variables ${\displaystyle X:(\Omega ,{\mathcal {F}},\operatorname {P} )\to \mathbb {R} }$ and ${\displaystyle Y:({\tilde {\Omega }},{\tilde {\mathcal {F}}},{\tilde {\operatorname {P} }})\to \mathbb {R} }$ (on two possibly different probability spaces) are equal in distribution (or law), denoted by ${\displaystyle X\,{\stackrel {\mathcal {D}}{=}}\,Y,}$ if they have the same cumulative distribution functions; that is, if ${\displaystyle F_{X}=F_{Y}.}$ The motivation for the definition stems from the observation that if ${\displaystyle F_{X}=F_{Y},}$ then that is exactly to say that ${\displaystyle {\mathcal {L}}_{X}}$ and ${\displaystyle {\mathcal {L}}_{Y}}$ agree on the π-system ${\displaystyle \{(-\infty ,a]:a\in \mathbb {R} \}}$ which generates ${\displaystyle {\mathcal {B}}(\mathbb {R} ),}$ and so by the example above: ${\displaystyle {\mathcal {L}}_{X}={\mathcal {L}}_{Y}.}$

A similar result holds for the joint distribution of a random vector. For example, suppose ${\displaystyle X}$ and ${\displaystyle Y}$ are two random variables defined on the same probability space ${\displaystyle (\Omega ,{\mathcal {F}},\operatorname {P} ),}$ with respectively generated π-systems ${\displaystyle {\mathcal {I}}_{X}}$ and ${\displaystyle {\mathcal {I}}_{Y}.}$ The joint cumulative distribution function of ${\displaystyle (X,Y)}$ is

$${\displaystyle F_{X,Y}(a,b)=\operatorname {P} [X\leq a,Y\leq b]=\operatorname {P} \left[X^{-1}((-\infty ,a])\cap Y^{-1}((-\infty ,b])\right],\quad {\text{ for all }}a,b\in \mathbb {R} .}$$

However, ${\displaystyle A=X^{-1}((-\infty ,a])\in {\mathcal {I}}_{X}}$ and ${\displaystyle B=Y^{-1}((-\infty ,b])\in {\mathcal {I}}_{Y}.}$ Because

$${\displaystyle {\mathcal {I}}_{X,Y}=\left\{A\cap B:A\in {\mathcal {I}}_{X},{\text{ and }}B\in {\mathcal {I}}_{Y}\right\}}$$

is a π-system generated by the random pair ${\displaystyle (X,Y),}$ the π-𝜆 theorem is used to show that the joint cumulative distribution function suffices to determine the joint law of ${\displaystyle (X,Y).}$ In other words, ${\displaystyle (X,Y)}$ and ${\displaystyle (W,Z)}$ have the same distribution if and only if they have the same joint cumulative distribution function.

In the theory of stochastic processes, two processes ${\displaystyle (X_{t})_{t\in T},(Y_{t})_{t\in T}}$ are known to be equal in distribution if and only if they agree on all finite-dimensional distributions; that is, for all ${\displaystyle t_{1},\ldots ,t_{n}\in T,\,n\in \mathbb {N} ,}$

$${\displaystyle \left(X_{t_{1}},\ldots ,X_{t_{n}}\right)\,{\stackrel {\mathcal {D}}{=}}\,\left(Y_{t_{1}},\ldots ,Y_{t_{n}}\right).}$$

The proof of this is another application of the π-𝜆 theorem.^[3]

Independent random variables

The theory of π-system plays an important role in the probabilistic notion of independence. If ${\displaystyle X}$ and ${\displaystyle Y}$ are two random variables defined on the same probability space ${\displaystyle (\Omega ,{\mathcal {F}},\operatorname {P} )}$ then the random variables are independent if and only if their π-systems ${\displaystyle {\mathcal {I}}_{X},{\mathcal {I}}_{Y}}$ satisfy for all ${\displaystyle A\in {\mathcal {I}}_{X}}$ and ${\displaystyle B\in {\mathcal {I}}_{Y},}$

$${\displaystyle \operatorname {P} [A\cap B]~=~\operatorname {P} [A]\operatorname {P} [B],}$$

which is to say that ${\displaystyle {\mathcal {I}}_{X},{\mathcal {I}}_{Y}}$ are independent. This actually is a special case of the use of π-systems for determining the distribution of ${\displaystyle (X,Y).}$

Example

Let ${\displaystyle Z=\left(Z_{1},Z_{2}\right),}$ where ${\displaystyle Z_{1},Z_{2}\sim {\mathcal {N}}(0,1)}$ are iid standard normal random variables. Define the radius and argument (arctan) variables

$${\displaystyle R={\sqrt {Z_{1}^{2}+Z_{2}^{2}}},\qquad \Theta =\tan ^{-1}\left(Z_{2}/Z_{1}\right).}$$

Then ${\displaystyle R}$ and ${\displaystyle \Theta }$ are independent random variables.

To prove this, it is sufficient to show that the π-systems ${\displaystyle {\mathcal {I}}_{R},{\mathcal {I}}_{\Theta }}$ are independent: that is, for all ${\displaystyle \rho \in [0,\infty )}$ and ${\displaystyle \theta \in [0,2\pi ],}$

$${\displaystyle \operatorname {P} [R\leq \rho ,\Theta \leq \theta ]=\operatorname {P} [R\leq \rho ]\operatorname {P} [\Theta \leq \theta ].}$$

Confirming that this is the case is an exercise in changing variables. Fix ${\displaystyle \rho \in [0,\infty )}$ and ${\displaystyle \theta \in [0,2\pi ],}$ then the probability can be expressed as an integral of the probability density function of ${\displaystyle Z.}$

$${\displaystyle {\begin{aligned}\operatorname {P} [R\leq \rho ,\Theta \leq \theta ]&=\int _{R\leq \rho ,\,\Theta \leq \theta }{\frac {1}{2\pi }}\exp \left({-{\frac {1}{2}}(z_{1}^{2}+z_{2}^{2})}\right)dz_{1}\,dz_{2}\\[5pt]&=\int _{0}^{\theta }\int _{0}^{\rho }{\frac {1}{2\pi }}e^{-{\frac {r^{2}}{2}}}\;r\,dr\,d{\tilde {\theta }}\\[5pt]&=\left(\int _{0}^{\theta }{\frac {1}{2\pi }}\,d{\tilde {\theta }}\right)\;\left(\int _{0}^{\rho }e^{-{\frac {r^{2}}{2}}}\;r\,dr\right)\\[5pt]&=\operatorname {P} [\Theta \leq \theta ]\operatorname {P} [R\leq \rho ].\end{aligned}}}$$