Chapter 4: E4.1 Solution

Working problems is a crucial part of learning mathematics. No one can learn topology merely by poring over the definitions, theorems, and examples that are worked out in the text. One must work part of it out for oneself. To provide that opportunity is the purpose of the exercises.

James R. Munkres

Let $(\Omega,\mathcal{F},\mathbb{P})$ be a probability triple. Let $\mathcal{I}_{1}$ , $\mathcal{I}_{2}$ and $\mathcal{I}_{3}$ be three $\pi$ -systems on $\Omega$ such that, for $k=1,2,3$ , $\mathcal{I}_{k}\subseteq\mathcal{F}\mbox{ and }\Omega\in\mathcal{I}_{k}.$ Prove that if $\mathbb{P}(I_{1}\cap I_{2}\cap I_{3})=\mathbb{P}(I_{1})\mathbb{P}(I_{2})\mathbb{P}(I_{3})$ whenever $I_{k}\in\mathcal{I}_{k}$ ( $k=1,2,3$ ), then $\sigma(\mathcal{I}_{1}),\sigma(\mathcal{I}_{2}),\sigma(\mathcal{I}_{3})$ are independent. Why did we require that $\Omega\in\mathcal{I}_{k}$ ?

By fixing $I_{2}\in\mathcal{I}_{2}$ and $I_{3}\in\mathcal{I}_{3}$ , the two measures on $\sigma(\mathcal{I}_{1})$ agree on $\mathcal{I}_{1}$ , and they have the same total mass $\mathbb{P}(I_{2}\cap I_{3})=\mathbb{P}(\Omega\cap I_{2}\cap I_{3})=\mathbb{P}(\Omega)\mathbb{P}(I_{2})\mathbb{P}(I_{3})=\mathbb{P}(I_{2})\mathbb{P}(I_{3}).$ Hence, they agree on $\sigma(\mathcal{I}_{1})$ .

By fixing $H_{1}\in\sigma(\mathcal{I}_{1})$ and $I_{3}\in\mathcal{I}_{3}$ , the two measures on $\sigma(\mathcal{I}_{2})$ agree on $\mathcal{I}_{2}$ , and they have the same total mass $\mathbb{P}(H_{1}\cap I_{3})=\mathbb{P}(H_{1}\cap\Omega\cap I_{3})=\mathbb{P}(H_{1})\mathbb{P}(\Omega)\mathbb{P}(I_{3})=\mathbb{P}(H_{1})\mathbb{P}(I_{3}).$ Hence, they agree on $\sigma(\mathcal{I}_{2})$ .

By fixing $H_{1}\in\sigma(\mathcal{I}_{1})$ and $H_{2}\in\sigma(\mathcal{I}_{2})$ , the two measures on $\sigma(\mathcal{I}_{3})$ agree on $\mathcal{I}_{3}$ , and they have the same total mass $\mathbb{P}(H_{1}\cap H_{2})=\mathbb{P}(H_{1}\cap H_{2}\cap\Omega)=\mathbb{P}(H_{1})\mathbb{P}(H_{2})\mathbb{P}(\Omega)=\mathbb{P}(H_{1})\mathbb{P}(H_{2}).$ Hence, they agree on $\sigma(\mathcal{I}_{3})$ .

We conclude that $\sigma(\mathcal{I}_{1}),\sigma(\mathcal{I}_{2}),\sigma(\mathcal{I}_{3})$ are independent.

We need $\Omega$ in the three $\pi$ -systems because we need the equation for the total mass in each case (otherwise, consider one $\pi$ -system being a set of measure $0$ ). Alternatively, we could have required pairwise independence of the $\pi$ -systems.

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Chapter 4: E4.1 Solution

Chapter 4: E4.1 Solution