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My Question: Let $(\Omega,\mathcal{F},\textbf{P})$ be a probability triple such that $\Omega$ is ${countable}$. Prove that it is impossible for there to exist a sequence $A_1,A_2,\ldots \in \mathcal{F}$ which is ${independent}$ such that $\textbf{P}(A_i)=\frac{1}{2}$ for each $i$. Hint: First prove that for each $\omega\in \Omega$, and for each $n\in \mathbb{N}$ we have $P(\{\omega\})\leq \frac{1}{2^n}$. Then derive a contradiction.

My Work: Let $\omega \in \Omega$ be arbitrary. Then since $\textbf{P}(\Omega)=\textbf{P}(\bigcup \{\omega\})=1$, and $(\Omega,\mathcal{F},\textbf{P})$ is a valid probability triple, then $\textbf{P}$ is countably additive, so that $\textbf{P}(\bigcup \{\omega\})=\sum\limits_{n=1}^\infty \textbf{P}(A_n)$. Here I am not sure where to go. My problem: I am really not seeing how to go about this problem. Any help is appreciated.

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Thank you for asking this. –  BCLC Aug 24 at 1:43

1 Answer 1

up vote 2 down vote accepted

Before we start, let's prove by induction that there exists $\mathcal{P}_{n}$ which partitions $\Omega$, only uses the first n elements of the sequence and such that for any $B \in \mathcal{P}_{n}$, $P(B) = 2^{-n}$.

Initial Case: Observe that $\mathcal{P}_{1} =\{A_{1}, A_{1}^{c}\}$ partitions $\Omega$ and only uses the 1st element of the sequence. Furthermore, any $B \in \mathcal{P}_{1}$ is such that $P(B) = 2^{-1}$.

Induction Case: Assume there exists $\mathcal{P}_{n}$ which partitions $\Omega$, only uses the first n elements of the sequence and such that for any $B \in \mathcal{P}_{n}$, $P(B) = 2^{-n}$. Let's construct $\mathcal{P}_{n+1}$.Consider $\mathcal{P}_{n+1} = \{B \cap A_{n+1} \text{ or } B \cap A^{c}_{n+1}: B \in \mathcal{P}_{n}\}$. Trivially, $\mathcal{P}_{n+1}$ partitions $\Omega$ and only uses the first $n+1$ elements of the sequence. By independence, for any $B \in \mathcal{P}_{n+1}$, $P(B) = 2^{-n-1}$.

Now, we are ready for the main proof.

Since $\mathcal{P}_{n}$ is a partition, for every $w \in \Omega$, there exist $B \in \mathcal{P}_{n}$ such that $w \in B$ and, thus, $P(w) \leq P(B)$. Since $\Omega$ is countable, $P(B) = \sum_{w' \in B}{P(w')}$. Hence, by the properties of $\mathcal{P}_{n}$, $\sum_{w' \in B}{P(w')} = 2^{-n}$. Conclude that, for every $w \in \Omega$, $P(w) < 2^{-n}$. Hence, for every $w \in \Omega$, $P(w) = 0$.

The contradiction is complete observing that, since $\Omega$ is countable:

$$0 = \sum_{w \in \Omega}{P(w)} = P(\cup_{w \in \Omega}{\{w\}} = P(\Omega) = 1$$

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Thank You for the insight and the great explanation. –  Mathstudent Feb 11 '13 at 6:42
    
Thank you for answering the question. –  BCLC Aug 24 at 1:52

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