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How can I prove the following: The following are equivalent

(a) A probability space is trivial

(b) A probabilty space does not support a binomially distributed random variable, i.e there does not exist a random variable on a probability space which has a binomial distribution.

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1 Answer 1

Denote by $(\Omega,\mathcal{A},\mathbb{P})$ the given probability space.

  • $\Rightarrow$: Let $X: \Omega \to \mathbb{R}$ a random variable. Since $X$ is measurable (i.e. $X^{-1}(B) \in \mathcal{A}$ for all $B \in \mathcal{B}(\mathbb{R})$), we have $$\mathbb{P}(X \in B) \in \{0,1\}$$ for all $B \in \mathcal{B}(\mathbb{R})$. In particular for $B:= \{k\}$: $$\mathbb{P}(X=k) \in \{0,1\}$$ Hence there cannot exist a binomial distributed random variable (because then we would have $0<\mathbb{P}(X=k) = {n \choose k} \cdot p^k \cdot (1-p)^{n-k}<1$ for $k=0,\ldots,n$ for some $n \in \mathbb{N}, p \in (0,1)$).
  • $\Leftarrow$: Assume $(\Omega,\mathcal{A},\mathbb{P})$ is not trivial. Then there exists $A \in \mathcal{A}$ such that $0<p:=\mathbb{P}(A)<1$. Define a random variable $X$ by $$X(\omega) := \begin{cases} 1 & \omega \in A \\ 0 & \omega \notin A \end{cases}$$ $X$ is measurable (since $A \in \mathcal{A}$) and one can easily see that $X \sim B(1,p)$. This means that there exists a binomial distributed random variable. Contradiction!
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