证明在等可能结果的概率空间Ω={0,1}³上可定义70个参数为1/2的伯努利随机变量

Let's break this down step by step to confirm we can define exactly 70 unique Bernoulli(1/2) random variables on this probability space.

Step 1: Lay out the basics of the probability space #

First, Ω={0,1}³ is the set of all 3-length binary sequences, so it has exactly 8 elements total:

• Ω = {(0,0,0), (0,0,1), (0,1,0), (0,1,1), (1,0,0), (1,0,1), (1,1,0), (1,1,1)}

Every outcome is equally likely, meaning each element has a probability of 1/8.

Step 2: Define what a Bernoulli(1/2) random variable requires here #

A random variable ( X: \Omega \to {0,1} ) counts as Bernoulli(1/2) if:

• ( \mathbb{P}(X=1) = \mathbb{P}(X=0) = 1/2 )

Since each outcome has a probability of 1/8, the set of outcomes where ( X=1 ) (let's call this set ( A = X^{-1}(1) )) must contain exactly 4 elements. Why? Because ( 4 \times 1/8 = 1/2 ), which matches the required probability for ( X=1 ).

Step 3: Count the number of valid subsets (and their corresponding random variables) #

Each unique 4-element subset of Ω maps directly to a unique Bernoulli(1/2) random variable: for a subset ( A ), define ( X_A(\omega) = 1 ) if ( \omega \in A ), and ( 0 ) otherwise.

To find how many such subsets exist, we calculate the combination number ( \binom{8}{4} )—the number of ways to choose 4 elements from 8:
[
\binom{8}{4} = \frac{8!}{4! \times (8-4)!} = \frac{8 \times 7 \times 6 \times 5}{4 \times 3 \times 2 \times 1} = 70
]

Step 4: Confirm all these random variables are distinct #

No two different 4-element subsets produce the same random variable. If ( A \neq B ), there's at least one outcome ( \omega ) that's in one subset but not the other. For that ( \omega ), ( X_A(\omega) \neq X_B(\omega) ), so the two variables are clearly distinct.

Final conclusion #

Since there are exactly 70 unique 4-element subsets of Ω, each corresponding to a unique Bernoulli(1/2) random variable, we've proven we can define 70 such distinct variables on the given probability space.

内容的提问来源于stack exchange，提问作者user519957