# Find a probability measure on the set of all finite binary sequences $\Omega$ such that $\mathbb{P}(\{\omega\}) > 0$

Let $$\Omega$$ be the set of all finite sequences of zeros and ones. Find a probability measure $$\mathbb{P}$$ on $$\mathcal{A}=\mathcal{P}(\Omega)$$, such that for each finite sequence $$\omega \in \Omega$$, $$\mathbb{P}(\{\omega\}) > 0$$

I know that, for a valid probability measure, I need to establish $$\sigma$$-additivity and make sure that $$\mathbb{P}(\Omega)=1$$.

My idea so far was to define $$\mathbb{P}(\{\omega\})$$ as "Probability that $$\omega$$ starts with $$0$$" which I thought to be $$\mathbb{P}(\{\omega\}) = \frac{1}{2}$$, but I don't know if this assumption is correct or how to mathematically argue for it.

## 2 Answers

What if you assigned probability mass $$2^{-2n}$$ to each unique sequence of length $$n$$? We can define $$\ell(\omega)$$ to be the length of $$\omega$$, and let $$\mathbb{P}(\{\omega\}) = 2^{-2\ell(\omega)}$$ and $$\mathbb{P} : 2^\Omega\to [0, \infty)$$ be defined by $$\mathbb{P}(S) = \sum_{\omega\in S} \mathbb{P}(\{\omega\})$$ (this is well-defined, as the sum is unconditionally convergent). Then, the total measure of $$\Omega$$ would be $$\sum_{n=1}^{\infty} \frac{\text{# of length-}n\text{ sequences}}{2^{2n}} = \sum_{n=1}^{\infty} \frac{2^n}{2^{2n}} = \sum_{n=1}^{\infty} 2^{-n} = 1$$ We have therefore defined $$\mathbb{P}$$ such that it is $$\sigma$$-additive and $$\mathbb{P}(\Omega) = 1$$.

Your example would not really work, if $$\omega$$ does not start with a 0 then $$\mathbb P(\{\omega\}) = 0$$.

The key thing to note is that the set of all finite sequences of ones and zeros, $$\Omega$$, is a countable set. This makes it much easier to define a probability measure. Can you think of another countable space with a probability measure on the space that give positive mass to singleton sets? If so, then can you see how can you use it to define a probability measure on $$\Omega$$?