absolutely continuous measure - probability theory

Question

Let $(\Omega, F, P)$ be a probability space.

My book says, that:

"In the case when $\Omega$ consists of finitely many elements and $P[{ω}] > 0$ for every $ω \in \Omega$, then for every probability measure $Q$ we have $Q << P$. Equivalence means $Q[{ω}] > 0$ for every $ω$".

Can someone explain me why this is true?

I know that $Q << P$ means, that $P[A] = 0 \Rightarrow Q[A] = 0$, for $A \in F$, but why does this relation hold? And why must $\Omega$ be finite?

Answer 1

The right definition is: $Q << P$ means, that $P[A] = 0 \Rightarrow \color{red}Q[A] = 0$, for $A \in F$. Since $P[A]$ is never zero except if $A$ is the empty set, the implication holds.

As far as finiteness.

If $\Omega$ was not finite then we couldn't say so easily that $P[\{ω\}] > 0$ for every $ω \in \Omega.$ If $\Omega$ is not finite but countable and $\sum_{\omega\in \Omega} P[\{\omega\}]=1$ then the same statement is true.

Answer 2

Proof of

In the case when $\Omega$ consists of finitely many elements and $P[{ω}] > 0$ for every $ω \in \Omega$, then for every probability measure $Q$ we have $Q << P$.

For any $A \in F$, we need to show that if $P(A)=0$ then $Q(A)=0$. So, for what $A$'s do we have that $P(A) = 0$? If we find those $A$'s, then we have to check the corresponding $Q(A)$'s.

If there's no such $A$, then, vacuously, 'If $P(A) = 0$, then $Q(A) = 0$' is true.

Actually, there is such an $A$: $P(A) = 0 \iff A=\emptyset$. (Otherwise, $A$ will contain some $\omega$'s which by definition have positive $P$-probability).

QED

Note: If you for some reason restrict your $A$'s to be the singleton's then that's fine because $P(A)$ will never be zero. Vacuously, 'If $A$ is a singleton and $P(A) = 0$, then $Q(A) = 0$' is true.

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Discussion of $P[{ω}] > 0$ assumption:

The thing is if there was some sample point $\omega_z$ with zero $P$-probability, i.e. $P(\omega_z) = 0$, then it's pretty simple to find construct $Q$ with if $Q(\omega_z) > 0$. Consider the following:

$\Omega = \{\omega_n\}_{n=0}^{k}$ where $k$ is some positive integer greater than $1$.

$P(\omega_n) = \frac{1}{k} \ \forall n = 1, \dots, k$

$P(\omega_0) = 0$

$Q(\omega_n) = \frac{1}{k+1} \ \forall n = 0, 1, \dots, k$

Here, $P(\omega_0) = 0$ but $Q(\omega_0) = \frac{1}{k+1}$. You can even let $P(\omega_n)$ be something else. Convince yourself.

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Proof of

Equivalence means $Q[{ω}] > 0$ for every $ω$".

$$P \ \text{and} \ Q \ \text{are equivalent}$$

$$\iff$$

$$P << Q \ \text{and} \ Q << P$$

$$\iff$$

$$(P(A) = 0 \iff Q(A) = 0) \forall A \in F$$

$$\iff$$

$$(P(A) > 0 \iff Q(A) > 0) \forall A \in F \tag{*}$$

Now since $\{\omega\} \in F$, choose $A=\{\omega\}$.

We are given that $P(\{\omega\}) > 0$. Thus, if $P$ and $Q$ are equivalent, then by $(*)$, $Q(\{\omega\}) > 0$.

QED

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Discussion of finiteness assumption:

It seems that the finiteness assumption is simply so that we can talk about the individual elements of $\Omega$ having positive probability. However, we can actually relax finite to countable, as $A=\emptyset$ will still be the only $A \in F$ that gives $P(A)=0$, and we can still talk about $P(\omega)$ even with $P(\omega) > 0$.

Another explanation could be that the next part of the text, which I guess is this, is talking about atoms in a finite probability space. It looks like the text is trying to explain the concept of atoms by using a finite probability space as an example: The explanation involves decreasing/increasing $\sigma$-algebras which leads to the next definition which is on filtrations.

Also see Example 1.1.14 in the text.