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If we assign probabilities to every elementary event such that they sum up to 1 and they're all positive, can we say that this function is a probability function with its domain equal to the set of all subsets of the union of those elementary events? Or should we also specify that P(A or B) = P(A) + P(B) for all A, B where A and B=Null set?

Edit:

$\Omega = \{\omega_1,\omega_2,\omega_3\}$

Let $f:\Omega \rightarrow[0,1]$ such that:

$f(\omega_1)=0.2$, $f(\omega_2)=0.3$, $f(\omega_3)=0.5$.

Let $F:2^\Omega\rightarrow[0,1]$ such that:

$F(\{\omega_i\})=f(\omega_i), i\in \{1,2,3\}$,

$F(\{\omega_1,\omega_2\})=f(\omega_1)+f(\omega_2)$

$F(\{\omega_1,\omega_3\})=f(\omega_1)+f(\omega_3)$

$F(\{\omega_2,\omega_3\})=f(\omega_2)+f(\omega_3)$

$F(\{\omega_1,\omega_2\,\omega_3\})=f(\omega_1)+f(\omega_2)+f(\omega_3)$

$F(\emptyset)=0$

Then we can say that F is a probability function, right? Can we say that this is the unique probability function generated by f (not speaking formally)?

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    $\begingroup$ Could you elaborate a little bit more? I don't understand what exactly is your question is about... $\endgroup$ – Tim Sep 28 '17 at 12:04
  • $\begingroup$ Sure. I edited my post. $\endgroup$ – user_anon Sep 28 '17 at 13:43
  • $\begingroup$ I think that by "probability function" you mean "probability measure". See en.wikipedia.org/wiki/Probability_function $\endgroup$ – Kodiologist Sep 28 '17 at 14:39
  • $\begingroup$ Yes. You're right $\endgroup$ – user_anon Sep 28 '17 at 14:43
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    $\begingroup$ This is the discrete sigma algebra: it's the largest sigma algebra that can be defined for $\Omega$. Note that when all the probabilities are positive, $\Omega$ must be finite or only countably infinite. Your question otherwise is unclear, because you have defined $F$ completely: it doesn't "generate" anything. Perhaps you are asking whether $F$ is uniquely determined by its values $F(\{\omega\})=f(\omega)$ on singleton sets? If so, your formulas make it clear that it is. There seems to be nothing left to add. $\endgroup$ – whuber Sep 28 '17 at 15:31
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If the number of elements in $\Omega$ is finite or countably infinite, then assigning probability $p_i > 0$ to element $\omega_i$ such that $$\sum_i p_i =\sum_i P(\omega_i)= 1\tag{1}$$ and defining that for any subset $A$ of $\Omega$, $$P(A) = \sum_{\omega \in A} P(\omega)\tag{2}$$ makes $P(\cdot)$ a probability measure meaning a map from $2^\Omega$ to $[0,1]$. Without $(2)$, you cannot extend $P(\cdot)$ to assign probability to anything other than the elementary events. You do need to assert that $(2)$ holds in addition to $(1)$ in order to claim that $P(\cdot)$ is a probability measure. You cannot extend $P(\cdot)$ as defined in $(1)$ to create a different map from $2^\Omega$ to $[0,1]$; the map specified by $(2)$ is unique.

If the number of elements in $\Omega$ is uncountable, then the assignment $(1)$ cannot hold and the whole enterprise is moot.

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