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I have a probability space $(\Omega, \mathcal{P}(\Omega), P)$. I want to know the probability of the empty set $\{\}$.

Intuitively, I would say this probability is zero. It certainly is for the uniform distribution because there $P(A) = \frac{|A|}{|\Omega|}$.

How do I see this for arbitrary probability measures? I tried $\Omega = \Omega \cup \{\}$ but by $1 = P(\Omega) = P(\Omega \cup \{\}) = P(\Omega) \cdot P(\{\})$ I am led to $P(\{\}) = 1$ which seems very odd.

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  • $\begingroup$ $P(\Omega\cup\{\})=P(\Omega)+P(\{\})$. Addition in stead of multiplication. You are dealing with the union of two disjoint sets. Not the intersection of two independent sets. $\endgroup$ – drhab Dec 18 '14 at 16:32
  • $\begingroup$ In the above (last line), when taking a disjoint union, you get a sum, not a product (product is for intersection of independent events). $\endgroup$ – Clement C. Dec 18 '14 at 16:32
  • $\begingroup$ Thanks, how could I overlook this ;-) $\endgroup$ – Marc Dec 18 '14 at 16:36
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Hint: Note that $ A\cup \emptyset = A$, thus:

$P(A\cup \emptyset) = P(A)+P(\emptyset)+P(A\cap\emptyset) = P(A)$, now $P(A)>0 \implies ?$

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Hint: $$P(\emptyset)=P(\emptyset\cup\emptyset)=P(\emptyset)+P(\emptyset)$$ This because $\emptyset\cap\emptyset=\emptyset$, i.e. the sets are disjoint.

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  • $\begingroup$ What about the other direction: If $P(A)=0$, can I infer $A = \{\}$? That's more a question about conventions because if $P(\{w\})=0$ for a $\omega \in \Omega$, it isn't "needed". But I guess such elements are allowed, aren't they? $\endgroup$ – Marc Dec 18 '14 at 16:41
  • $\begingroup$ @Marc no you cannot. There are events with probability $0$ that are not empty. For example $P(X=2),X\sim Normal$ $\endgroup$ – user76844 Dec 18 '14 at 17:11
  • $\begingroup$ Ah, ok. I hadn't thought about continuous distributions. So elements with probability zero are allowed. $\endgroup$ – Marc Dec 18 '14 at 17:37
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For any measure space $(X,\mathcal M,\mu)$, $\mu(\varnothing)=0$ by definition. Since a probability measure is just a special case where $\mu(X)=1$, we still have $\mu(\varnothing)=0$.

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