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Let $A, B$ be two event. My question is as follows:

Will the following relation holds: $$A \to B \Rightarrow \Pr(A) \le\Pr(B) $$

And why?

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Note that events are subsets of your sample space. I suppose what you really mean by $A \to B$ is that $A \subset B$. Then $P(A) \leqslant P(B)$ follows by the monotonicity of the probability measure. – ZulfiqarIII Jan 4 '12 at 15:54
That somebody who asked this can ask that somehow escapes me. – Did Jan 4 '12 at 15:54
up vote 3 down vote accepted

In terms of intuition, the fact that some event $A$ implies some event $B$ means that whenever $A$ happens, $B$ happens. But if the event $B$ happens, we might not have event $A$. So in other words, we have that $\mathbb P(A) \le \mathbb P(B)$ because the probability that $A$ happens is also "the probability that $B$ happens because of $A$", which is less than "the probability that $B$ happens" with no constrains on it. (Note that there are cases with equality, for instance when $A \Rightarrow B$ and $B \Rightarrow A$. )

From a theoretical point of view though (this is a little bit more advanced for a beginner course in probability though), the expression "$A \Rightarrow B$" doesn't make much sense, since the way probabilities are defined is that the "$\mathbb P$" is actually a function from something we call an space of possible events to the interval of real numbers $[0,1]$. The possible events are sets, and the right way to say $A \Rightarrow B$ in this system would be that the set $A$ is included in the set $B$, i.e. $A \subseteq B$. (You need to think about those sets as "a regrouping of possibilities", for instance if the event space is all subsets of $\{1,2,3,4,5,6\}$ in the context where we roll a fair dice, an example of event would be the possibility that "the roll is even" or "the roll is a $1$ or a $4$".) Since in the construction of probabilities, one most common axiom is that a probability function is countably additive, or in other words $$ \mathbb P \left( \bigcup_{i=0}^{\infty} A_i \right) = \sum_{i=0}^{\infty} \,\mathbb P (A_i) $$ when the sets $A_i$ are pairwise disjoint, and another axiom would be that $P(\varnothing) = 0$, we can deduce from that that \begin{align} \mathbb P(B) = \mathbb P( (A \cap B) \cup (A^c \cap B)) & = \mathbb P( (A \cap B) \cup (A^C \cap B) \cup \varnothing \cup \dots) \\ & = \mathbb P(A \cap B) + \mathbb P(A^c \cap B) + \mathbb P (\varnothing) + \mathbb P (\varnothing) + \dots \\ & = \mathbb P(A) + \mathbb P(A^c \cap B) + 0 + 0 + \dots \\ & \ge \mathbb P(A). \end{align}

Note that the reason I added this is because I used the axiom "countably additive" and not "finitely additive". The way to show that countably additive implies finitely additive is by adding plenty of $\varnothing$'s after the finitely many sets.

There are many possible axioms you can add/choose that are equivalent though. I just took my favorite.

Hope that helps,

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Many treatments take $P(\Omega)=1$ to be an axiom and deduce $P(\varnothing) = 0$ as a theorem using countable additivity: $$1=P(\Omega)=P(\Omega\cup\varnothing\cup\varnothing\cdots) = P(\Omega)+\sum P(\varnothing)$$ – Dilip Sarwate Jan 4 '12 at 17:28
Yes, but I would've needed to describe $\Omega$ then, and an appropriate explanation of what $\Omega$ means is that it is the dude from which you build the $\sigma$-algebra over. I didn't wanna get into that. – Patrick Da Silva Jan 4 '12 at 17:34

Yes. Since events are sets of states, $A\implies B$ means $A\subseteq B$, which is equivalent to $B=A\cup(B\backslash A)$, a disjoint union. So $P(B)=P(A)+P(B\backslash A)\geq P(A)$.

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