Let's have fun on probability theory, here is my first problem set in the said subject.

Problems It was noted that statisticians who follow the deFinetti school do not accept the Axiom of Countable Additivity, instead adhering to the Axiom of Finite Additivity. Show that the Axiom of Countable Additivity implies Finite Additivity.Although, by itself, the Axiom of Finite Additivity does not imply Countable Additivity, suppose we supplement it with the following. Let $A_1\supset A_2\supset\cdots\supset A_n\supset \cdots$ be an infinite sequence of nested sets whose limit is the empty set, which we denote by $A_n\downarrow\emptyset$. Consider the following:

Axiom of Continuity: If $A_n\downarrow\emptyset$, then $P(A_n)\rightarrow 0$

Prove that the Axiom of Continuity and the Axiom of Finite Additivity imply Countable Additivity. Prove each of the following statements. (Assume that any conditioning event has positive probability.) If $P(B)=1$, then $P(A|B)=P(A)$ for any $A$. If $A\…

Problems It was noted that statisticians who follow the deFinetti school do not accept the Axiom of Countable Additivity, instead adhering to the Axiom of Finite Additivity. Show that the Axiom of Countable Additivity implies Finite Additivity.Although, by itself, the Axiom of Finite Additivity does not imply Countable Additivity, suppose we supplement it with the following. Let $A_1\supset A_2\supset\cdots\supset A_n\supset \cdots$ be an infinite sequence of nested sets whose limit is the empty set, which we denote by $A_n\downarrow\emptyset$. Consider the following:

Axiom of Continuity: If $A_n\downarrow\emptyset$, then $P(A_n)\rightarrow 0$

Prove that the Axiom of Continuity and the Axiom of Finite Additivity imply Countable Additivity. Prove each of the following statements. (Assume that any conditioning event has positive probability.) If $P(B)=1$, then $P(A|B)=P(A)$ for any $A$. If $A\…