What follows is a guide of solved exercises where the objective is to demonstrate some useful theorems for probability theory. Try to solve them and then compare your results >:D

1. Prove that P(A^c) = 1 -P(A) and, based on this, make a deduction to argue that P(\emptyset) = 0
   SHOW SOLUTION

   From the definition of probability measure it follows that, if A and B are any measurable events, then it will hold that

   [a]	0\leq P(A) \leq 1
   [b]	A\cap B = \emptyset \rightarrow P(A\cup B) = P(A) + P(B)
   	P(\Omega) = 1

   Now, since A\cap A^c = \emptyset, from part [b], it will hold that:

   [d]	A\cap A^c = \emptyset \rightarrow P(A\cup A^c) = P(A) + P(A^c)

   Since A\cap A^c = \emptyset is always true and A\cup A^c = \Omega, it will then hold that

   1=P(\Omega) = P(A\cup A^c) = P(A) + P(A^c)

   And therefore

   P(A^c) = 1-P(A)

   Which is what was to be proven.

   To prove that P(\emptyset)=0, it is enough to take A=\Omega in the relation just proven and it will hold that

   P(\emptyset) = P(\Omega^c) =1 - P(\Omega) = 1-1 = 0

2. a) Prove that A\subseteq B \rightarrow P(B\setminus A) = P(B) - P(A), is the equality generally true?b) Prove that A\subseteq B \rightarrow P(B)\leq P(A)SHOW SOLUTION a)

   (1)	A\subseteq B; Premise
   	\equiv A\cap B = A
   (2)	B\setminus A = B\cap A^c; Definition of Set Theory
   (3)	B= (B\cap A) \cup (B\cap A^c); Property of sets
   (4)	(B\cap A)\cap (B\cap A^c)=\emptyset; Property of sets
   (5)	P(B)= P[(B\cap A) \cup (B\cap A^c)]; From (3)
   	P(B)= P (B\cap A) + P(B\cap A^c); From (4) + Definition, Probability Measure
   	P(B)= P (B\cap A) + P(B\setminus A); From (2)
   	P(B\setminus A) =P(B) - P (B\cap A)
   	{P(B\setminus A) =P(B) - P (A) }; From (1)

   Therefore {\{A\subseteq B\}\vdash P(B\setminus A) = P(B) - P(A)}.

   Note that this equality is not generally true, as it depends on A\subseteq B being satisfied. From these developments, it can be seen that if this is not satisfied, then it will hold that P(B\setminus A) = P(B) - P(A\cap B).

   SHOW SOLUTION b)

   Based on the reasoning in a) it follows that:

   \{A\subseteq B\}\vdash P(B\setminus A) = P(B) - P(A)

   \equiv \{A\subseteq B\}\vdash P(A) + P(B\setminus A) = P(B)

   Finally, since P is a probability measure, it holds that \forall X (P(X)\geq 0), so, therefore:

   {\{A\subseteq B\}\vdash P(A) \leq P(B)}.

    

3. a) Prove that P(A\cup B) = P(A) + P(B) - P(A\cup B). Accompany the proof with a diagram.b) Using the previous result, prove that P(A\cup B) \leq P(A) + P(B)SHOW SOLUTION PART a)

   [/latex]	A\cup B = (A\triangle B) \cup (A\cap B); Property of sets
   [/latex]	A\triangle B := (A\setminus B) \cup (B\setminus A); Definition of symmetric difference
   [/latex]	(A\triangle B)\cap ( A \cap B) = \emptyset; property of sets
   [/latex]	(A\setminus B)\cap (B\setminus A) = \emptyset; property of sets
   [/latex]	P(A\cup B) = P[(A\triangle B) \cup (A\cap B)]; From (1)
   	P(A\cup B) = P(A\triangle B) + P(A\cap B)]; From (3)
   	P(A\cup B) = P(A\setminus B) + P(B\setminus A) + P(A\cap B)]; From (2,4)
   [/latex]	P(B\setminus A) = P(B) - P(A\cap B); Applying result from exercise 2
   [/latex]	P(A\setminus B) = P(A) - P(A\cap B); Same as (6)
   [/latex]	P(A\cup B) = P(A) - P(A\cap B) + P(B) - P(A\cap B) + P(A\cap B); from (5,6,7)
   	{P(A\cup B) = P(A) + P(B) - P(A\cap B)}.

   SHOW SOLUTION PART b)

   \vdash P(A\cup B) = P(A) + P(B) - P(A\cap B); result from part a)

   \equiv\; \vdash P(A\cup B) + P(A\cap B) = P(A) + P(B)

   \equiv\; \vdash {P(A\cup B) \leq P(A) + P(B)}

    

4. If \{E_n\} is an infinite family of events such that E_1\supseteq E_2 \supseteq E_2 \supseteq \cdots \supseteq E_n \supseteq E_{n+1}\supseteq \cdots. Prove that

   \displaystyle P\left( \bigcap_{n=1}^\infty E_n \right) = \lim_{n\to \infty}P(E_n) SHOW SOLUTION

   (1)	E_n \supseteq E_{n+1}; Hypothesis
   (2)	E_n^c \subseteq E_{n+1}^c; By complementation from (1)
   (3)	\displaystyle(E_n^c \subseteq E_{n+1}^c) \rightarrow P\left( \bigcup_{n=1}^\infty E_n^c \right)= \lim_{n\to\infty}P(E_n^c); Continuity Property
   (4)	\displaystyle \bigcup_{n=1}^\infty E_n^c = \left( \bigcap_{n=1}^\infty E_n \right)^c; DeMorgan’s Laws
   (5)	P\left(E_n^c\right) = 1 - P(E_n); Proven in exercise 1
   (6)	\displaystyle P\left(\left[ \bigcap_{n=1}^\infty E_n\right]^c \right) = \lim_{n\to\infty}[1-P(E_n)] = 1 - \lim_{n\to\infty}P(E_n); from (2,4,5) applied on (3)
   	\displaystyle 1 - P\left( \bigcap_{n=1}^\infty E_n \right) = 1 - \lim_{n\to\infty}P(E_n)
   	{\displaystyle P\left( \bigcap_{n=1}^\infty E_n \right) = \lim_{n\to\infty}P(E_n)}

   \displaystyle\therefore\{E_n \supseteq E_{n+1}\}\vdash P\left( \bigcap_{n=1}^\infty E_n \right) = \lim_{n\to\infty}P(E_n).