Answer to Question #131323 in Statistics and Probability for Ajay

Question

Proof Boole's inequality
for n events A1,A2,...An.

Answer 1

Let $B_i=A_i-\cup_{k=1}^{i-1}A_k$

then $B_i\subseteq A_i\ and\ P(B_i)\leq P(A_i).$

Let $i\neq j$ , without loss of generality, assume $i>j$, then

$B_i\cap B_j=(A_i-\cup_{k=1}^{i-1}A_k)\cap (A_j-\cup_{k=1}^{j-1}A_k)\subseteq\\ \subseteq(A_i-\cup_{k=1}^{i-1}A_k)\cap A_j,$

since $j<i$, then $A_j\subseteq \cup_{k=1}^{i-1}A_k$, hence $(A_i-\cup_{k=1}^{i-1}A_k)\cap A_j=\empty,$ and

$B_i\cap B_j=\empty.$

Since $B_i\cap B_j=\empty$, then, by countable additivity,

$P(\cup_{k=1}^{n}B_k)=\sum_{k=1}^nP(B_k)$.

$\cup_{k=1}^{n}B_k=\\ A_1\cup (A_2-A_1)\cup(A_3-\cup_{k=1}^{2}A_k)\cup...\\ ...\cup (A_n-\cup_{k=1}^{n-1}A_k)=\\ (A_1\cup A_2)\cup (A_3-\cup_{k=1}^{2}A_k)\cup...\\ ...\cup (A_n-\cup_{k=1}^{n-1}A_k)=\\ (A_1\cup A_2\cup A_3)\cup... \cup (A_n-\cup_{k=1}^{n-1}A_k)=...\\ =(A_1\cup A_2\cup...\cup A_{n-1}\cup (A_n-\cup_{k=1}^{n-1}A_k)=\\ \cup_{k=1}^{n}A_k)$

therefore,

$\cup_{k=1}^{n}B_k=\cup_{k=1}^{n}A_k\ and\ \\ P(\cup_{k=1}^{n}A_k)=P(\cup_{k=1}^{n}B_k)=\\ \sum_{k=1}^nP(B_k)\leq \sum_{k=1}^nP(A_k)$,

thus,

$P(\cup_{k=1}^{n}A_k) \leq \sum_{k=1}^nP(A_k).$