Q. 2.14 - A First Course in Probability

Question

Prove Boole’s inequality:

$P\left(\underset{i=1}{\overset{\infty }{\cup }}{A}_i\right)\le \sum _{i=1}^{\infty }P\left(A_i\right)$

Step-by-Step Solution

Answer

Proof by mathematical induction:

Assume that equality stands for some$n$, and it follows that inequality stands for$n+1$.

Step 1 Given Information.

Prove that for all events$A_1 ,A_{2 },...A_n$.

$P\left(\underset{k=1}{\overset{n}{\cup }}{A}_k\right)\le \sum _{k=1}^{n}P\left(A_k\right)$

Step 2 Explanation.

Proof by mathematical induction:

For$n=1$ the statement is true

$P\left(A_1\right)\le P\left(A_1\right)$

Hypothesis:$P\left(\underset{k=1}{\overset{n}{\cup }}{A}_k\right)\le \sum _{k=1}^{n}P\left(A_k\right)$ this is true for any$n$ event.

Then for any $n+1$events:

$P\left(\underset{k=1}{\overset{n+1}{\cup }}{A}_k\right)=P\left[\left(\underset{k=1}{\overset{n}{\cup }}{A}_k\right)\cup A_{n+1}\right] \text{ associative law }$

&$$\begin{gathered} =P\left(\underset{k=1}{\overset{n}{\cup }}{A}_k\right)+P\left(A_{n+1}\right)-P\left[\left(\underset{k=1}{\overset{n}{\cup }}{A}_k\right)\cap A_{n+1}\right] \\ \le \sum _{k=1}^{n}P\left(A_k\right)+P\left(A_{n+1}\right)-P\left[\left(\underset{k=1}{\overset{n}{\cup }}{A}_k\right)\cap A_{n+1}\right] \\ \le \sum _{k=1}^{n+1}P\left(A_k\right)-P\left[\left(\underset{k=1}{\overset{n}{\cup }}{A}_k\right)\cap A_{n+1}\right] \\ \le \sum _{k=1}^{n+1}P\left(A_k\right) \end{gathered}$$ 

 

$P\left(\stackrel{n+1}{\cup }{A}_k\right)\le \sum _{k=1}^{n+1}P\left(A_k\right)$

This proves the statement, by the principle of mathematical induction.