Question

Question: (Requires calculus) Show that if $\mathbf{}{\mathit{E}}_{\mathbf{1}}\mathbf{,}{\mathit{E}}_{\mathbf{2}}\mathbf{,}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{,}{\mathit{E}}_{\mathbf{n}}$ is an infinite sequence of pair wise disjoint events in a sample space S, then$\mathit{p}\left({\cup }_{i=1}^{\infty }{E}_i\right)\mathbf{=}{\mathbf{\sum }}_{\mathbf{i}\mathbf{=}\mathbf{1}}^{\mathbf{\infty }}\mathit{p}\left(E_i\right)$ [ .[Hint: Use Exercise 36 and take limits.]

Short answer

Answer:

It is proved that $p\left({\cup }_{i=1}^{\infty }{E}_i\right)=\sum _{i=1}^{\infty }p\left(E_i\right)$.

Step-by-step solution

Given Information

It is given that that $E_1,E_2,...,E_n$ is a sequence of $n$ pair wise disjoint events in a sample space $S$, here $n$is a positive integer

Definition and formula to be used

Principle of mathematical Induction states that “Let be a property of positive integers such that,

Basic Step: $\mathit{p}\mathbf{\left(}\mathbf{1}\mathbf{\right)}$ is true

Inductive Step: if $\mathit{p}\mathbf{\left(}\mathbf{n}\mathbf{\right)}$is true, then$\mathit{p}\left(n+1\right)$ is true

Then,$\mathit{p}\mathbf{\left(}\mathbf{n}\mathbf{\right)}$is true for all positive integers.”

Apply the principle of mathematical induction

By using mathematical induction prove the result for $n=1$

For $n=1$

$p\left(E_1\right)=p\left(E_1\right)$

For $n=2$

$p\left(E_1{\cup E}_2\right)=p\left(E_1\right)+p\left(E_2\right)-p\left(E_1{\cap E}_2\right)$

As are disjoint events $E_1{\cap E}_2=\varphi$

$p\left(E_1{\cap E}_2\right)=0$

Applying the result in equation

$P\left(E_1{\cup E}_2\right)=P\left(E_1\right)+P\left(E_2\right)$

Thus, the result is true for $n=2$

Assume it is true for $n=k$

$p\left(E_1{\cup E}_2{\cup E}_3...{\cup E}_K\right)=p\left(E_1\right)+p\left(E_2\right)+p\left(E_3\right)+...+p\left(E_K\right)$

Prove the result for $n=k+1$

$$\begin{gathered} p\left(E_1{\cup E}_2{\cup E}_3...{\cup E}_{K+1}\right)=p\left(E_1\right)+p\left(E_2\right)+p\left(E_3\right)+...+p\left(E_{K+1}\right) \\ p\left(E_1{\cup E}_2{\cup E}_3...{\cup E}_K{\cup E}_{K+1}\right)=p\left(\left(E_1{\cup E}_2{\cup E}_3...{\cup E}_K\right)\cup \left(E_{K+1}\right)\right) \end{gathered}$$

Use equation $\left(2\right)$and $\left(3\right)$in the above equation

$p\left(E_1{\cup E}_2{\cup E}_3...{\cup E}_{K+1}\right)=p\left(E_1\right)+p\left(E_2\right)+p\left(E_3\right)+...+p\left(E_K\right)+p\left(E_{K+1}\right)$

It is true for $n=k+1$

So, $p\left({\cup }_{i=1}^n{E}_i\right)=\sum _{i=1}^{n}p\left(E_i\right)$

Here $n$ is a positive integer.

Take limit n→∞  on both sides 

Use limit $n\to \infty$ in equation

$$\begin{gathered} \underset{n\to \infty }{\lim }p\left({\cup }_{i=1}^{\infty }{E}_i\right)=\underset{n\to \infty }{\lim }\sum _{i=1}^{\infty }p\left(E_i\right) \\ p\left({\cup }_{i=1}^{\infty }{E}_i\right)=\sum _{i=1}^{\infty }p\left(E_i\right) \end{gathered}$$

Thus, it is proved that