beatricalwu

2022-07-21

Let $\mathcal{A}$=event space. Assume that $\{{A}_{n}\}$ is a monotone nondecreasing sequence of events in A$\mathcal{A}$. Let $\{{A}_{n}\}$ be a sequence of sets defined as follows:

${B}_{1}={A}_{1};{B}_{i}={A}_{i}-{A}_{i-1}\text{}\text{for}\text{}i\in \{2,3,4,\dots \}$

Prove that $\{{B}_{n}\}$ is a sequence of mutually exclusive events in $\mathcal{A}$. Justify each line of proof.

I know that what I need to show here is ${B}_{i}\cap {B}_{j}=\mathrm{\varnothing}$ for any $i<j$. I've divided it to be proof by cases where case $1$ is $i=1\text{}\text{and}\text{}j\in \{2,3,4,\dots \}$ and case $2$ is $i\in \{2,3,4,\dots \}$ and $j\in \{2,3,4,\dots \}$ where $i<j$ so that ${B}_{i}={A}_{i}-{A}_{i-1}\text{}\text{and}\text{}{B}_{j}={A}_{j}-{A}_{j-1}$ and ${B}_{i}={A}_{i}-{A}_{i-1}\text{}\text{and}\text{}{B}_{j}={A}_{j}-{A}_{j-1}$ where $i\in \{2,3,4\dots \}$ and $j\in \{i+1,i+2,\dots \}$.

${B}_{1}={A}_{1};{B}_{i}={A}_{i}-{A}_{i-1}\text{}\text{for}\text{}i\in \{2,3,4,\dots \}$

Prove that $\{{B}_{n}\}$ is a sequence of mutually exclusive events in $\mathcal{A}$. Justify each line of proof.

I know that what I need to show here is ${B}_{i}\cap {B}_{j}=\mathrm{\varnothing}$ for any $i<j$. I've divided it to be proof by cases where case $1$ is $i=1\text{}\text{and}\text{}j\in \{2,3,4,\dots \}$ and case $2$ is $i\in \{2,3,4,\dots \}$ and $j\in \{2,3,4,\dots \}$ where $i<j$ so that ${B}_{i}={A}_{i}-{A}_{i-1}\text{}\text{and}\text{}{B}_{j}={A}_{j}-{A}_{j-1}$ and ${B}_{i}={A}_{i}-{A}_{i-1}\text{}\text{and}\text{}{B}_{j}={A}_{j}-{A}_{j-1}$ where $i\in \{2,3,4\dots \}$ and $j\in \{i+1,i+2,\dots \}$.

tykoyz

Beginner2022-07-22Added 17 answers

Your idea is correct. As a first step, take an element $e\in {B}_{j}$. You have to prove that $e\notin {B}_{i}$. For this, use the definition ${B}_{j}={A}_{j}-{A}_{j-1}$. Since $e\in {A}_{j}-{A}_{j-1}$, what does this tell you about $e$?

Kenya Leonard

Beginner2022-07-23Added 6 answers

If $i<j$ then ${B}_{i}\subseteq {A}_{i}\subseteq {A}_{j-1}$ and ${A}_{j-1}\cap {B}_{j}=\varnothing $.

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