MY-ASSIGNMENTEXPERT™可以为您提供ucsd.edu Math280A Probability Theory概率论课程的代写代考和辅导服务！
Math280A课程简介
Coursework: There will be weekly homework assignments due on Mondays (starting in Week 2); they are posted below. There will be 5 quizzes, in weeks 1, 3, 5, 7, and 9 of the quarter; they will take place during the scheduled Thursday lecture time, with an alternate sitting available in the late evening to accommodate those in distant time-zones. And there will be a take-home final exam during exam week. Timing and due dates for all courses assessments can be found below.
Prerequisites
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Math280A Probability Theory HELP（EXAM HELP， ONLINE TUTOR）
(Exercise 4.4 in Driver) Let $A, B, C \subseteq \Omega$. Recall that the symmetric difference of sets is $A \triangle B=\left(A \cap B^c\right) \cup\left(B \cap A^c\right)$.
(a) Show that $A \cap C^c \subseteq\left(A \cap B^c\right) \cup\left(B \cap C^c\right)$.
(b) Use part (a) to show that
$$
A \triangle C \subseteq(A \triangle B) \cup(B \triangle C)
$$
(c) Now, let $v: 2^{\Omega} \rightarrow[0, \infty)$ be an outer measure. Show that the function $d: 2^{\Omega} \times 2^{\Omega} \rightarrow$ $[0, \infty)$ defined by $d(A, B)=v(A \triangle B)$ satisfies the triangle inequality:
$$
d(A, C) \leq d(A, B)+d(B, C)
$$
(a) Suppose $x \in A \cap C^c$. Then $x \in A$ and $x \notin C$. There are two cases:
- If $x \notin B$, then $x \in A \cap B^c$, and therefore $x \in (A \cap B^c) \cup (B \cap C^c)$.
- If $x \in B$, then $x \in B \cap C^c$, and therefore $x \in (A \cap B^c) \cup (B \cap C^c)$. Therefore, in either case, $x \in (A \cap B^c) \cup (B \cap C^c)$, which shows that $A \cap C^c \subseteq (A \cap B^c) \cup (B \cap C^c)$.
(b) Using part (a), we have \begin{align*} A \triangle C &= (A \cap C^c) \cup (C \cap A^c) \ &\subseteq (A \cap B^c) \cup (B \cap C^c) \cup (C \cap B^c) \cup (B \cap A^c) \ &= (A \triangle B) \cup (B \triangle C). \end{align*}
(c) We need to show that for any $\$ A, B, C \backslash$ subseteq $\backslash O m e g a \$$,
$$
d(A, C)=v(A \triangle C) \leq v(A \triangle B)+v(B \triangle C)=d(A, B)+d(B, C) .
$$
Let $\$ \backslash$ epsilon $>0 \$$. By definition of outer measure, there exist sets $\$ A_{-} 1, A_{-} 2$, $\backslash$ dots $\$$ and $\$ B_{-} 1, B _2, \backslash$ dots $\$$ such that $\$ A \backslash$ subseteq $\backslash$ bigcup_ ${i=1}^{\wedge} \backslash$ infty A_ $\$ \$$, $\$ B \backslash$ subseteq $\backslash$ bigcup_ ${i=1}^{\wedge} \backslash$ infty $B$ _ $i$, and
$$
v(A)+\epsilon>\sum_{i=1}^{\infty} v\left(A_i\right), \quad v(B)+\epsilon>\sum_{i=1}^{\infty} v\left(B_i\right) .
$$
Let $\mathcal{A}$ be a field over $\Omega$, and let $\mathbb{P}$ be a probability measure on $(\Omega, \sigma(\mathcal{A}))$. Let $B \in \sigma(\mathcal{A})$. Prove that for any $\epsilon>0$, there is a set $A \in \mathcal{A}$ such that $\mathbb{P}(A \triangle B)<\epsilon$. I.e. $\mathcal{A}$ is “dense” in $\sigma(\mathcal{A})$. [Hint: show that the collection of all sets $B$ satisfying this property is a $\sigma$-field.]
Let $\mathcal{B}$ be the collection of all sets $B \in \sigma(\mathcal{A})$ such that for any $\epsilon > 0$, there is a set $A \in \mathcal{A}$ satisfying $\mathbb{P}(A \triangle B) < \epsilon$. We want to show that $\mathcal{B}$ is a $\sigma$-field.
First, note that $\Omega \in \mathcal{A}$, so $\Omega \in \mathcal{B}$. Also, if $B \in \mathcal{B}$, then for any $\epsilon > 0$, there exists $A \in \mathcal{A}$ such that $\mathbb{P}(A \triangle B) < \epsilon$. But then $\mathbb{P}((A \triangle B)^c) = \mathbb{P}(A \cap B^c) + \mathbb{P}(A^c \cap B) \leq 2\epsilon$, since $A$ and $B$ differ only on a set of measure less than $\epsilon$. Thus, $(A \triangle B)^c \in \mathcal{B}$.
Finally, let $(B_n)$ be a sequence of sets in $\mathcal{B}$. For any $\epsilon > 0$, there exist sets $A_n \in \mathcal{A}$ such that $\mathbb{P}(A_n \triangle B_n) < \frac{\epsilon}{2^n}$ for all $n$. Then, \begin{align*} \mathbb{P}\left(\bigtriangleup_{n=1}^\infty B_n\right) &= \mathbb{P}\left(\bigcap_{n=1}^\infty (B_n \setminus \bigcup_{k=1}^{n-1} B_k)\right) \ &= \lim_{n \to \infty} \mathbb{P}(B_n \setminus \bigcup_{k=1}^{n-1} B_k) \ &\leq \lim_{n \to \infty} \mathbb{P}(B_n \triangle A_n) + \mathbb{P}(A_n \triangle \bigcup_{k=1}^{n-1} B_k) \ &\leq \lim_{n \to \infty} \left(\frac{\epsilon}{2^n} + \sum_{k=1}^{n-1} \frac{\epsilon}{2^k}\right) \ &= \epsilon. \end{align*} Therefore, $\bigtriangleup_{n=1}^\infty B_n \in \mathcal{B}$. This shows that $\mathcal{B}$ is a $\sigma$-field.
Since $\mathcal{A}$ is a field and $\sigma(\mathcal{A})$ is the smallest $\sigma$-field containing $\mathcal{A}$, we have $\mathcal{A} \subseteq \mathcal{B} \subseteq \sigma(\mathcal{A})$. But since $\mathcal{B}$ is a $\sigma$-field, it contains $\mathcal{A}$ if and only if it is equal to $\sigma(\mathcal{A})$. Thus, $\mathcal{A}$ is dense in $\sigma(\mathcal{A})$, as desired.
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