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In the text "All of Statistics A Concise Course in Statistical Inference" by Larry Wassermann,and I'm inquiring of my solution to $\text{Exercise (1)}$ ? Due to the fact that resolving the exercise requires one to be familiar with the proof it has been provided for the readers convince.

$\text{Exercise (1)}$

Fill in the details of the proof of $\text{Theorem (1.8)}$. Also prove the monotone deceasing case.

$\text{Theorem (1.8) (Continuity of Probabilities).}$ If $A_{n} \rightarrow A$ then

$$ \lim_{n \rightarrow \infty} \big( \mathcal{P}(A_{n}) \big) \rightarrow \mathcal{P}(A)$$

Suppose that $A_{n}$ is monotone increasing so that $A_{1} \subset A_{2} \subset $ Then let$A = \lim_{n \rightarrow \infty} A_{n} = \bigcup_{i=1}^{\infty} A_{i}.$ Define $B_{1} = A_{1}$ ,and also define the set $B_{2} = \big\{ \omega \in \Omega: \omega \in A_{2} , \omega \notin A_{1}\big\} $ $B_{3} = \big\{ \omega \in \Omega: \, \omega \in A_{3}, \omega \notin A_{2}, \omega \notin A_{1} \big\}, ... $ It can be shown that $B_{1}$, $B_{2}$, … are disjoint, $A_{n} = \bigcup_{i=1}^{n}A_{i} = \bigcup_{i=1}^{n}B_{i}$ for each $n$ and $\bigcup_{i=1}^{\infty}B_{i} = \bigcup_{i=1}^{\infty} A_{i}$ $\big( \text{Exercise (1)} \big)$ From Axiom $(3)$, one can say that

$$\mathcal{P(A_{n})} = \mathcal{P} \bigg( \bigcup_{i = 1}^{n} B_{i} \bigg) = \sum_{i = 1} \mathcal{P}(B_{i}) $$

Abusing axiom $(3)$ again,

$$\lim_{n \rightarrow \infty} \mathcal{P}(A_{n}) = \lim_{n \rightarrow \infty} \sum_{i = 1}^{n} \mathcal{P}(B_{i}) = \sum_{i = 1}^{ \infty} \mathcal{P}(B_{i}) = \mathcal{P} \bigg( \bigcup_{i=1}^{\infty} B_{i} \bigg) = \mathcal{P}(A).$$

$\text{Solution}$

$\text{Proof of the Monotone Increasing Case}$

Recall from the proof of $(1.8)$ that the author defines $B_{1} = A_{1}$, and also the respective sets $B_{2}$, $B_{3}$. The author claims that $B_{1}, B_{2}, ...$ are disjoint. To verify this one can observe that $\big(A_{i} \big)_{i \in I}$ is a family of sets, it's easy to see that

$$\bigcap_{i\in I} A_{i} = \emptyset.$$

The author also claims that for each $n$.

$(1)$

$$A_{n} = \bigcup_{i=1}^{n}A_{i} = \bigcup_{i=1}^{n}B_{i}$$

and he also claims that,

$(2)$

$$B_{n}= \bigcup_{i=1}^{\infty}B_{i} = \bigcup_{i=1}^{\infty} A_{i}$$

To prove $(1)$ and $(2)$ take,

$$ \lim_{ n \rightarrow \infty} (A_{n}) = \lim_{ n \rightarrow \infty} \bigg(\bigcup_{i=1}^{n}A_{i} \bigg) = \lim_{ n \rightarrow \infty} \bigg( \bigcup_{i=1}^{n}B_{i} \bigg) \rightarrow A. $$

As well as,

$$\lim_{ n \rightarrow \infty} (B_{n}) = \lim_{ n \rightarrow \infty} \bigg(\bigcup_{i=1}^{n}B_{i} \bigg) = \lim_{ n \rightarrow \infty} \bigg( \bigcup_{i=1}^{n}A_{i} \bigg) \rightarrow B .$$

QED.

$\text{Proof of the monotone decreasing case }$

Suppose that $A_{n}$ is monotone decreasingso that $A_{1} \subset A_{2} \subset $ Then let$A = \lim_{n \rightarrow \infty} A_{n} = \bigcap_{i=1}^{\infty} A_{i}.$ Define $B_{1} = A_{1}$ ,and also define the set $B_{2} = \big\{ \omega \in \Omega: \omega \in A_{2} , \omega \notin A_{1}\big\} $ $B_{3} = \big\{ \omega \in \Omega: \, \omega \in A_{3}, \omega \notin A_{2}, \omega \notin A_{1} \big\}, ... $ It can be shown that $B_{1}$, $B_{2}$, … are disjoint, $A_{n} = \bigcap_{i=1}^{n}A_{i} = \bigcap_{i=1}^{n}B_{i}$ for each $n$ and $\bigcap_{i=1}^{\infty}B_{i} = \bigcap_{i=1}^{\infty} A_{i}$ $\big( \text{Exercise (1)} \big)$ From Axiom $(3)$, one can say that

$$\mathcal{P(A_{n})} = \mathcal{P} \bigg( \bigcap_{i = 1}^{n} B_{i} \bigg) = \sum_{i = 1} \mathcal{P}(B_{i}) $$

Abusing $\text{Axiom (3)}$ again,

$$\lim_{n \rightarrow \infty} \mathcal{P}(A_{n}) = \lim_{n \rightarrow \infty} \sum_{i = 1}^{n} \mathcal{P}(B_{i}) = \sum_{i = 1}^{ \infty} \mathcal{P}(B_{i}) = \mathcal{P} \bigg( \bigcap_{i=1}^{\infty} B_{i} \bigg) = \mathcal{P}(A). $$

Recall from the proof of $(1.8)$ that the author defines $B_{1} = A_{1}$, and also the respective sets $B_{2}$, $B_{3}$. The author claims that $B_{1}, B_{2}, ...$ are disjoint. To verify this one can observe that $\big(A_{i} \big)_{i \in I}$ is a family of sets, it's easy to see that

$$\bigcap_{i\in I} A_{i} = \emptyset.$$

The author also claims that for each $n$.

$(1)$

$$A_{n} = \bigcap_{i=1}^{n}A_{i} = \bigcap_{i=1}^{n}B_{i}$$

and he also claims that,

$(2)$

$$B_{n}= \bigcap_{i=1}^{\infty}B_{i} = \bigcap_{i=1}^{\infty} A_{i}$$

To prove $(1)$ and $(2)$ take,

$$ \lim_{ n \rightarrow \infty} (A_{n}) = \lim_{ n \rightarrow \infty} \bigg(\bigcap_{i=1}^{n}A_{i} \bigg) = \lim_{ n \rightarrow \infty} \bigg( \bigcap_{i=1}^{n}B_{i} \bigg) \rightarrow A. $$

As well as,

$$\lim_{ n \rightarrow \infty} (B_{n}) = \lim_{ n \rightarrow \infty} \bigg(\bigcap_{i=1}^{n}B_{i} \bigg) = \lim_{ n \rightarrow \infty} \bigg( \bigcap_{i=1}^{n}A_{i} \bigg) \rightarrow B .$$

QED.

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1 Answer 1

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it's easy to see that $$\bigcap_{i\in I} A_{i} = \emptyset.$$

This is false. Since $A_i$ is increasing, $\bigcap_{i=1}^\infty A_i = A_1$. Unless $A_1 = \emptyset$, your statement is not true. Fortunately, it was also pointless, as you do not use it anywhere.

he also claims that,

(2) $$B_{n}= \bigcup_{i=1}^{\infty}B_{i} = \bigcup_{i=1}^{\infty} A_{i}$$

This is not equal to $B_n$ for any $n$. It is also not equal to "$B$" (as you use later), as you have never defined a set $B$. Nor do you have any need to.

To prove (1) and (2)

take, $$\lim_{ n \to \infty} (A_{n}) = \lim_{ n \to\infty} \bigg(\bigcup_{i=1}^{n}A_{i} \bigg) = \lim_{ n \to\infty} \bigg( \bigcup_{i=1}^{n}B_{i} \bigg) \to A.$$

You are also massively mixing up notations here. Note:

  • $\{A_n\}_{n=1}^{\infty}$ is a sequence (of sets in this case).
  • If the sequence converges, $\lim_{n\to\infty} A_n$ is the value that the sequence converges to. It is NOT the sequence. It is a single value. Therefore, saying $\lim_{n\to\infty} A_n \to A$ makes as much sense as saying $2 \to 2$. Sequences converge to something. Single values are just that one value. The following two statements are equivalent: $$\lim_{n\to\infty} A_n = A$$ and $$A_n \to A$$ But $\lim_{n\to\infty} A_n \to A$ is nonsensical.
  • Despite the presence of $\infty$ in the notation, $$\bigcup_{i=1}^\infty A_i := \{x \mid \exists i, x \in A_i\}$$ is not defined as a limit. Part of what you need to prove is that $\lim_{n\to\infty} \bigcup_{i=1}^n A_i = \bigcup_{i=1}^\infty A_i$
  • You are making use of (1) here in an attempt to prove (2), but you haven't offered anything at all towards a proof of (1).

As well as, $$\lim_{ n \rightarrow \infty} (B_{n}) = \lim_{ n\to \infty} \bigg(\bigcup_{i=1}^{n}B_{i} \bigg) = \lim_{ n \to \infty} \bigg( \bigcup_{i=1}^{n}A_{i} \bigg) \to B.$$

$B_n \ne \bigcup_{i=1}^n B_i$ and I've already mentioned that there is no set "$B$". This adds nothing to your proof.

Proof of the monotone decreasing case

Suppose that $A_n$ is monotone decreasing so that $A_1\subset A_2\subset$

$A_1\subset A_2\subset\ ...$ is the definition of monotone increasing. Decreasing would be $A_1\supset A_2\supset\ ...$

The rest of this is just a copy-paste of the first part with unions converted to intersections. It makes no sense. Since $A_1$ is the superset of all later sets, all of your $B_i$ in this case are empty, except $B_1$.

Finally, even if your proofs were correct, you have only examined the cases where $A_n$ is monotone. What if $A_n$ is neither increasing nor decreasing?

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  • $\begingroup$ To fix the second issue you mentioned would it be tentative to define $A_{n} = B_{n}$ ? $\endgroup$
    – Zophikel
    Jun 2, 2019 at 23:27
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    $\begingroup$ No. $A_n$ and $B_n$ are already defined in the original proof outline. You should not change those definitions. And there is no reason to. You don't need to label this set anything other than the expression. There is no need for $B_n =$ in that statement (or anything else). These are sets. To prove sets are equal, you show that every element of one is also in the other, and vice versa. $\endgroup$ Jun 3, 2019 at 1:17
  • $\begingroup$ Ahh ok I understand where I missed up, for the last part the original problem said only to inspect the case where $A_{n}$ is monotone decreasing $\endgroup$
    – Zophikel
    Jun 3, 2019 at 13:23
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    $\begingroup$ That's good. The problem with these proof critiques is that they always come off so negative. But I actually want to encourage you, not discourage. However, in order to learn how to do it right, you've got to break away from the things you are doing wrong. $\endgroup$ Jun 3, 2019 at 15:45
  • $\begingroup$ I'll try to see if I can go back through the proof and reattempt it I'm reworking through the definitions. I might try some other easier problems before coming back to this one. $\endgroup$
    – Zophikel
    Jun 3, 2019 at 17:16

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