### 统计代写|概率论作业代写Probability and Statistics代考5CCM241A|Conditional probability

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• Statistical Inference 统计推断
• Statistical Computing 统计计算
• (Generalized) Linear Models 广义线性模型
• Statistical Machine Learning 统计机器学习
• Longitudinal Data Analysis 纵向数据分析
• Foundations of Data Science 数据科学基础

## 统计代写|概率论作业代写Probability and Statistics代考5CCM241A|The formula for multiplying probabilities

Let a sample space $(\Omega, \mathcal{F}, P)$ be given. Let’s consider the following problem: if it is known that an event $B \in \mathcal{F}$ occurred $(P(B)>0)$, then, by using this additional information, how to find the probability of some (different from $B$ ) event $A \in \mathcal{F}$ ?

Under these conditions, it is natural to regard not $\Omega$ but $B$ as the space of elementary events, since the fact that $B$ occurred means that we are talking only about those elementary events that belong to $B$. In general case, an event $A B$ implies $B(A B \subseteq B)$, but if it is known that an event $B$ occured, then (under this condition) those and only those elementary events that belong to $A B$, imply $A$. Since we identify the event $A$ and the set of outcomes that lead to the event $A$, we now need to identify the event $A$ with the event $A_{s}=A B$. We can say that an event (set) $A_{\Delta}=A B$ is an event $A$, viewed from the point of view, according to which the space of elementary events is an event $B$.
In the new sample space, the $\sigma$-algebra of events $\mathcal{F}{B}$ is defined (or, as they say, induced) by the $\sigma$-algebra of events $\mathcal{F}$ (namely, $\mathcal{F}{B}$ consists of events of the form $\left.A_{B}=A B\right)$. Check, that $\mathcal{F}{B}$ is really a $\sigma$-algebra, i.e. $\mathcal{F}{B}$ satisfies the conditions $\mathbf{A 1}, \mathbf{A} \mathbf{2}^{\prime}$,
A3 from $\S$ 1, point 1.1: $\mathcal{F}{B}=\left{A{B}=A B: A \in \mathcal{F}\right}$.
Note that $\mathcal{F}{B}$ is called a $\sigma$-algebra generated by an event $A$. Let’s define on $\left(B, \mathcal{F}{B}\right)$ a function $P_{B}$ through the probabilistic function $P$ as follows: for $A_{B} \in \mathcal{F}{B}$ we put $$P{B}\left(A_{B}\right)=\frac{P\left(A_{B}\right)}{P(B)}=\frac{P(A B)}{P(B)} .$$
It follows from this definition (1) that for any event $A_{B} \in \mathcal{F}{B}$ its probability $P{B}\left(A_{B}\right) \geq 0, P_{B}(B)=1$ and for $A_{B}^{i}=A^{i} B \in \mathcal{F}, A^{i} B \cap A^{j} B=\varnothing(i \neq j)$
$$P_{B}\left(\sum_{i=1}^{\infty} A^{i} B\right)=\sum_{i=1}^{\infty} P_{B}\left(A^{i} B\right)$$
The function $P_{B}$ thus introduced satisfies all the axioms $\mathbf{P} 1, \mathbf{P} 2, \mathbf{P} 3$ of the probabilistic function, hence is a probability (probabilistic function) on $\left(B, \mathcal{F}_{B}\right)$.

So, we built a new probability space $\left(B, \mathcal{F}{B}, P{B}\right)$ by the event $B \in \mathcal{F}, P(B)>0$.
This sample space is called the probability space generated by the event $B$.
We will explain the meaning of the probability $P_{B}(\cdot)$ with the help of the classical definition of probability.
In this case $\Omega=\left{\omega_{1}, \omega_{2}, \ldots, \omega_{n}\right},|\Omega|=n<\infty$ and
$$P\left(\omega_{i}\right)=\frac{1}{n}=\frac{1}{|\Omega|} . \quad i=1,2, \ldots, n$$

## 统计代写|概率论作业代写Probability and Statistics代考5CCM241A|Independence

The concept of independence of two or more events (or trials), in a certain sense, occupies a central place in the probability theory. From the mathematical point of view, this concept defined the uniqueness that distinguishes the probability theory in the general theory, dealing with the study of measurable spaces with measure. We should also note that one of the founders of the probability theory, the outstanding scientist A. Kolmogorov, paid special attention to the fundamental nature of the concept of independence in the probability theory in the thirties of the last century (see [12]) A. Kolmogorov, Basic concepts of the probability theory. – Moscow, ed. «Science», 1974).

Below we first dwell on the concepts of independence of events, after extending this notion to partitions and to the algebra of sets, in conclusion we will consider the independence of trials and $\sigma$-algebras.

## 统计代写|概率论作业代写Probability and Statistics代考5CCM241A|Independence of events

Let a sample space $(\Omega, \mathcal{F}, P)$ and events $A, B \in \mathcal{F}$ be given. If a conditional probability of an event $A$ under the condition that an event $B(P(B)>0)$ occurred is equal to (unconditional) probability of an event $A$, i.e.
$$P(A / B)=P(A),$$
then it is natural to assume that the event $A$ does not depend on the event $B$. If this is so (i.e, $\left(3^{\prime}\right.$ ) takes place), then from the formula of conditional probability (3) we obtain the formula
$$P(A B)=P(A) P(B)$$

Now let $P(A)>0$ and condition (3′) be satisfied. Then, in view of the fact that (4) holds, we obtain
$$P(B / A)=\frac{P(B A)}{P(A)}=\frac{P(B) P(A)}{P(A)}=P(B),$$
i.e. the event $B$ does not depend on the event $A$.
From what has been said, we come to the following conclusion: the concept of independence of two events is a symmetric concept – if the event $A$ does not depend on the event $B$, then the event $B$ does not depend on the event $A$. But the above formulas $\left(3^{\prime}\right)$, (5) have one drawback – they require the condition of strict positiveness of the probabilities that stand in the denominators in formulas (3′) and (5): $P(A)>0$ or $P(B)>0$. But, as we noted earlier ( $\S$, point $1.2$ ), the event can have a probability of 0 (zero), but it can happen. Therefore, in the formulas (3′) and (5), the requirements $P(A)>0$ or $P(B)>0$ restrict the domains of applicability of these formulas and the concept of independence of events. Therefore, relation (4), which is a consequence of definitions (3′) and (5), but which does not require conditions $P(A)>0$ or $P(B)>0$, is taken for the definition of independence.

Definition 2. If the probability of the product of events $A$ and $B$ is equal to the product of the probabilities of events $A$ and $B$, i.e. if relation (4) is satisfied, then the events $A$ and $B$ are called independent events.

We obtain from the definition, that if $P(A)=0$, then for any $B, P(A B)=0=$ $=P(A) P(B)$ (because $A B \subseteq A$, therefore $0 \leq P(A B) \leq P(A)=0$, i.e. $P(A B)=0$ ) i.e. (4) takes place. In other words, if $P(A)=0$, then $A$ and any event $B$ are independent.
We now formulate several assertions related to independence in the form of a theorem.

Theorem 2. a) If $P(B)>0$, then independence of events $A$ and $B$, i.e. ratio (4), is equivalent to condition $P(A / B)=P(A)$.
b) If $A$ and $B$ are independent events, then events $\bar{A}, B(A, \bar{B})$ and $\bar{A}, \bar{B}$ are also independent events;
c) If $P(A)=0$ or $P(A)=1$, then $A$ and any event $B$ are independent;
d) If $A$ and $B_{1}$ are independent events, $A$ and $B_{2}$ are independent events, while $B_{1} B_{2}=\varnothing$, then $A$ and $B_{1}+B_{2}$ are independent events.

Proof. a) In this case (3′) implies (4) (we saw this above). If, however, (4) holds, then
$$P(A / B)=\frac{P(A B)}{P(B)}=\frac{P(A) P(B)}{P(B)}=P(A)$$
i.e. the formula (3′) is correct.
b) It suffices to show that condition (4) implies relations
$$P(\bar{A} B)=P(\bar{A}) P(B), \quad P(\bar{A} \bar{B})=P(\bar{A}) P(\bar{B}) .$$

## 统计代写|概率论作业代写Probability and Statistics代考5CCM241A|The formula for multiplying probabilities

A3 从§§1、点1.1：\mathcal{F}{B}=\left{A{B}=A B: A \in \mathcal{F}\right}\mathcal{F}{B}=\left{A{B}=A B: A \in \mathcal{F}\right}.

## 统计代写|概率论作业代写Probability and Statistics代考5CCM241A|Independence of events

b) 如果一种和乙是独立事件，然后是事件一种¯,乙(一种,乙¯)和一种¯,乙¯也是独立事件；
c) 如果磷(一种)=0或者磷(一种)=1， 然后一种和任何事件乙是独立的；
d) 如果一种和乙1是独立事件，一种和乙2是独立事件，而乙1乙2=∅， 然后一种和乙1+乙2是独立事件。

b) 证明条件 (4) 蕴含关系就足够了

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