### 统计代写|随机信号处理作业代写Statistical Signal Processing代考|Independence

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• Statistical Inference 统计推断
• Statistical Computing 统计计算
• Advanced Probability Theory 高等概率论
• Advanced Mathematical Statistics 高等数理统计学
• (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|随机信号处理作业代写Statistical Signal Processing代考|Independence

Given a probability space $(\Omega, F, P)$, two events $F$ and $G$ are defined to be independent if $P(F \cap G)=P(F) P(G)$. A collection of events $\left{F_{i} ; i=\right.$ $0,1, \ldots, k-1}$ is said to be independent or mutually independent if for any distinct subcollection $\left{F_{I_{i}} ; i=0,1, \ldots, m-1\right}, l_{m} \leq k$, we have that
$$P\left(\prod_{i=0}^{m-1} F_{l_{i}}\right)=\prod_{i=0}^{m-1} P\left(F_{l_{i}}\right) .$$
In words: the probability of the intersection of any subcollection of the given events equals the product of the probabilities of the separate events. Unfortunately it is not enough to simply require that $P\left(\bigcap_{i=0}^{k-1} F_{i}\right)=\prod_{i=0}^{k-1} P\left(F_{i}\right)$

as this does not imply a similar result for all possible subcollections of events, which is what will be needed. For example, consider the following case where $P(F \cap G \cap H)=P(F) P(G) P(H)$ for three events $F$, $G$, and $H$, yet it is not true that $P(F \cap G)=P(F) P(G)$
\begin{aligned} P(F) &=P(G)=P(H)=\frac{1}{3} \ P(F \cap G \cap H) &=\frac{1}{27}=P(F) P(G) P(H) \ P(F \cap G) &=P(G \cap H)=P(F \cap H)=\frac{1}{27} \neq P(F) P(G) . \end{aligned}
The example places zero probability on the overlap $F \cap G$ except where it also overlaps $H$, i.e., $P\left(F \cap G \cap H^{c}\right)=0$. Thus in this case $P(F \cap G \cap H)=$ $P(F) P(G) P(H)=1 / 27$, but $P(F \cap G)=1 / 27 \neq P(F) P(G)=1 / 9$.

The concept of independence in the probabilistic sense we have defined relates easily to the intuitive idea of independence of physical events. For example, if a fair die is rolled twice, one would expect the second roll to be unrelated to the first roll because there is no physical connection between the individual outcomes. Independence in the probabilistic sense is reflected in this experiment. The probability of any given outcome for either of the individual rolls is $1 / 6$. The probability of any given pair of outcomes is $(1 / 6)^{2}=1 / 36$ – the addition of a second outcome diminishes the overall probability by exactly the probability of the individual event, viz., $1 / 6$. Note that the probabilities are not added – the probability of two successive outcomes cannot reasonably be greater than the probability of either of the outcomes alone. Do not, however, confuse the concept of independence with the concept of disjoint or mutually exclusive events. If you roll the die once, the event the roll is a one is not independent of the event the roll is a six. Given one event, the other cannot happen they are neither physically nor probabilistically independent. These are mutually exclusive events.

## 统计代写|随机信号处理作业代写Statistical Signal Processing代考|Elementary Conditional Probability

Intuitively, independence of two events means that the occurrence of one event should not affect the occurrence of the other. For example, the knowledge of the outcome of the first roll of a die should not change the probabilities for the outcome of the second roll of the die if the die has no memory. To be more precise, the notion of conditional probability is required. Consider the following motivation. Suppose that $(\Omega, \mathcal{F}, P)$ is a probability space and that an observer is told that an event $G$ has already occurred. The

observer thus has a posteriori knowledge of the experiment. The observer is then asked to calculate the probability of another event $F$ given this information. We will denote this probability of $F$ given $G$ by $P(F \mid G)$. Thus instead of the a priori or unconditional probability $P(F)$, the observer must compute the a posteriori or conditional probability $P(F \mid G)$, read as “the probability that event $F$ occurs given that the event $G$ occurred.” For a fixed $G$ the observer should be able to find $P(F \mid G)$ for all events $F$, thus the observer is in fact being asked to describe a new probability measure, say $P_{G}$, on $(\Omega, \mathcal{F})$. How should this be defined? Intuition will lead to a useful definition and this definition will indeed provide a useful interpretation of independence.

First, since the observer has been told that $G$ has occurred and hence $\omega \in G$, clearly the new probability measure $P_{G}$ must assign zero probability to the set of all $\omega$ outside of $G$, that is, we should have
$$P\left(G^{c} \mid G\right)=0$$
or, equivalently,
$$P(G \mid G)=1 .$$
Eq. (2.91) plus the axioms of probability in turn imply that
$$P(F \mid G)=P\left(F \cap\left(G \cup G^{c}\right) \mid G\right)=P(F \cap G \mid G) .$$
Second, there is no reason to suspect that the relative probabilities within $G$ should change because of the conditioning. For example, if an event $F \subset G$ is twice as probable as an event $H \subset G$ with respect to $P$, then the same should be true with respect to $P_{G}$. For arbitrary events $F$ and $H$, the events $F \cap G$ and $H \cap G$ are both in $G$, and hence this preservation of relative probability implies that
$$\frac{P(F \cap G \mid G)}{P(H \cap G \mid G)}=\frac{P(F \cap G)}{P(H \cap G)} .$$
But if we take $H=\Omega$ in this formula and use (2.92)-(2.93), we have that
$$P(F \mid G)=P(F \cap G \mid G)=\frac{P(F \cap G)}{P(G)},$$
which is in fact the formula we now use to define the conditional probability of the event $F$ given the event $G$. The conditional probability can be interpreted as “cutting down” the original probability space to a probability space with the smaller sample space $G$ and with probabilities equal to the renormalized probabilities of the intersection of events with the given event $G$ on the original space.

## 统计代写|随机信号处理作业代写Statistical Signal Processing代考|Problems

1. Suppose that you have a set function $P$ defined for all subsets $F \subset \Omega$ of a sample space $\Omega$ and suppose that you know that this set function satisfies (2.7-2.9). Show that for arbitrary (not necessarily disjoint) events.
$$P(F \cup G)=P(F)+P(G)-P(F \cap G) .$$
2. Describe the sigma-field of subsets of $\Re$ generated by the points or singleton sets. Does this sigma-field contain intervals of the form $(a, b)$ for $b>a$ ?
3. Given a finite subset $A$ of the real line $\Re$, prove that the power set of $A$ and $\mathcal{B}(A)$ are the same. Repeat for a countably infinite subset of $\Re$.
4. Given that the discrete sample space $\Omega$ has $n$ elements, show that the power set of $\Omega$ consists of $2^{n}$ elements.
5. ${ }^{*}$ Let $\Omega=\Re$, the real line, and consider the collection $\mathcal{F}$ of subsets of $\Re$ defined as all sets of the form
$$\bigcup_{i=0}^{k}\left(a_{i}, b_{i}\right] \cup \bigcup_{j=0}^{m}\left(c_{j}, d_{j}\right]^{e}$$
for all possible choices of nonnegative integers $k$ and $m$ and all possible choices of real numbers $a_{i}<b_{i}, c_{i}<d_{i}$. If $k$ or $m$ is 0 , then the respective unions are defined to be empty so that the empty set itself has the form given. In other words, $\mathcal{F}$ contains all possible finite unions of half-open intervals of this form and complements of such half-open intervals. Every set of this form is in $\mathcal{F}$ and every set in $\mathcal{F}$ has this form. Prove that $\mathcal{F}$ is a field of subsets of $\Omega$. Does $\mathcal{F}$ contain the points? For example, is the singleton set ${0}$ in $\mathcal{F}$ ? Is $\mathcal{F}$ a sigma-field?
6. Let $\Omega=[0, \infty)$ be a sample space and let $\mathcal{F}$ be the sigma-field of subsets of $\Omega$ generated by all sets of the form $(n, n+1)$ for $n=1,2, \ldots$

## 统计代写|随机信号处理作业代写Statistical Signal Processing代考|Problems

1. 假设你有一个集合函数磷为所有子集定义F⊂Ω一个样本空间Ω并假设你知道这个集合函数满足（2.7-2.9）。证明对于任意（不一定是不相交的）事件。
磷(F∪G)=磷(F)+磷(G)−磷(F∩G).
2. 描述子集的 sigma-fieldℜ由点或单例集生成。这个 sigma-field 是否包含形式的间隔(一种,b)为了b>一种 ?
3. 给定一个有限子集一种实线的ℜ，证明幂集一种和乙(一种)是相同的。重复一个可数无限的子集ℜ.
4. 鉴于离散样本空间Ω拥有n元素，表明幂集Ω由组成2n元素。
5. ∗让Ω=ℜ，实线，并考虑集合F的子集ℜ定义为所有形式的集合
⋃一世=0ķ(一种一世,b一世]∪⋃j=0米(Cj,dj]和
对于所有可能的非负整数选择ķ和米以及所有可能的实数选择一种一世<b一世,C一世<d一世. 如果ķ或者米为 0 ，则相应的联合被定义为空，因此空集本身具有给定的形式。换句话说，F包含这种形式的半开区间的所有可能的有限并集和这种半开区间的补集. 此表格的每一组都在F和每一组F有这种形式。证明F是一个子集的域Ω. 做F包含点？例如，是单例集0在F? 是F西格玛场？
6. 让Ω=[0,∞)是一个样本空间，让F是子集的 sigma 域Ω由所有形式的集合生成(n,n+1)为了n=1,2,…

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## MATLAB代写

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