### 统计代写|统计推断代写Statistical inference代考|STAT2004

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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|统计推断代写Statistical inference代考|Permutations and combinations

In this section we start by defining permutations and combinations. Our real interest lies in counting permutations and combinations. We will show how the same general counting rules can be applied to a number of apparently different problems. Our eventual aim is to use these rules to evaluate probabilities.
Definition 2.3.3 (Permutations and combinations)
Suppose that we have a set $Q$ of $n$ distinct objects.
i. A permutation of length $k \leq n$ is an ordered subset of $Q$ containing $k$ elements.

ii. A combination of length $k \leq n$ is an unordered subset of $Q$ containing $k$ elements.

We distinguish between these two cases by using (…) to denote permutation and ${\ldots}$ to denote combination.
Claim 2.3.4 (Number of permutations and number of combinations)
i. If the number of permutations of length $k$ that can be formed from $n$ distinct elements is denoted ${ }^{n} P_{k}$, then
$${ }^{n} P_{k}=\frac{n !}{(n-k) !} .$$
ii. If the number of combinations of length $k$ that can be formed from $n$ distinct elements is denoted ${ }^{n} C_{k}$, then
$${ }^{n} C_{k}=\frac{n !}{k !(n-k) !}$$
The number of permutations, ${ }^{n} P_{k}=n \times(n-1) \times \ldots \times(n-k+1)$, is a direct consequence of the multiplication rule. Note that one implication of this is that number of ways of ordering all $n$ elements is ${ }^{n} P_{n}=n$ !. The general expression for the number of combinations requires a little more thought.

Suppose that we know the number of combinations of length $k$, that is, we know ${ }^{n} C_{k}$. By the above argument, the number of ways of ordering each one of these combinations is $k !$. The multiplication rule then tells us that ${ }^{n} P_{k}=k !^{n} C_{k}$. By rearranging we arrive at the general result, ${ }^{n} C_{k}={ }^{n} P_{k} / k !$.

## 统计代写|统计推断代写Statistical inference代考|Conditional probability and independence

We start with the by-now-familiar setup of an experiment, a probability space $(\Omega, \mathcal{F}, \mathrm{P})$, and an event of interest $A$. The probability $\mathrm{P}(A)$ gives us an indication of how likely it is that the outcome of the experiment, $\omega \in \Omega$, is in $A$, that is, how likely the event $A$ is to occur. Now suppose that we know that event $B$ has occurred. This will alter our perception of how probable $A$ is since we are now only interested in outcomes that are in $B$. In effect we have shrunk the sample space from $\Omega$ to $B$. In these circumstances the appropriate measure is given by conditional probability.
Definition 2.4.1 (Conditional probability)
Consider the probability space $(\Omega, \mathcal{F}, \mathrm{P})$ and $A, B \in \mathcal{F}$ with $\mathrm{P}(B)>0$. The conditional probability of $A$ given $B$ is the probability that $A$ will occur given that $B$ has occurred,
$$\mathrm{P}(A \mid B)=\frac{\mathrm{P}(A \cap B)}{\mathrm{P}(B)}$$
Note the importance of the statement that $\mathrm{P}(B)>0$. We cannot condition on events with zero probability. This makes sense intuitively; it is only reasonable to condition

on events that have some chance of happening. The following example illustrates the use of Definition 2.4.1.
Example 2.4.2 (Rolling two dice again)
Consider the setup in Example 2.1.1. We have shown that, if we roll two fair dice and $A$ is the event that the sum of the two values is greater than 10 , then $\mathrm{P}(A)=\frac{1}{12}$. Now suppose that event $B$ is the event that the value on the second die is a 6 . The situation is illustrated in Figure 2.2. By looking at the sample space, we can see that $|B|=6$ and $|A \cap B|=2$, so $\mathrm{P}(B)=\frac{1}{6}$ and $\mathrm{P}(A \cap B)=\frac{1}{18}$. By Definition 2.4.1, $\mathrm{P}(A \mid B)=\frac{1}{18} / \frac{1}{6}=\frac{1}{3}$
\begin{tabular}{ccccc}
\multicolumn{5}{c}{ Second die roll } \
2 & 3 & 4 & 5 & 6 \
$\cdot$ & $\cdot$ & $\cdot$ & $\cdot$ & $B$ \
$\cdot$ & $\cdot$ & $\cdot$ & $\cdot$ & $B$ \
$\cdot$ & $\cdot$ & $\cdot$ & $\cdot$ & $B$ \
$\cdot$ & $\cdot$ & $\cdot$ & $\cdot$ & $B$ \
$\cdot$ & $\cdot$ & $\cdot$ & $\cdot$ & $A, B$ \
$\cdot$ & $\cdot$ & $\cdot$ & $A$ & $A, B$
\end{tabular}
Figure $2.2$ Sample space for experiment of throwing two fair dice, with events $A$ and $B$ marked.
When we use the intuitive definition given by (2.1), the terms in $|\Omega|$ cancel out:
$$\mathrm{P}(A \mid B)=\frac{\mathrm{P}(A \cap B)}{\mathrm{P}(B)}=\frac{|A \cap B|}{|\Omega|} \frac{|\Omega|}{|B|}=\frac{|A \cap B|}{|B|}$$
By conditioning on $B$, we are making $B$ take on the role of the whole sample space in our probability calculations. This makes sense since we know that the outcome of the experiment is in $B$.

## 统计代写|统计推断代写Statistical inference代考|Independence

Before we give a precise definition, it is worthwhile briefly exploring intuitive notions about independence. The Concise Oxford Dictionary’s definition is “free from outside control; not subject to another’s authority”. This is actually reasonably close to the mathematical use of the term. We will define independence in terms of events. An intuitively appealing definition is that events $A$ and $B$ are independent if the probability that $A$ occurs is unaffected by whether or not $B$ has occurred. Assuming $\mathrm{P}(B)>0$, we could write this as $\mathrm{P}(A \mid B)=\mathrm{P}(A)$. Substituting from Definition $2.4$.1 of conditional probability, $\mathrm{P}(A \cap B) / \mathrm{P}(B)=\mathrm{P}(A)$, and hence, $\mathrm{P}(A \cap B)=\mathrm{P}(A) \mathrm{P}(B)$. This is the basis of our definition of independence.
Definition 2.4.11 (Independence)
Consider a probability space $(\Omega, \mathcal{F}, \mathrm{P})$ with $A, B \in \mathcal{F} . A$ and $B$ are said to be independent, denoted $A \perp B$, if and only if $\mathrm{P}(A \cap B)=\mathrm{P}(A) \mathrm{P}(B)$.

Note that this definition, unlike our conditional probability statements, does not exclude the possibility that $\mathrm{P}(B)=0$. It is important not to confuse the properties of mutual exclusivity with independence. If $A$ and $B$ are mutually exclusive then $\mathrm{P}(A \cap B)=0$. Thus, mutually exclusive events will not be independent unless $\mathrm{P}(A)=0$ or $\mathrm{P}(B)=0$ or both.

## 统计代写|统计推断代写Statistical inference代考|Permutations and combinations

ii. 长度组合ķ≤n是的无序子集问包含ķ元素。

i．如果长度的排列数ķ可以由n表示不同的元素n磷ķ， 然后

n磷ķ=n!(n−ķ)!.
ii. 如果长度的组合数ķ可以由n表示不同的元素nCķ， 然后

nCķ=n!ķ!(n−ķ)!

## 统计代写|统计推断代写Statistical inference代考|Conditional probability and independence

\begin{tabular}{ccccc} \multicolumn{5}{c}{ 第二次掷骰 } \ 2 & 3 & 4 & 5 & 6 \ $\cdot$ & $\cdot$ & $\cdot$ & $\cdot$ & $B$ \ $\cdot$ & $\cdot$ & $\cdot$ & $\cdot$ & $B$ \ $\cdot$ & $\cdot$ & $\cdot$ & $\cdot$ & $B$ \ $\cdot$ & $\cdot$ & $\cdot$ & $\cdot$ & $B$ \ $\cdot$ & $\cdot$ & $\cdot$ & $\cdot$ & $A , B$ \ $\cdot$ & $\cdot$ & $\cdot$ & $A$ & $A, B$ \end{表格}\begin{tabular}{ccccc} \multicolumn{5}{c}{ 第二次掷骰 } \ 2 & 3 & 4 & 5 & 6 \ $\cdot$ & $\cdot$ & $\cdot$ & $\cdot$ & $B$ \ $\cdot$ & $\cdot$ & $\cdot$ & $\cdot$ & $B$ \ $\cdot$ & $\cdot$ & $\cdot$ & $\cdot$ & $B$ \ $\cdot$ & $\cdot$ & $\cdot$ & $\cdot$ & $B$ \ $\cdot$ & $\cdot$ & $\cdot$ & $\cdot$ & $A , B$ \ $\cdot$ & $\cdot$ & $\cdot$ & $A$ & $A, B$ \end{表格}

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## MATLAB代写

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