数理统计学基础与应用

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"Mathematical Statistics数理统计" 数理统计是统计学的一个核心分支，它主要研究如何收集、分析、解释和展示数据。这门学科不仅涵盖了概率论的基础理论，还涉及统计推断、估计理论、假设检验和回归分析等多个方面。在实际应用中，数理统计方法广泛应用于社会科学、自然科学、工程、医学、经济和金融等领域。《Mathematical Statistics》这本书可能是对这些概念进行深入探讨的专业著作，可能包括以下主要内容： 1. **概率论基础**：书中可能首先介绍概率论的基本概念，如随机变量、概率分布（如正态分布、二项分布、泊松分布等）、条件概率和贝叶斯定理，这些都是数理统计的基础。 2. **大数定律与中心极限定理**：这两个定理是统计学中的基石，它们解释了样本均值趋向于总体均值和样本分布趋近于正态分布的现象，为统计推断提供了理论依据。 3. **统计推断**：这部分可能会涵盖参数估计（点估计和区间估计）以及假设检验（如t检验、卡方检验、F检验等），这些都是在数据基础上对总体参数做出判断的方法。 4. **线性模型与回归分析**：书中可能详细讨论了线性回归模型，包括最小二乘法、残差分析和多元线性回归，以及如何处理离群值和多重共线性等问题。 5. **假设检验与置信区间**：这部分会讲解如何通过样本数据来验证或拒绝关于总体的假设，以及如何建立总体参数的置信区间。 6. **非参数统计**：对于不满足特定分布假设的数据，非参数统计方法如秩检验和符号秩检验可能是重要的工具。 7. **蒙特卡洛模拟**：在计算复杂或理论上难以解决的问题时，蒙特卡洛方法可以模拟随机过程来获取近似结果。 8. **贝叶斯统计**：书可能还会涉及贝叶斯方法，这是一种统计推理框架，允许利用先验信息更新对未知参数的信念。 9. **统计软件的应用**：书中可能包含如何使用R、SAS、SPSS等统计软件进行数据分析的实例。 10. **实例分析与案例研究**：为了帮助读者理解理论知识，书里可能会提供真实世界的数据集和案例，让读者实践应用所学的方法。数理统计是一门理论与实践并重的学科，它要求读者既有扎实的数学基础，又具备解决实际问题的能力。通过对《Mathematical Statistics》的学习，读者能够掌握统计学的核心原理，并学会运用这些原理解决实际问题。

). Since A and B ∩A

are disjoint, we have

P (A ∪ B)=P (A)+P (B ∩ A

)

and

P (B ∩ A

)=P (B) − P (B ∩ A).

Thus P (A ∪ B)=P (A)+P (B) − P (A ∩ B).

(d) Deﬁne B

= A

and B

= A

∩A

k−1

for k ≥ 2. Then B

, ···

are disjoint and



k=1

and A =

∞



k=1

Hence

P (A)=

∞



k=1

P (B

) = lim

n→∞



k=1

P (B

) = lim

n→∞

P (A

(e) First, it follows from (c) that

P (A

∪ A

) ≤ P (A

)+P (A

)

and so for any n<∞,wehave





k=1



≤



k=1

P (A

Now let B



k=1

, B =



∞

k=1

and note that B

⊂ B

n+1

Thus applying (d),

P (B) = lim

n→∞

P (B

) ≤ lim

n→∞



k=1

P (A

∞



k=1

P (A

)

which completes the proof.

EXAMPLE 1.1: For any events A

, A

, and A

,wehave

P (A

∪ A

P (A

)+P (A

)

−P (A

∩ A

) − P (A

∩ A

) − P (A

∩ A

)

+P (A

∩ A

To see this, note that

P (A

∪ A

)=P (A

∪ A

)+P (A

) − P (A

∩ (A

∪ A

))

 2000 by Chapman & Hall/CRC

Many experiments can be broken down into a sequence of sub-

experiments for which the number of outcomes is easily counted.

For example, consider dealing 5 cards from a deck of 52 cards.

This particular “experiment” is a sequence of 5 sub-experiments

corresponding to the 5 cards being dealt.

EXAMPLE 1.3: Consider a simple experiment that consists of

rolling a pair of dice. The outcomes can be denoted by an ordered

pair (i, j) with i representing the outcome of the ﬁrst die and j

representing the outcome of the second; these can be thought of

as the outcome of two sub-experiments. We can then represent the

sample space by

Ω={(1, 1), (1, 2), ···, (1, 6), (2, 1), ···, (2, 6), ···, (6, 1), ···, (6, 6)}.

For each outcome of the ﬁrst die, there are 6 outcomes of the second

die. Since there are 6 outcomes for the ﬁrst die, it follows that the

sample space has 6 × 6 = 36 outcomes. ✸

Example 1.3 suggests that if an experiment consists of a sequence

of sub-experiments E

, ···, E

having, respectively, n

, ···,n

possi-

ble outcomes then the total number of outcomes is N = n

×···×n

However, some care should be taken in deﬁning the numbers

, ···,n

. In many cases, the outcome of sub-experiment E

de-

pends on the outcome of the previous i − 1 sub-experiments. For

example, if we are dealing 5 cards from a deck of 52 cards, once the

ﬁrst card is dealt, there are only 51 possible outcomes for the second

card dealt; however, there are in fact 52 possible outcomes for the

second card if we ignore the outcome of the ﬁrst card. In deﬁning n

for i ≥ 2, we must take into account the outcomes of the previous

i − 1 sub-experiments. A general rule for counting the number of

outcomes in an experiment can be described as follows: Suppose a

random experiment E consists of sub-experiments E

, ···, E

where

the outcome of sub-experiment E

may depend on the outcomes of

sub-experiments E

, ···, E

i−1

but the number of outcomes of E

independent of the outcomes of E

, ···, E

i−1

. Then the number

of possible outcomes of E is

N = n

×···×n

This is sometimes called the product rule for determining the

number of outcomes in the sample space.

EXAMPLE 1.4: Consider an urn consisting of the integers 0 to

 2000 by Chapman & Hall/CRC

9. We will select three integers from the urn to form a three digit

number. The integers can be selected in two possible ways: “with

replacement” or “without replacement”. If the integers are selected

with replacement, an integer is replaced in the urn after it has been

selected and so can be selected again in subsequent draws. On the

other hand, if the integers are selected without replacement, an

integer is removed from the urn after its selection and therefore

cannot be selected subsequently. If we draw the integers with

replacement, we have 10 × 10 × 10 = 1000 possible three digit

numbers since we have 10 choices at each stage. On the other hand,

if the selection is done without replacement, we have 10×9×8 = 720

possible sequences since we have removed one integer from the urn

at the second stage and two integers at the third stage. ✸

In many cases, the outcomes of an experiment can be represented

as sets of k elements drawn without replacement from a set of n

elements; such sets are called either permutations or combinations

depending on whether we distinguish between distinct orderings

of the elements of the sets. Using the product rule stated above,

we can determine the number of permutations and combinations.

Consider a set consisting of n distinct elements; for convenience, we

will represent this set by S = {1, 2, ···,n}. Suppose we look at all

sequences of k ≤ n elements drawn (without replacement) from S;

each of these sequences is called a permutation of length k from S.

Using the product rule, it follows that the number of permutations

of length k from S is

n × (n − 1) ×···×(n − k +1)=

(n − k)!

We can also consider all subsets of size k in S; such a subset is called

a combination of size k. A combination diﬀers from a permutation in

the sense that the ordering of elements is unimportant; that is, the

sets {1, 2, 3}, {2, 1, 3} and {3, 1, 2} represent the same combination

but three distinct permutations. Note that for each subset (or

combination) of size k,wehavek! permutations of length k and

so the number of combinations of size k in S is

k!(n − k)!





We can also think of a combination of size k from S as splitting S

into two disjoint subsets, one of size k and one of size n − k.Thus

 2000 by Chapman & Hall/CRC

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