这份讲义涵盖主题 6:概率统计(Probability & Statistics)2。它增加泊松模型、组合随机变量、连续分布,以及估计和检验的思想。
概率与统计2
A-Level 数学 · 第 6 主题
6.1
泊松分布
大纲
| Candidates should be able to: | Notes and examples |
|---|---|
| • use formulae to calculate probabilities for the distribution $\text{Po}(\lambda)$ | |
| • use the fact that if $X \sim \text{Po}(\lambda)$ then the mean and variance of $X$ are each equal to $\lambda$ | Proofs are not required. |
| • understand the relevance of the Poisson distribution to the distribution of random events, and use the Poisson distribution as a model | |
| • use the Poisson distribution as an approximation to the binomial distribution where appropriate | The conditions that $n$ is large and $p$ is small should be known; $n > 50$ and $np < 5$, approximately. |
| • use the normal distribution, with continuity correction, as an approximation to the Poisson distribution where appropriate. | The condition that $\lambda$ is large should be known; $\lambda > 15$, approximately. |
来源:剑桥国际大纲
泊松分布(Poisson distribution)$X \sim \mathrm{Po}(\lambda)$ 为一个固定区间中随机事件的数目建模,当事件以一个稳定的平均速率 $\lambda$ 发生时:
$$P(X = r) = e^{-\lambda}\frac{\lambda^r}{r!}.$$
对于一个泊松变量,平均值和方差都等于 $\lambda$。当 $n$ 大且 $p$ 小时,泊松分布是二项分布的一个好近似(approximation)。当 $\lambda$ 大时,正态分布(带连续性校正)近似泊松。
例题。 $X \sim \mathrm{Po}(3)$。求 $P(X = 2)$。
$$P(X = 2) = e^{-3}\frac{3^2}{2!} = e^{-3}\times 4.5 = 0.224.$$
探索
The Poisson distribution
Change the mean λ. Poisson models the number of random events in a fixed interval — rare events give a skewed shape.
词汇表
训练
| 英文 | 中文 | 拼音 |
|---|---|---|
| Poisson distribution | 泊松分布 | pō sōng fēn bù |
| approximation | 近似 | jìn sì |
6.2
随机变量的线性组合
大纲
| Candidates should be able to: | Notes and examples |
|---|---|
| • use, when solving problems, the results that – $\text{E}(aX + b) = a\text{E}(X) + b$ and $\text{Var}(aX + b) = a^2\text{Var}(X)$ – $\text{E}(aX + bY) = a\text{E}(X) + b\text{E}(Y)$ – $\text{Var}(aX + bY) = a^2\text{Var}(X) + b^2\text{Var}(Y)$ for independent $X$ and $Y$ – if $X$ has a normal distribution then so does $aX + b$ – if $X$ and $Y$ have independent normal distributions then $aX + bY$ has a normal distribution – if $X$ and $Y$ have independent Poisson distributions then $X + Y$ has a Poisson distribution. | Proofs of these results are not required. |
来源:剑桥国际大纲
当你用一个线性规则改变一个变量时,期望(expectation,平均)和方差(variance)遵循这些规则:
$$E(aX + b) = aE(X) + b, \qquad \mathrm{Var}(aX + b) = a^2\,\mathrm{Var}(X).$$
对于两个独立变量 $X$ 和 $Y$:
$$E(aX + bY) = aE(X) + bE(Y), \qquad \mathrm{Var}(aX + bY) = a^2\,\mathrm{Var}(X) + b^2\,\mathrm{Var}(Y).$$
两个有用的事实:若 $X$ 有一个正态分布(normal distribution)则 $aX + b$ 也有;而独立泊松变量的和又是泊松。
例题。 $X$ 有平均值 $5$ 和方差 $4$。求 $E(3X - 1)$ 和 $\mathrm{Var}(3X - 1)$。
$$E(3X - 1) = 3(5) - 1 = 14, \qquad \mathrm{Var}(3X - 1) = 3^2 \times 4 = 36.$$
探索
Linear combination lab
E(aX + b) = aE(X) + b
Change a scaling factor and see how the expected value scales.
词汇表
训练
| 英文 | 中文 | 拼音 |
|---|---|---|
| expectation | 期望 | qī wàng |
| variance | 方差 | fāng chà |
| normal distribution | 正态分布 | zhèng tài fēn bù |
6.3
连续随机变量
大纲
| Candidates should be able to: | Notes and examples |
|---|---|
| • understand the concept of a continuous random variable, and recall and use properties of a probability density function | For density functions defined over a single interval only; the domain may be infinite, e.g. $\frac{3}{x^4}$ for $x \geqslant 1$. |
| • use a probability density function to solve problems involving probabilities, and to calculate the mean and variance of a distribution. | Including location of the median or other percentiles of a distribution by direct consideration of an area using the density function. Explicit knowledge of the cumulative distribution function is not included. |
来源:剑桥国际大纲
一个连续型随机变量(continuous random variable)能取一个范围内的任何值。它的概率来自一个概率密度函数(probability density function)$f(x)$,有两个关键的性质:
$$f(x) \geqslant 0, \qquad \int_{-\infty}^{\infty} f(x)\,dx = 1.$$
一个概率是 $f$ 下的面积,而平均值通过积分求得:
$$P(a < X < b) = \int_a^b f(x)\,dx, \qquad E(X) = \int_{-\infty}^{\infty} x\,f(x)\,dx.$$
累积分布函数(cumulative distribution function)是 $F(x) = P(X \leqslant x) = \int_{-\infty}^{x} f(t)\,dt$;中位数(median)解 $F(m) = 0.5$,而其他百分位数(percentiles)解 $F(x) = p$。
例题。 一个连续变量对 $0 \leqslant x \leqslant 2$ 有 $f(x) = \tfrac12 x$(其他地方为 $0$)。求 $E(X)$。
$$E(X) = \int_0^2 x\cdot\tfrac12 x\,dx = \int_0^2 \tfrac12 x^2\,dx = \left[\tfrac{x^3}{6}\right]_0^2 = \frac{8}{6} = \frac{4}{3}.$$
方差(variance)用同样的思路,$\mathrm{Var}(X)=\displaystyle\int_{-\infty}^{\infty}x^2 f(x)\,dx-\big(E(X)\big)^2$。对同样的 $f(x)=\tfrac12 x$ 在 $[0,2]$ 上:$\displaystyle\int_0^2 x^2\cdot\tfrac12 x\,dx=\left[\tfrac{x^4}{8}\right]_0^2=2$,所以 $\mathrm{Var}(X)=2-\left(\tfrac43\right)^2=2-\tfrac{16}{9}=\tfrac{2}{9}$。
探索
Area = probability
P(a < X < b) = ∫ f(x) dx
For a continuous variable, probability is the area under the density curve between two values.
词汇表
训练
| 英文 | 中文 | 拼音 |
|---|---|---|
| continuous random variable | 连续型随机变量 | lián xù xíng suí jī biàn liàng |
| probability density function | 概率密度函数 | gài lǜ mì dù hán shù |
| cumulative distribution function | 累积分布函数 | lěi jī fēn bù hán shù |
| median | 中位数 | zhōng wèi shù |
| percentiles | 百分位数 | bǎi fēn wèi shù |
6.4
抽样与估计
大纲
| Candidates should be able to: | Notes and examples |
|---|---|
| • understand the distinction between a sample and a population, and appreciate the necessity for randomness in choosing samples | |
| • explain in simple terms why a given sampling method may be unsatisfactory | Including an elementary understanding of the use of random numbers in producing random samples. Knowledge of particular sampling methods, such as quota or stratified sampling, is not required. |
| • recognise that a sample mean can be regarded as a random variable, and use the facts that $\text{E}(\overline{X}) = \mu$ and that $\text{Var}(\overline{X}) = \frac{\sigma^2}{n}$ | |
| • use the fact that $\overline{X}$ has a normal distribution if $X$ has a normal distribution | |
| • use the Central Limit Theorem where appropriate | Only an informal understanding of the Central Limit Theorem (CLT) is required; for large sample sizes, the distribution of a sample mean is approximately normal. |
| • calculate unbiased estimates of the population mean and variance from a sample, using either raw or summarised data | Only a simple understanding of the term 'unbiased' is required, e.g. that although individual estimates will vary the process gives an accurate result 'on average'. |
| • determine and interpret a confidence interval for a population mean in cases where the population is normally distributed with known variance or where a large sample is used | |
| • determine, from a large sample, an approximate confidence interval for a population proportion. |
来源:剑桥国际大纲
一个样本(sample)是从整个总体(population)选出的一个小群。一个随机样本需要随机性(randomness),使每个成员有一个被选中的公平机会。
有些方法不令人满意,因为它们是有偏(biased)的:只抽志愿者,或最先到达的 20 个人,会过度代表某些类型的人。一个真正随机的样本用随机数——给总体的每个成员编号,然后抽号(从一张表或一个生成器)来决定谁在样本里。
样本平均值 $\bar{X}$ 本身是一个随机变量,带
$$E(\bar{X}) = \mu, \qquad \mathrm{Var}(\bar{X}) = \frac{\sigma^2}{n}.$$
由中心极限定理(Central Limit Theorem),对于一个大样本 $\bar{X}$ 近似正态,无论总体的形状如何。
从一个样本你能找到总体平均值和方差的无偏估计(unbiased estimates)。一个置信区间(confidence interval)给出一个可能包含真实平均值的范围。当总体是正态、$\sigma$ 已知(或样本大)时,一个 $95\%$ 的区间是
$$\bar{x} \pm 1.96\,\frac{\sigma}{\sqrt{n}}.$$
你也能从一个大样本找到一个总体比例(population proportion)的置信区间。
例题。 一个 $n = 64$ 的样本有平均值 $\bar{x} = 50$,来自一个 $\sigma = 8$ 的总体。求总体平均值的一个 $95\%$ 置信区间。
$$50 \pm 1.96\times\frac{8}{\sqrt{64}} = 50 \pm 1.96 \;\Rightarrow\; (48.0,\ 52.0).$$
探索
The sampling distribution
X̄ ~ N(μ, σ²/n)
By the Central Limit Theorem, sample means follow a normal curve — narrower for bigger samples.
词汇表
训练
| 英文 | 中文 | 拼音 |
|---|---|---|
| sample | 样本 | yàng běn |
| population | 总体 | zǒng tǐ |
| randomness | 随机性 | suí jī xìng |
| biased | 有偏 | yǒu piān |
| Central Limit Theorem | 中心极限定理 | zhōng xīn jí xiàn dìng lǐ |
| unbiased estimates | 无偏估计 | wú piān gū jì |
| confidence interval | 置信区间 | zhì xìn qū jiān |
| population proportion | 总体比例 | zǒng tǐ bǐ lì |
6.5
假设检验
大纲
| Candidates should be able to: | Notes and examples |
|---|---|
| • understand the nature of a hypothesis test, the difference between one-tailed and two-tailed tests, and the terms null hypothesis, alternative hypothesis, significance level, rejection region (or critical region), acceptance region and test statistic | Outcomes of hypothesis tests are expected to be interpreted in terms of the contexts in which questions are set. |
| • formulate hypotheses and carry out a hypothesis test in the context of a single observation from a population which has a binomial or Poisson distribution, using – direct evaluation of probabilities – a normal approximation to the binomial or the Poisson distribution, where appropriate | |
| • formulate hypotheses and carry out a hypothesis test concerning the population mean in cases where the population is normally distributed with known variance or where a large sample is used | |
| • understand the terms Type I error and Type II error in relation to hypothesis tests | |
| • calculate the probabilities of making Type I and Type II errors in specific situations involving tests based on a normal distribution or direct evaluation of binomial or Poisson probabilities. |
来源:剑桥国际大纲
一个假设检验(hypothesis test)用样本数据来判断一个断言。你设立两个陈述:原假设(null hypothesis)$H_0$(被检验的断言,通常是"没有变化")和备择假设(alternative hypothesis)$H_1$(你反而怀疑的)。若 $H_1$ 指向一个方向(例如 $\mu > 50$)测试是单尾(one-tailed),若它允许两个方向($\mu \neq 50$)则是双尾(two-tailed)。
你固定一个显著性水平(significance level,常常 $5\%$)、从数据算出一个检验统计量(test statistic),并看它是否落入拒绝域(rejection region,也叫临界域);若是你拒绝 $H_0$,否则统计量在接受域中。
两个错误可能发生:一个第一类错误(Type I error)是当 $H_0$ 实际为真时拒绝它;一个第二类错误(Type II error)是当 $H_0$ 实际为假时接受它。你能从拒绝域求它们的概率:$P(\text{第一类})=P(\text{统计量在拒绝域}\mid H_0)$——这等于显著性水平——而 $P(\text{第二类})=P(\text{统计量在接受域}\mid H_1$ 在一个给定值处为真$)$,从二项、泊松或正态分布计算。
例题。 一个总体被断言有平均值 $50$,带 $\sigma = 8$。一个 $n = 64$ 的样本给出 $\bar{x} = 52$。在 $5\%$ 水平检验平均值是否已改变。
$H_0\!: \mu = 50$ 且 $H_1\!: \mu \neq 50$(双尾)。检验统计量是
$$z = \frac{\bar{x} - \mu}{\sigma/\sqrt{n}} = \frac{52 - 50}{8/8} = 2.$$
$5\%$(双尾)的临界值是 $1.96$。因为 $2 > 1.96$,你拒绝 $H_0$:有证据表明平均值已改变。
例题(一个二项检验)。 一枚硬币被声称是公平的,但被怀疑正面落得太少:$H_0\!:p=0.5$,$H_1\!:p<0.5$。在 $n=30$ 次投掷里你看到 $X=9$ 个正面。在 $H_0$ 下,$X\sim B(30,0.5)$,所以单尾尾概率是
$$P(X\leqslant 9)=\sum_{k=0}^{9}\binom{30}{k}(0.5)^{30}\approx 0.021.$$
因为 $0.021<0.05$,拒绝 $H_0$:硬币看来对正面有偏。(对大的 $n$,二项用一个正态近似;对稀有事件一个泊松检验同样工作。而这里,若规则是"当 $X\leqslant 9$ 时拒绝",那么 $P(\text{第一类})=P(X\leqslant 9\mid p=0.5)\approx0.021$。)
探索
The rejection region
reject H₀ if z < −z*
The shaded tail is the rejection region — if the test statistic lands there, reject H₀.
词汇表
训练
| 英文 | 中文 | 拼音 |
|---|---|---|
| hypothesis test | 假设检验 | jiǎ shè jiǎn yàn |
| null hypothesis | 原假设 | yuán jiǎ shè |
| alternative hypothesis | 备择假设 | bèi zé jiǎ shè |
| one-tailed | 单尾 | dān wěi |
| two-tailed | 双尾 | shuāng wěi |
| significance level | 显著性水平 | xiǎn zhù xìng shuǐ píng |
| test statistic | 检验统计量 | jiǎn yàn tǒng jì liàng |
| rejection region | 拒绝域 | jù jué yù |
| Type I error | 第一类错误 | dì yī lèi cuò wù |
| Type II error | 第二类错误 | dì èr lèi cuò wù |
| Probability & Statistics | 概率统计 | gài lǜ tǒng jì |
6.5
考试技巧
- 对稀有的、随机的、独立的事件用泊松分布;它的平均值等于它的方差($= \lambda$)。
- 当组合独立随机变量时,方差相加(它们从不相减)。
- 对于一个假设检验,陈述 $H_0$ 和 $H_1$、显著性水平、检验统计量,以及一个结合上下文的结论。
- 对于一个置信区间,用正确的 $z$(或 $t$)值并用文字解释它。
本主题的互动课程
逐步学习,并即时检测练习。