进阶概率与统计

A continuous variable $X$ is described by a probability density function 概率密度函数 $f(x)$, which may be defined piecewise. The probability over a range is the area under $f$, and the mean of any function of $X$ is

The cumulative distribution function 累积分布函数 $F(x) = P(X \leqslant x)$ is the running total: $F(x) = \displaystyle\int_{-\infty}^{x} f(t)\,dt$, and $f(x) = F'(x)$ wherever the density is continuous. Use $F$ to find probabilities and percentiles 百分位数 (for example, the median 中位数 solves $F(x) = 0.5$).

Worked example. A variable has $f(x) = \tfrac12 x$ for $0 \leqslant x \leqslant 2$. Find the median.

The cumulative distribution function is $F(x) = \displaystyle\int_0^x \tfrac12 t\,dt = \tfrac14 x^2$. Set $F(m) = 0.5$:

Worked example: mean, variance and $E(g(X))$. For the same $f(x)=\tfrac12 x$ on $[0,2]$:

The same rule handles any $g$: $E\!\left(\tfrac1X\right)=\displaystyle\int_0^2 \frac1x\cdot\tfrac12 x\,dx=\int_0^2\tfrac12\,dx=1$. Note that $E\!\left(\tfrac1X\right)=1$ is not $\dfrac{1}{E(X)}=\tfrac34$.

The distribution of a related variable. If $Y=g(X)$, find $Y$'s distribution through its cumulative function. For $Y=X^2$: $F_Y(y)=P(X^2\leqslant y)=P(-\sqrt y\leqslant X\leqslant\sqrt y)=F_X(\sqrt y)-F_X(-\sqrt y)$, then differentiate for $f_Y=F_Y'$. When $g$ is monotonic increasing there is a shortcut, $F_Y(y)=F_X\big(g^{-1}(y)\big)$.

Worked example. With $f_X(x)=\tfrac12 x$ on $[0,2]$ (so $F_X(x)=\tfrac14 x^2$), let $Y=X^2$. For $0\leqslant y\leqslant 4$, $F_Y(y)=F_X(\sqrt y)=\tfrac14 y$, so $f_Y(y)=F_Y'(y)=\tfrac14$ – that is, $Y$ is uniform on $[0,4]$.

For $Y=X^3$ (the syllabus example) the range becomes $0\leqslant y\leqslant 8$, and $F_Y(y)=F_X\big(y^{1/3}\big)=\tfrac14 y^{2/3}$, so $f_Y(y)=\tfrac16 y^{-1/3}$. Always state the new range of $Y$ first.

For an independent random sample from a normal population with unknown variance, base your hypothesis test 假设检验 for the mean on the $t$-distribution. This assumption is especially important for a small sample. The same idea gives a confidence interval 置信区间 for the mean:

where $t$ comes from the $t$-tables with $n - 1$ degrees of freedom. To compare two populations, use a two-sample (2-sample) or paired-sample $t$-test, after finding a pooled estimate 合并估计 of the shared variance when appropriate.

Worked example. An independent random sample of $n = 10$ from a normal population has mean $\bar{x} = 50$ and sample standard deviation $s = 4$ (using the $n-1$ divisor). Find a $95\%$ confidence interval for the mean (use $t = 2.262$ for $9$ degrees of freedom).

Worked example: one-sample $t$-test. The same sample is used to test a claim that the population mean is $52$, at the $5\%$ level, assuming the population is normal. $H_0:\mu=52$, $H_1:\mu\neq52$.

The critical values are $\pm2.262$ ($9$ degrees of freedom, two-tailed $5\%$). Since $|-1.58|<2.262$, do not reject $H_0$: there is insufficient evidence that the mean differs from $52$. This agrees with the interval, which contains $52$.

Worked example: two-sample $t$-test with a pooled variance. Two independent samples from normal populations with a common variance: sample A has $n_A=8$, $\bar x_A=24.5$, $s_A^2=4.2$; sample B has $n_B=10$, $\bar x_B=22.1$, $s_B^2=5.0$ (unbiased estimates). Test $H_0:\mu_A=\mu_B$ against $H_1:\mu_A\neq\mu_B$ at the $5\%$ level.

The pooled estimate weights each variance by its degrees of freedom:

With $16$ degrees of freedom the two-tailed $5\%$ critical value is $2.120$. Since $2.35>2.120$, reject $H_0$: there is evidence that the population means differ. The matching $95\%$ confidence interval for $\mu_A-\mu_B$ is $2.4\pm2.120(1.023)$, which is $(0.23,\ 4.57)$, and it excludes $0$.

Worked example: paired $t$-test. Six runners are timed before and after a training plan. The improvements (before minus after, in seconds) are $1.2$, $0.8$, $-0.3$, $1.5$, $0.9$, $0.6$. The same runners appear twice, so work with the differences as one sample: $\bar d=0.783$ and $s_d=0.618$. With $H_0:\mu_d=0$ and $H_1:\mu_d>0$,

The one-tailed $5\%$ critical value for $5$ degrees of freedom is $2.015$. Since $3.11>2.015$, reject $H_0$: there is evidence that the plan improves times, assuming the differences are normally distributed. A two-sample test here would be wrong, because the two sets of times are not independent.

Choosing the test. Same individuals measured twice → paired $t$. Two independent small samples from normal populations with a common variance → pooled two-sample $t$. Large samples (or known variances) → a test using the normal distribution, with $\sqrt{\tfrac{s_A^2}{n_A}+\tfrac{s_B^2}{n_B}}$ for independent large samples. If the population variances are known, use those variances in place of the sample estimates; exact small-sample normal inference also needs normal populations.

A $\chi^2$-test (chi-squared test 卡方检验) compares observed counts $O$ with expected counts $E$ from a theoretical distribution 理论分布:

Use independent observations, with positive expected counts large enough for the chi-squared approximation; combine suitable classes if needed. Compare with a table value for the right number of degrees of freedom 自由度. Two uses: a goodness of fit 拟合优度 test (does the data follow the proposed model?), and a test for independence 独立性 of two variables in a contingency table 列联表.

Worked example. Four equally likely categories give observed counts $20, 30, 25, 25$ (so each expected count is $25$). Test the fit at the $5\%$ level.

With $4 - 1 = 3$ degrees of freedom the table value is $7.815$. Since $2 < 7.815$, do not reject the model.

Worked example: fitting a prescribed distribution. $80$ families with three children are counted by number of girls. Test, at the $5\%$ level, whether $B(3,\,0.5)$ fits.

Girls	$0$	$1$	$2$	$3$
$O$	$6$	$28$	$34$	$12$
$E=80\times P$	$80\times\tfrac18=10$	$80\times\tfrac38=30$	$30$	$10$

All expected values are at least $5$, so no classes are combined. The degrees of freedom are $4-1=3$ (critical value $7.815$); since $2.67<7.815$, do not reject $B(3,\,0.5)$. If $p$ had been estimated from the data, one more degree of freedom is lost: $\nu=4-1-1=2$.

Worked example: contingency table. $100$ students are classified by course and by whether they passed.

	Pass	Fail	Total
Course A	$30$	$20$	$50$
Course B	$20$	$30$	$50$
Total	$50$	$50$	$100$

Each expected value is $\dfrac{\text{row total}\times\text{column total}}{\text{grand total}}=\dfrac{50\times50}{100}=25$, so

There is $(2-1)(2-1)=1$ degree of freedom, and the $5\%$ critical value is $3.841$. Since $4>3.841$, reject $H_0$: there is evidence that the result is not independent of the course. (Yates' correction is not required in this syllabus.)

A non-parametric test 非参数检验 does not require a normal population, but still has sampling assumptions; the signed-rank test also needs a symmetric distribution of differences to test its centre. The basic ones are:

• the sign test 符号检验: count how many values fall above and below a proposed median, and test those counts with a binomial model;
• the Wilcoxon signed-rank test 威尔科克森符号秩检验 (the matched-pairs test for paired data), which also uses the sizes of the differences, not just their signs;
• the Wilcoxon rank-sum test 威尔科克森秩和检验, for comparing two separate samples.

Worked example. Test whether a median is $5$. In a sample of $10$ values (none equal to $5$), $9$ lie above $5$ and $1$ lies below. Test at the $5\%$ level (two-tailed).

Under $H_0$ (median $= 5$) the number above follows $B(10, 0.5)$. The observed result ($9$ above) is extreme, so find $P(X \geq 9) = \binom{10}{9}(0.5)^{10} + (0.5)^{10} = \dfrac{11}{1024} = 0.0107$. For a two-tailed test compare with $\tfrac{1}{2}(5\%) = 0.025$. Since $0.0107 < 0.025$, reject $H_0$: there is evidence the median is not $5$.

Worked example: Wilcoxon signed-rank test. A machine is set so that the median mass of a bag is $20$ g. Eight bags have masses $23.1$, $18.4$, $26.0$, $21.7$, $19.5$, $27.3$, $24.8$, $22.2$ g. Test at the $5\%$ level whether the median is greater than $20$ g, assuming the distribution is symmetric.

Subtract $20$, then rank the differences by size, ignoring sign:

Difference	$-0.5$	$-1.6$	$1.7$	$2.2$	$3.1$	$4.8$	$6.0$	$7.3$
Rank	$1$	$2$	$3$	$4$	$5$	$6$	$7$	$8$

The sum of the negative ranks is $P_-=1+2=3$; the positive ranks sum to $36-3=33$. Under $H_1$ (median $>20$) negative differences should be rare, so the test statistic is $T=3$. For $n=8$ the one-tailed $5\%$ critical value is $5$, and $3\leqslant5$, so reject $H_0$: there is evidence that the median mass is above $20$ g. A sign test on the same data ($6$ of $8$ above) gives $P(X\geqslant6)=0.145$ and would not reject: using the sizes can give the Wilcoxon test greater power under its symmetry assumption; this one sample does not establish a general power comparison.

For $n$ larger than the tables, a sum of positive ranks (or a sum of negative ranks) has the following approximate normal distribution under the null, when there are no zero differences or tied absolute differences. Use a continuity correction; ties need adjusted variance. The minimum of both rank sums is not itself distributed this way:

Worked example: Wilcoxon rank-sum test. Two independent samples of reaction times (ms). A: $12$, $15$, $18$, $21$, $25$. B: $14$, $19$, $22$, $26$, $28$, $30$. Test at the $5\%$ level (two-tailed) whether the populations are identical.

Rank all $11$ values together: $12(1)$, $14(2)$, $15(3)$, $18(4)$, $19(5)$, $21(6)$, $22(7)$, $25(8)$, $26(9)$, $28(10)$, $30(11)$. The smaller sample is A ($m=5$, $n=6$); its rank sum is $R_m=1+3+4+6+8=22$. Also find $m(m+n+1)-R_m=5(12)-22=38$, and take the smaller: $W=22$.

For $m=5$, $n=6$ the two-tailed $5\%$ critical value is $18$. Since $22>18$, do not reject $H_0$: there is insufficient evidence of a difference between the populations.

The probability generating function 概率母函数 of a non-negative integer-valued variable $X$ is

It packs the whole distribution into one function. When the corresponding moments are finite, the mean and variance come from its derivatives at $t = 1$: $E(X) = G'(1)$ and $\mathrm{Var}(X) = G''(1) + G'(1) - \big(G'(1)\big)^2$. Also, the PGF of a sum of independent variables is the product of their PGFs.

Worked example. $X$ has $P(X=0) = 0.5$, $P(X=1) = 0.3$, $P(X=2) = 0.2$. Find $E(X)$ using the PGF.

Here $G(t) = 0.5 + 0.3t + 0.2t^2$, so $G'(t) = 0.3 + 0.4t$ and

Differentiate again for the variance: $G''(t)=0.4$, so

(Check directly: $E(X^2)=0.3+0.8=1.1$ and $1.1-0.7^2=0.61$.)

Standard PGFs (with $q=1-p$): discrete uniform on $1,\dots,n$: $\dfrac{t(1-t^n)}{n(1-t)}$; binomial $B(n,p)$: $(q+pt)^n$; geometric $\mathrm{Geo}(p)$ on $1,2,\dots$: $\dfrac{pt}{1-qt}$; Poisson $\mathrm{Po}(\lambda)$: $e^{\lambda(t-1)}$.

Worked example: Poisson mean and variance. For $G(t)=e^{\lambda(t-1)}$: $G'(t)=\lambda e^{\lambda(t-1)}$ and $G''(t)=\lambda^2 e^{\lambda(t-1)}$, so $G'(1)=\lambda$ and $G''(1)=\lambda^2$. Then $E(X)=\lambda$ and $\mathrm{Var}(X)=\lambda^2+\lambda-\lambda^2=\lambda$.

Worked example: sum of independent variables. $X_1$ and $X_2$ are independent copies of the variable above. The PGF of $Y=X_1+X_2$ is the product:

So, for example, $P(Y=2)=0.29$: the $t^2$ coefficient collects $0.5\times0.2$ twice and $0.3\times0.3$ once. The coefficients add to $1$, a quick check.

• For a continuous random variable, the pdf integrates to $1$ over its range, and $E(X) = \int x f(x)\,dx$.
• Use the $t$-distribution when the sample is small and the population variance is unknown; state the degrees of freedom.
• For a chi-squared test compute $\sum (O-E)^2/E$, compare with the critical value at the right degrees of freedom, and combine classes with $E < 5$.
• State $H_0$ and $H_1$ and give the conclusion in context for every test.