因果推断：X对y影响的测量

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"《因果推断：衡量X对y的影响》 Austin Nichols 著" 《因果推断：衡量X对y的影响》是Austin Nichols撰写的一本深入探讨因果关系的著作，书中详细阐述了如何在统计模型中度量一个变量X对另一个变量Y的影响。因果关系是科学研究中的核心概念，它描述了事件A（原因）如何导致事件B（结果）的发生。本书适合对英文有一定理解能力的读者，尤其是那些在社会科学、经济学或统计学领域工作的专业人士。书中的内容涵盖了多个关键主题： 1. 因果效应与处理效应：处理效应是指给定一个特定的处理（如政策、药物等）对结果变量的平均影响。书中介绍了如何设计实验和观察研究来估计这些效应，包括随机对照试验和自然实验。 2. 统计模型：书中讨论了各种统计模型在因果推断中的应用，例如线性回归、逻辑斯蒂回归（logit）、Probit模型以及线性概率模型（LPM）。这些模型帮助我们从数据中提取因果信息，同时考虑潜在的偏差和误差。 3. 方法论示例：书中通过图表（如局部线性回归、核选择、偏效应图等）展示了不同的建模技术，帮助读者理解如何近似和估计因果效应。比如，图1.5对比了Logit、Probit和LPM模型的预测曲线。 4. 偏差与方差的权衡：书中探讨了估计器的偏差-方差 tradeoff，如图1.7和1.8所示，解释了为何在实际应用中，我们需要找到一个在偏差和方差之间平衡的估计方法，以最小化均方误差（MSE）。 5. 实证研究案例：通过具体的研究实例，如需求与供给曲线的观测（图1.9）、回归断点设计（图5.1和5.2）以及选举投票的RD分析（图5.3），作者向读者展示了如何在真实世界情境中应用因果推断理论。 6. 数学基础：附录部分提供了关于向量几何、导数作为线性逼近的斜率以及概率事件等数学概念的介绍，为理解复杂的统计方法提供了必要的数学背景。《因果推断：衡量X对y的影响》是一本深入浅出的教材，旨在帮助读者掌握因果推断的基本原理和实用技巧，以便在研究中准确评估变量间的影响关系。通过对书中内容的深入学习，读者可以更好地理解和应用统计方法来解决现实世界的因果问题。

6 CHAPTER 1. BASIC IDEAS AND METHODS

mix of k ones and n − k zeros is

P (k successes in n trials) =

(1 − p)

n−k

and we can calculate the probability of observing 60 or more ones or 40 or fewer

ones. The hypothesis test is easy in Stata: bitesti 100 60 .5 gives a p-

value of 0.057 so we fail to reject the null.

A better approximate test for these situations, that easily generalizes to com-

plex survey data, is the svy:tab construction, followed by test commands as

desired (exhibit 1.1.5). If you don’t have complex survey data, you can still use

the svy commands by specifying svyset, srs.

Failure to reject the null is not the same as accepting that the null hypothesis is

true; in fact, rejecting the null is not the same as asserting that the null hypothesis

is deﬁnitely false (since we have picked α to be wrong a ﬁxed proportion of the

time). When we fail to reject the null, it may be that we simply do not have

enough data, or a design with sufﬁcient statistical power, to distinguish among

alternatives. If, in the tab above, the proportion continued to be roughly the same

as we collected more data, we would soon reject that the true proportion was one

half. We simply cannot say much (given the data we have) about what the true

proportion is, but what we can say is often usefully summarized by a conﬁdence

interval.

1.1.4 Conﬁdence Intervals

If we conceive a hypothesis test, there is a conﬁdence interval that corresponds

to the range of hypothesized values for which we do not reject the null. For the

asymptotically valid statistic for a single proportion

Z =

X − µ

µ(1 − µ)/n

the values of µ for which we would fail to reject the null produce Z in the range

(-1.96,1.96), since values in that range give p-values greater than 0.05 (conﬁdence

less than 95 percent). The endpoints are the values that give p-values of exactly

0.05 or conﬁdence of 95 percent, so we call it a 95 percent conﬁdence interval (for

any other choice of α we would construct a 1 − α conﬁdence interval instead).

1.1. BASIC STATISTICS 7

We can solve for the endpoints of that range:



X − µ

µ(1 − µ)/n



= 1.96 ⇒ µ = X ± 1.96

µ(1 − µ)/n

so for 60 successes in 100 trials we can assert that the true proportion lies in the

range (0.502,0.698) with 95 percent conﬁdence. In this case, however, we have

already decided that the asymptotically valid statistic is not good enough for our

sample of 100.

Using the binomial probability distribution to create a conﬁdence interval is

a little trickier, since there is no guarantee that any hypothesized µ will produce

a p-value of exactly 5 percent (Brown et al. 2001, Agresti and Coull 1998), but

svy:tab with the ci option is a good approximation. In the example in Exhibit

1.1.5, the conﬁdence interval includes one half, but the test output indicates that

we would reject the hypothesis that the true proportion is one half.

Exhibit 1.1.5 Using svy:tab.

clear

set obs 100

g support=_n>40

la def s 0 "0 (No or false)" 1 "1 (Yes or true)"

la val support s

la var support "Support the Proposal"

svyset, srs

svy:tab support, ci

test _b[p2]=.5

1.1.6 Crosstabs

The crosstab is the simplest and most intuitive way to compare two distributions.

For example, we might compare outcomes for two medical procedures in a simple

two by two table. Suppose you have kidney stones and you can pick from two

equal-cost treatment options, P or O

Published data on outcomes by treatment

(Charig et al. 1986) indicates a success rate for P of 83 percent and a success

rate for O of 78 percent, shown as row proportions in the tab of treatment versus

success in Exhibit 1.1.7.

In fact, O (open surgery) is a substantially higher-cost option than P (percutaneous

nephrolithotomy), both in terms of pecuniary cost and morbidity associated with treatment. For

the purposes of the illustration, we will pretend they cost the same. There are additional treatment

options, including extracorporeal shockwave lithotripsy (ESWL) that have better success rates and

lower costs in many cases, which we will ignore for this example.

8 CHAPTER 1. BASIC IDEAS AND METHODS

Exhibit 1.1.7 A simple crosstab.

. use kidneyst, clear

. ta T S, row nokey

| Success

Treatment | 0 1 | Total

-----------+----------------------+----------

P | 61 289 | 350

| 17.43 82.57 | 100.00

-----------+----------------------+----------

O | 77 273 | 350

| 22.00 78.00 | 100.00

-----------+----------------------+----------

Total | 138 562 | 700

| 19.71 80.29 | 100.00

We could calculate column proportions as well, but those proportions make

little sense in this context. They answer the question, what is the chance that a

successful procedure was of type P or of type O? This is not what we care about

if we are suffering from kidney stones. We want to know which procedure has a

higher success rate, which is given by comparing row proportions, and P seems

the clear winner at 83 percent. The comparison of success rates naturally invites

a causal conclusion—who would choose treatment O given this information?

If we wanted to determine whether such a difference in success rates could

have arisen merely from chance, we could use svy: tab with the row and se

options, then test the hypothesis that success rates are identical, and ﬁnd that

the apparent advantage of treatment P is not “statistically signiﬁcant” (i.e. fail to

reject the null that success rates are identical). For the purposes of this illustration,

we will pretend we have 10 times the data shown above and the F statistic is 23

instead of 2.3 (by artiﬁcially manufacturing 7000 observations from 700), which

changes the p-value from ten percent to one and a half millionths. This is merely

to abstract from worries about statistical signiﬁcance and focus on the ranking of

treatments.

What happens to our ranking of treatments when we compare the effective-

ness of the two treatments P and O for kidney stones in two size categories large

(at least 2cm) and not large? For both larger and smaller stones, treatment O

dominates, with success rates of 73 and 93 percent compared to 69 and 87 percent

(Exhibit 1.1.8). The ranking of treatments according to success rate reverses when

we make the comparison in each of two categories.

Exhibit 1.1.8 Success rate reversal.

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