-安慰剂检验介绍、操作及举例.doc
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安慰剂检验介绍(Placebo test) 安慰剂是一种附加实证检验的思路,并不存在一个具体的特定的操作方法。一般存在两种寻找安慰剂变量的方法。 比如,在已有的实证检验中,发现自变量Xi会影响自变量Zi与因变量Yi之间存在相关关系。在其后的实证检验中,采用其他主体(国家,省份,公司)的Xj变量作为安慰剂变量,检验Xj是否影响Zi与Yi之间的相关关系。如果不存在类似于Xi的影响,即可排除Xi的安慰剂效应,使得结果更为稳健。 另一种寻找安慰剂变量的方法。已知,Xi是虚拟变量,Xi=1,if t>T;Xi=0 if t<T;Xi对Zi对Yi的影响的影响在T时前后有显著差异(DID)。在其后的实证检验中,将Xi`设定为Xi`=1,if t>T+n;Xi`=0 if t<T+n,其中n根据实际情况取值,可正可负。检验Xi`是否影响Zi与Yi之间的相关关系。如果不存在类似于Xi的影响,即可排除Xi的安慰剂效应,使得结果更为稳健。 举例: 以美国市场某种政策冲击识别策略的因果关系考察,在最后部分选取英国同期的因变量,检验是否有类似的特征,就是安慰剂检验。 以中国2007年所得税改革作为减税的政策冲击以验证减税对企业创新的影响。亦可以通过把虚拟的政策实施时间往前往后推几年,作为虚拟的政策时点,如果检验发现没有类似的因果,文章的主要结论就更加可信了。 以下是详细的例题,安慰剂检验在最后。 Surviving Graduate Econometrics with R: Difference-in-Differences Estimation — 2 of 8 The following replication exercise closely follows the homework assignment #2 in ECNS 562. The data for this exercise can be found here. The data is about the expansion of the Earned Income Tax Credit. This is a legislation aimed at providing a tax break for low income individuals. For some background on the subject, see Eissa, Nada, and Jeffrey B. Liebman. 1996. Labor Supply Responses to the Earned Income Tax Credit. Quarterly Journal of Economics. 111(2): 605-637. The homework questions (abbreviated): 1. Describe and summarize data. 2. Calculate the sample means of all variables for (a) single women with no children, (b) single women with 1 child, and (c) single women with 2+ children. 3. Create a new variable with earnings conditional on working (missing for non-employed) and calculate the means of this by group as well. 4. Construct a variable for the “treatment” called ANYKIDS and a variable for after the expansion (called POST93—should be 1 for 1994 and later). 5. Create a graph which plots mean annual employment rates by year (1991-1996) for single women with children (treatment) and without children (control). 6. Calculate the unconditional difference-in-difference estimates of the effect of the 1993 EITC expansion on employment of single women. 7. Now run a regression to estimate the conditional difference-in-difference estimate of the effect of the EITC. Use all women with children as the treatment group. 8. Reestimate this model including demographic characteristics. 9. Add the state unemployment rate and allow its effect to vary by the presence of children. 10. Allow the treatment effect to vary by those with 1 or 2+ children. 11. Estimate a “placebo” treatment model. Take data from only the pre-reform period. Use the same treatment and control groups. Introduce a placebo policy that begins in 1992 (so 1992 and 1993 both have this fake policy). A review: Loading your data Recall the code for importing your data: STATA: /*Last modified 1/11/2011 */ ************************************************************************* *The following block of commands go at the start of nearly all do files*/ *Bracket comments with /* */ or just use an asterisk at line beginning clear /*Clears memory*/ set mem 50m /*Adjust this for your particular dataset*/ cd "C:\DATA\Econ 562\homework" /*Change this for your file structure*/ log using stata_assign2.log, replace /*Log file records all commands & results*/ display "$S_DATE $S_TIME" set more off insheet using eitc.dta, clear ************************************************************************* R: 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 # Kevin Goulding # ECNS 562 - Assignment 2 ########################################################################## # Load the foreign package require(foreign) # Import data from web site # update: first download the file eitc.dta from this link: # # Then import from your hard drive: eitc = read.dta("C:/link/to/my/download/folder/eitc.dta")</pre> Note that any comments can be embedded into R code, simply by putting a <code> # </code> to the left of your comments (e.g. anything to the right of <code> # </code> will be ignored by R). Alternately, you can download the data file, and import it from your hard drive: eitc = read.dta("C:\DATA\Courses\Econ 562\homework\eitc.dta") Describe and summarize your data Recall from part 1 of this series, the following code to describe and summarize your data: STATA: des sum R: In R, each column of your data is assigned a class which will determine how your data is treated in various functions. To see what class R has interpreted for all your variables, run the following code: 1 2 3 4 sapply(eitc,class) summary(eitc) source('sumstats.r') sumstats(eitc) To output the summary statistics table to LaTeX, use the following code: 1 2 require(xtable) # xtable package helps create LaTeX code from R. xtable(sumstats(eitc)) Note: You will need to re-run the code for sumstats() which you can find in an earlier post. Calculate Conditional Sample Means STATA: summarize if children==0 summarize if children == 1 summarize if children >=1 summarize if children >=1 & year == 1994 mean work if post93 == 0 & anykids == 1 R: 1 2 3 4 5 6 7 8 9 10 11 12 13 14 # The following code utilizes the sumstats function (you will need to re-run this code) sumstats(eitc[eitc$children == 0, ]) sumstats(eitc[eitc$children == 1, ]) sumstats(eitc[eitc$children >= 1, ]) sumstats(eitc[eitc$children >= 1 & eitc$year == 1994, ]) # Alternately, you can use the built-in summary function summary(eitc[eitc$children == 0, ]) summary(eitc[eitc$children == 1, ]) summary(eitc[eitc$children >= 1, ]) summary(eitc[eitc$children >= 1 & eitc$year == 1994, ]) # Another example: Summarize variable 'work' for women with one child from 1993 onwards. summary(subset(eitc, year >= 1993 & children == 1, select=work)) The code above includes all summary statistics – but say you are only interested in the mean. You could then be more specific in your coding, like this: 1 2 3 mean(eitc[eitc$children == 0, 'work']) mean(eitc[eitc$children == 1, 'work']) mean(eitc[eitc$children >= 1, 'work']) Try out any of the other headings within the summary output, they should also work: min() for minimum value, max() for maximum value, stdev() for standard deviation, and others. Create a New Variable To create a new variable called “c.earn” equal to earnings conditional on working (if “work” = 1), “NA” otherwise (“work” = 0) – use the following code: STATA: gen cearn = earn if work == 1 R: 1 2 3 4 5 6 7 eitc$c.earn=eitc$earn*eitc$work z = names(eitc) X = as.data.frame(eitc$c.earn) X[] = lapply(X, function(x){replace(x, x == 0, NA)}) eitc = cbind(eitc,X) eitc$c.earn = NULL names(eitc) = z Construct a Treatment Variable Construct a variable for the treatment called “anykids” = 1 for treated individual (has at least one child); and a variable for after the expansion called “post93” = 1 for 1994 and later. STATA: gen anykids = (children >= 1) gen post93 = (year >= 1994) R: 1 2 eitc$post93 = as.numeric(eitc$year >= 1994) eitc$anykids = as.numeric(eitc$children > 0) Create a plot Create a graph which plots mean annual employment rates by year (1991-1996) for single women with children (treatment) and without children (control). STATA: preserve collapse work, by(year anykids) gen work0 = work if anykids==0 label var work0 "Single women, no children" gen work1 = work if anykids==1 label var work1 "Single women, children" twoway (line work0 year, sort) (line work1 year, sort), ytitle(Labor Force Participation Rates) graph save Graph "homework\eitc1.gph", replace R: 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 # Take average value of 'work' by year, conditional on anykids minfo = aggregate(eitc$work, list(eitc$year,eitc$anykids == 1), mean) # rename column headings (variables) names(minfo) = c("YR","Treatment","LFPR") # Attach a new column with labels minfo$Group[1:6] = "Single women, no children" minfo$Group[7:12] = "Single women, children" minfo require(ggplot2) #package for creating nice plots qplot(YR, LFPR, data=minfo, geom=c("point","line"), colour=Group, xlab="Year", ylab="Labor Force Participation Rate") The ggplot2 package produces some nice looking charts. Calculate the D-I-D Estimate of the Treatment Effect Calculate the unconditional difference-in-difference estimates of the effect of the 1993 EITC expansion on employment of single women. STATA: mean work if post93==0 & anykids==0 mean work if post93==0 & anykids==1 mean work if post93==1 & anykids==0 mean work if post93==1 & anykids==1 R: 1 2 3 4 5 a = colMeans(subset(eitc, post93 == 0 & anykids == 0, select=work)) b = colMeans(subset(eitc, post93 == 0 & anykids == 1, select=work)) c = colMeans(subset(eitc, post93 == 1 & anykids == 0, select=work)) d = colMeans(subset(eitc, post93 == 1 & anykids == 1, select=work)) (d-c)-(b-a) Run a simple D-I-D Regression Now we will run a regression to estimate the conditional difference-in-difference estimate of the effect of the Earned Income Tax Credit on “work”, using all women with children as the treatment group. The regression equation is as follows: Where is the white noise error term. STATA: gen interaction = post93*anykids reg work post93 anykids interaction R: 1 2 reg1 = lm(work ~ post93 + anykids + post93*anykids, data = eitc) summary(reg1) Include Relevant Demographics in Regression Adding additional variables is a matter of including them in your coded regression equation, as follows: STATA: gen age2 = age^2 /*Create age-squared variable*/ gen nonlaborinc = finc - earn /*Non-labor income*/ reg work post93 anykids interaction nonwhite age age2 ed finc nonlaborinc R: 1 2 3 reg2 = lm(work ~ anykids + post93 + post93*anykids + nonwhite + age + I(age^2) + ed + finc + I(finc-earn), data = eitc) summary(reg2) Create some new variables We will create two new interaction variables: 1. The state unemployment rate interacted with number of children. 2. The treatment term interacted with individuals with one child, or more than one child. STATA: gen interu = urate*anykids gen onekid = (children==1) gen twokid = (children>=2) gen postXone = post93*onekid gen postXtwo = post93*twokid R: 1 2 3 4 5 6 7 8 9 10 11 12 # The state unemployment rate interacted with number of children eitc$urate.int = eitc$urate*eitc$anykids ## # Creating a new treatment term: # First, we'll create a new dummy variable to distinguish between one child and 2+. eitc$manykids = as.numeric(eitc$children >= 2) # Next, we'll create a new variable by interacting the new dummy # variable with the original interaction term. eitc$tr2 = eitc$p93kids.interaction*eitc$manykids Estimate a Placebo Model Testing a placebo model is when you arbitrarily choose a treatment time before your actual treatment time, and test to see if you get a significant treatment effect. STATA: gen placebo = (year >= 1992) gen placeboXany = anykids*placebo reg work anykids placebo placeboXany if year<1994 In R, first we’ll subset the data to exclude the time period after the real treatment (1993 and later). Next, we’ll create a new treatment dummy variable, and run a regression as before on our data subset. R: 1 2 3 4 5 6 7 8 9 10 # sub set the data, including only years before 1994. eitc.sub = eitc[eitc$year <= 1993,] # Create a new "after treatment" dummy variable # and interaction term eitc.sub$post91 = as.numeric(eitc.sub$year >= 1992) # Run a placebo regression where placebo treatment = post91*anykids reg3 <- lm(work ~ anykids + post91 + post91*anykids, data = eitc.sub) summary(reg3) The entire code for this post is available here (File –> Save As). If you have any questions or find problems with my code, you can e-mail me directly at kevingoulding {at} gmail [dot] com. To continue on to Part 3 of our series, Fixed Effects estimation, click here.展开阅读全文
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