Consider the following potential outcome framework. $Y_i (1)$: the log wage of person $i$ if person $i$ attends a job training program $Y_i (0)$: the log wage of person $i$ if person $i$ does not a job training program Suppose that $Y_i (1)=Y_i (0)-0.5$ for each person $i$. Set $n=203$. Implement the following steps:Generate $Y_i (0)=10+u_i$ for $i=1,…,n$, where $u_i$ is i.i.d. from $N(0, 5)$. Generate $Y_i (1)=Y_i (0)-0.5$ for $i=1,…,n$ Generate $X_i $from the Bernoulli distribution with mean 0.73 Compute $Y_i=Y_i (1) \cdot X_i+Y_i (0) \cdot (1-X_i)$. Run OLS regression: $Y_i=\beta_0+\beta_1 X_i+u_i$ Construct a 90% confidence interval for $ \beta_1 $. Record whether or not the confidence interval captures the treatment effect -0.5. Repeat the above steps 300 times. Use set.seed(904). What is the probability that the 90% confidence interval captures the treatment effect -0.5? (25 points)

Question

BlackTom AI · Accepted Answer

This is a Monte Carlo exercise about generating data under a fixed data-generating process and then evaluating how often a 90% CI for the OLS coefficient contains the true treatment effect.

Step 1: Understand the data-generating process
- We have Y_i(0) = 10 + u_i with u_i ~ N(0, 5). Therefore Y_i(0) is Normal with mean 10 and variance 5.
- The potential treated outcome is defined as Y_i(1) = Y_i(0) - 0.5, so the treatment reduces the log wage by 0.5 on the treated potential outcome scale.
- The observed outcome is Y_i = Y_i(1) * X_i + Y_i(0) * (1 - X_i). Since X_i ~ Bernoulli(0.73), this simplifies to Y_i = Y_i(0) - 0.5 * X_i. Intuitively, when treated (X_i = 1), you observe Y_i = Y_i(0) - 0.5; when not treated (X_i = 0), you observe Y_i = Y_i(0).
- The covariate X_i is independent of u_i and drawn from Bernoulli with p = 0.73, so X_i is a valid instrument-like binary regressor for the OLS regression we will run.

Step 2: What regression are we estimating?
- We run OLS on the model Y_i = β_0 + β_1 X_i + ε_i, where ε_i captures any remaining variation not explained by X_i.
- Given the data-generating form Y_i = (Y_i(0)) - 0.5 X_i and Y_i(0) independent of X_i, the conditional mean of Y_i given X_i is E[Y_i | X_i] = E[Y_i(0)] - 0.5 X_i. Hence the population regression coefficient on X_i is exactly β_1 = -0.5, and β_0 corresponds to E[Y_i(0)].
- In large samples, the OLS estimator β̂_1 should be unbiased for -0.5, and its standard error will depend on the variance of ε_i = u_i (since the only extra randomness beyond X_i is u_i embedded in Y_i(0)). The homoskedasticity assumption holds here because Var(Y_i | X_i) = Var(Y_i(0)) = 5, which does not depend on X_i.

Step 3: Confidence interval construction
- For each replication, after generating Y_i and X_i, we estimate the regression Y_i = β_0 + β_1 X_i + ε_i and construct a 90% CI for β_1 using the usual t-distribution approach: β̂_1 ± t_{0.95, n-2} · SE(β̂_1).
- The t-quantile t_{0.95, n-2} depends on the residual degrees of freedom (n - 2) with n = 203, so it’s close to the normal quantile but slightly larger in finite samples.
- The target true effect we want the CI to cover is the treatment effect on the observed outcome, which in this setup equals -0.5 (as shown by the derivation Y_i = Y_i(0) - 0.5 X_i).

Step 4: Repetition and randomness control
- We perform the entire sequence (data generation, regression, CI construction) 300 times to gauge the empirical coverage probability.
- The random seed is fixed at set.seed(904) to ensure replicability of the results.

Step 5: Interpretation of the reported probability
- For each replication, check whether the 90% CI for β_1 includes the value -0.5.
- The estimated probability is the proportion of replications in which -0.5 lies within the constructed CI.

Step 6: What should you expect conceptually?
- Since Y_i = Y_i(0) - 0.5 X_i with X_i independent of Y_i(0), the OLS regression is correctly specified for the conditional mean, and β_1 truly equals -0.5 in the population.
- The sampling variability (through u_i in Y_i(0)) will cause the estimated β̂_1 to deviate from -0.5 in finite samples, but the 90% CI should, on average, cover the true value about 90% of the time if the model assumptions are met and normal theory approximations are reasonable.
- Because u_i is Gaussian and the errors are homoskedastic, the coverage probability should be close to 0.90 in this setup, albeit with some sampling fluctuation across the 300 replications.

Step 7: How to report or compute the final probability
- After running all replications, count the number of times the 90% CI for β_1 contains -0.5 and divide by 300. That fraction is the Monte Carlo estimate of the coverage probability.
- Since the seed is fixed, you can reproduce the exact number if you run the same code with identical specifications.

In summary, you are evaluating how often a correctly specified 90% CI for the OLS coefficient β_1 captures the true effect of -0.5 under the stated data-generating process, with the expectation that the empirical coverage will be near 0.90 given the normal errors and adequate sample size.

类似问题

For stock A, we have 𝛽 𝑖 = 0.70. Suppose the expected market risk premium next year is 9% and the risk-free rate is 3%. What is the expected return of this stock based on the CAPM? (Please answer in % and round to 2 decimal places. If the answer is 8.057%, then in the box, write 8.06)

The following graph shows the Security Market Line in the current economy. If a stock has a β of 1.2, what is the expected return of this stock based on the CAPM? (Please answer in % and round to 2 decimal places. If the answer is 8.057%, then in the box, write 8.06)

Suppose Apple publicly announces earnings today and the announcement is above market expectations. If markets are semistrong-form efficient, then:

Evidence that investors who trade more often earn lower net returns is most directly related to

Behavioral finance argues that EMH failures require

更多留学生实用工具

智能学习助手

风格化写作助手

论文查重助手

文献引用助手

课堂转译助手

课堂笔记助手

Quiz搜索助手

学校历年真题

智能刷题助手

智能匹配练习

希望你的学习变得更简单