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)

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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.

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