Intermediate Econometrics

• Effect of a treatment on a relevant outcome. Examples:

1. Effect of microcredit access on the economic outcomes of households in developing countries
2. Effects of a reduction of class size on the academic outcomes of pupils
3. Effects of a youth job training program on employment status

• Let’s consider the simplest case where

1. Outcome: a continuous variable \(Y\)
2. Treatment: bynary variable \(D\) (assuming values 1 for treatment, 0 otherwise)

• The causal impact of \(D\) on \(Y\) is measured by the difference between the value of \(Y\) when \(D=1\) (\(y_1\)) and the value of \(Y\) when \(D=0\) (\(y_0\)).

• To an individual for which

1. \(D=0\), \(y_0\) is the observed or factual situation
2. The value of \(y\) if the individual had received the treatment is \(y_1\) and is called the counterfactual

• The fundamental problem is that the counterfactual is unobserved: potential outcome model

• Let’s suppose to have a sample of size \(n\), where we record a characteristic \(Y\) and there is a binary variable distinguishing two groups:

1. On the left what we observed
2. On the right, a matrix padded with the potential outcomes

\[\begin{pmatrix}Y & D \\ y_1 & 0\\ y_2 & 1 \\ y_3 & 0 \\ \vdots & \vdots \\ y_n & 1\end{pmatrix}\implies \begin{pmatrix} subject & Y_{(0)} & Y_{(1)} & D \\ 1 & \bf{y_{1(0)}} & {y_{1(1)}} & 0\\ 2 & y_{2(0)} & \bf{y_{2(1)}} & 1\\ 3 & \bf{y_{3(0)}} & {y_{3(1)}} & 0\\ \vdots & \vdots & \vdots & \vdots\\ n & y_{n(0)} & \bf{y_{n(1)}} & 1 \end{pmatrix}\]

• The natural estimator of the effect of the treatment would be: \[\frac{1}{n}\sum_{i=1}^{n}(y_{i(1)} - y_{i(0)})\]

• However, it is unfeasible for a missing data problem: either \(y_{0i}\) or \(y_{1i}\) is observed, but not both, as an individual cannot be at the same time treated and untreated

• In real settings, we have a sample that contains:

1. subsample of individuals who received the treatment (\(D=1\)) called the treatment group (denoted by \(T\))
2. subsample of individuals who didn’t receive the treatment (\(D=0\)) called the control group (\(C\))

• An estimator is obtained using the difference between the mean values of the outcome in the treatment and in the control group:

\[\frac{1}{n_T}\sum_{i=1}^{n_T}y_{i} - \frac{1}{n_C}\sum_{i=1}^{n_C}y_{i}\]

• Data can be either experimental or observational

1. Experimetal: the treatment is randomly assigned to some individuals
2. Observational: we are just observing, without assigning the treatment to the two groups

• With experimental data \[\frac{1}{n_T}\sum_{i=1}^{n_T}y_{i} - \frac{1}{n_C}\sum_{i=1}^{n_C}y_{i}\] should be a reliable estimator of the effect of the treatment.

• With observational data, the observed and unobserved characteristics of the individuals in the two groups may be different and therefore, and the estimator may include partly these differences and therefore may be biased. To overcome these difficulties, different estimators have been suggested for observational data, which rely on different assumptions

• Ideal setting to analyze treatment effects
• The treatment and the control groups are composed of individuals who are randomly drawn from the same population, and therefore the average observable and unobserved characteristics in both groups should be similar
• The first step of the analysis consists of checking that the observable characteristics are similar in the treatment and in the control group:

1. For numerical variables, this can be performed using a t test of equal means
2. Balance between groups of factors can be tested using Pearson’s \(\chi^2\) test

• Study of Burde and Linden (2013)
• The sample comprises villages from the Ghor province in northwestern Afghanistan:

1. 31 villages were selected and formed 11 village groups.
2. Five of them received a village-based school in summer 2007

• In fall 2007, a survey was conducted in the 31 villages
• We will focus on one of the two outcomes of interest considered in that study, that is the score obtained to a short test covering math and language skills.

• Assume that a variable \(X\) (a characteristic) is drawn from a normal distribution, with potentially a different mean for the treatment (\(\mu_T\)) and for the control group (\(\mu_C\)), but the same variance \(\sigma^2_X\) (we assume independence between the two groups)
• Let \(\bar X_T = \frac{1}{n_T}\sum_{i=1}^{n_T} X_i\) and \(\bar X_C = \frac{1}{n_C}\sum_{i=1}^{n_C} X_i\), the sample means for the treatment and control group
• We know that \(\bar X_T \sim \mathcal{N}(\mu_T, \sigma^2_X/n_T)\) and \(\bar X_C \sim \mathcal{N}(\mu_C, \sigma^2_X/n_C)\)
• Thus \[\bar X_T - \bar X_C \sim \mathcal{N}(\mu_T - \mu_c, \sigma^2_X/n_T+ \sigma^2_X/n_C) \implies \frac{\bar X_T - \bar X_C - (\mu_T - \mu_C)}{\sigma_X\sqrt{1/n_T + 1/n_C}} \sim \mathcal{N}(0,1)\]

• However, \(\sigma_X\) is unknown and we must consider an estimator: By using the pooled variance estimator we can use the estimate \[\hat \sigma^2_X = \frac{\sum_{i=1}^{n_T}(x_i - \bar x_T)^2 + \sum_{i=1}^{n_C}(x_i - \bar x_C)^2}{n_T + n_C - 2} \]

• Having absumed the equality of variances in the two groups (that can be testesd), the test statistic under \(H_0: \mu_T = \mu_C\), is distributed according to a student’s t distribution with \(n_T + n_C - 2\) degrees of freedom, that is

\[T = \frac{\bar X_T - \bar X_C}{\hat \sigma_X\sqrt{1/n_T + 1/n_C}} \sim t_{n_T + N_C - 2}\]