Tom_Leavitt_difference_in_means_estimator.tex

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\begin{document}

\begin{titlepage}
\title{ICPSR Causal Inference:\\Unbiasedness of Difference-in-Means Estimator under Complete Random Assignment}
\author{Thomas Leavitt}
\date{\today}
\maketitle
\end{titlepage}

\begin{lem} \label{lem: exp val Z}
Under complete, uniform random assignment, $\E_{\Omega}\left[Z_i\right] = \frac{n_t}{n}$ for all $i \in \left\{1, \dots , n\right\}$ units.
\end{lem}
\begin{proof}

First note that in general there are $\binom{n}{k} = \frac{n!}{\left(n - k\right)!k!}$ ways to choose a subset of $k$ elements from a fixed set of $n$ elements. Therefore, the number of ways to choose a subset of $n_t$ treated units from a fixed population of $n$ units is $\binom{n}{n_t} = \frac{n!}{\left(n - n_t\right)!n_t!}$. Notice that the $n - n_t$ units that are not selected for treatment are assigned to the control condition; hence, $n - n_t = n_c$ and $\binom{n}{n_t} = \frac{n!}{n_c!n_t!}$. 

Now let's condition on the event that an arbitrary unit $i$ is assigned to the treatment condition. The number of ways in which one could select $n_t - 1$ units from the remaining population of $n - 1$ units is $\binom{n - 1}{n_t - 1}$. Hence, the proportion of treatment assignment permutations in which unit $i$ is in the treatment condition is $\frac{\binom{n - 1}{n_t - 1}}{\binom{n}{n_t}}$.
First note that we can write $\frac{\binom{n - 1}{n_t - 1}}{\binom{n}{n_t}}$ as $\frac{\left(\frac{\left(n - 1\right)!}{\left(n - 1 - n_t - 1\right)!\left(n_t - 1\right)!}\right)}{\left(\frac{n!}{n_c!n_t!}\right)}$ based on the definition of the binomial coefficient, $\binom{n}{k}$, given above.

Since $n_c = n - n_t$ and we are conditioning on the event that one arbitrary unit $i$ has been selected into the treatment condition, in which case there is one fewer number of treated units, $n_t$, and one fewer unit in the population, $n$, it follows that $n_c = n - 1 - n_t - 1$. We can therefore substitute $n_c$ for $n - 1 - n_t - 1$:
\begin{align*}
\frac{\binom{n - 1}{n_t - 1}}{\binom{n}{n_t}} \\ 
& = \frac{\left(\frac{\left(n - 1\right)!}{\left(n - 1 - n_t - 1\right)!\left(n_t - 1\right)!}\right)}{\left(\frac{n!}{n_c!n_t!}\right)} \\
& = \frac{\left(\frac{\left(n - 1\right)!}{n_c!\left(n_t - 1\right)!}\right)}{\left(\frac{n!}{n_c!n_t!}\right)} \\
\end{align*}
Now we can simply manipulate the expression above and cancel terms until we are left with $\frac{n_t}{n}$:
\begin{align*}
& = \left(\frac{\left(n - 1\right)!}{n_c!\left(n_t - 1\right)!}\right)\left(\frac{n_c!n_t!}{n!}\right) \\
& = \left(\frac{\left(n - 1\right)\left(n - 2\right) \dots 1}{n_c\left(n_c - 1\right) \dots 1 \left(n_t - 1\right)\left(n_t - 2\right) \dots 1}\right) \left(\frac{n_c\left(n_c - 1\right) \dots 1 n_t \left(n_t - 1\right) \dots 1}{n\left(n - 1\right) \dots 1}\right) \\
& = \frac{\left(n - 1\right)\left(n - 2\right) \dots 2 n_c\left(n_c - 1\right) \dots 2 n_t \left(n_t - 1\right) \dots 2}{n_c\left(n_c - 1\right) \dots 2 \left(n_t - 1\right)\left(n_t - 2\right) \dots 2 n\left(n - 1\right) \dots 2} \\
& = \frac{n_t}{n}.
\end{align*}
Therefore, exactly $\frac{n_t}{n}$ out of all treatment assignment permutations will be those in which unit $i$ is in the treatment condition. 

Assuming a uniform distribution on the set of treatment assignment permutations, $\Omega$, then the probability of each permutation is $\frac{1}{\left\lvert\Omega\right\rvert}$ and the total probability of all permutations in which $i$ is treated is simply the sum of the probabilities of those permutations in which unit $i$ is in the treatment condition. Under uniform random assignment, the probability of each assignment permutation is $\frac{1}{\left\lvert\Omega\right\rvert}$. Thus, the probability that unit $i$ is treated is as follows:
\begin{align*}
\left(\frac{1}{\left\lvert\Omega\right\rvert}\right) \left(\left(\frac{n_t}{n}\right)\left(\left\lvert\Omega\right\rvert\right)\right) \\ 
& = \left(\frac{1}{\left\lvert\Omega\right\rvert}\right) \left(\frac{n_t\left\lvert\Omega\right\rvert}{n}\right) \\
& = \left(\frac{n_t\left\lvert\Omega\right\rvert}{\left\lvert\Omega\right\rvert n}\right) \\ 
& = \frac{n_t}{n}  
 \end{align*}
Since $\Pr\left(Z_i = 1\right) = \frac{n_t}{n}$ for all $i \in \left\{1, \dots , n\right\}$ units, it follows that the expected value of $Z_i \in \left\{0, 1\right\}$ is $\E_{\Omega}\left[Z_i\right] = 1\left(\frac{n_t}{n}\right) + 0\left(1 - \frac{n_t}{n}\right) = \frac{n_t}{n}$.
\end{proof}

\begin{prop} \label{prop: complete ran assign}
Under complete, uniform random assignment, $\mathbb{E}_{\Omega} \left[\frac{\mathbf{Z}^{\prime}\mathbf{Y}}{\mathbf{Z}^{\prime}\mathbf{Z}} - \frac{(\mathbf{1} - \mathbf{Z})^{\prime} \mathbf{Y}}{(\mathbf{1} - \mathbf{Z})^{\prime}(\mathbf{1} - \mathbf{Z})}\right] = \overline{y_{t}} - \overline{y_{c}} = \overline{\tau}$.
\end{prop}
\begin{proof}
\begin{align*}
\mathbb{E}_{\Omega} \left[\frac{\mathbf{Z}^{\prime}\mathbf{Y}}{\mathbf{Z}^{\prime}\mathbf{Z}} - \frac{(\mathbf{1} - \mathbf{Z})^{\prime} \mathbf{Y}}{(\mathbf{1} - \mathbf{Z})^{\prime}(\mathbf{1} - \mathbf{Z})}\right] \\
= \mathbb{E}_{\Omega} \left[\frac{\mathbf{Z}^{\prime}\mathbf{Y}}{\mathbf{Z}^{\prime}\mathbf{Z}}\right] - \mathbb{E}_{\Omega}\left[\frac{(\mathbf{1} - \mathbf{Z})^{\prime} \mathbf{Y}}{(\mathbf{1} - \mathbf{Z})^{\prime}(\mathbf{1} - \mathbf{Z})}\right] \\
\end{align*}
Notice that $\mathbf{Z}^{\prime}\mathbf{Y}$ is simply the sum of observed outcomes among treated units (i.e., the sum of observed outcomes among units for which $Z_i = 1$). The quantity $\mathbf{Z}^{\prime}\mathbf{y_t}$ is the sum of treatment potential outcomes among treated units. Since the observed outcomes for treated units is equal to those units' treatment potential outcomes, we can substitute $\mathbf{Z}^{\prime}\mathbf{y_t}$ for $\mathbf{Z}^{\prime}\mathbf{Y}$. Analogously, we can substitute $\left(\mathbf{1} - \mathbf{Z}\right)^{\prime}\mathbf{y_c}$ for $\left(\mathbf{1} - \mathbf{Z}\right)^{\prime}\mathbf{Y}$.
Substituting $\mathbf{Z}^{\prime}\mathbf{y_t}$ for $\mathbf{Z}^{\prime}\mathbf{Y}$ and $\left(\mathbf{1} - \mathbf{Z}\right)^{\prime}\mathbf{y_c}$ for $\left(\mathbf{1} - \mathbf{Z}\right)^{\prime}\mathbf{Y}$, we are left with:
\begin{align*}
= \mathbb{E}_{\Omega} \left[\frac{\mathbf{Z}^{\prime} \mathbf{y_t}}{\mathbf{Z}^{\prime}\mathbf{Z}}\right] - \mathbb{E}_{\Omega}\left[\frac{(\mathbf{1} - \mathbf{Z})^{\prime} \mathbf{y_c}}{(\mathbf{1} - \mathbf{Z})^{\prime}(\mathbf{1} - \mathbf{Z})}\right].
\end{align*}
Notice that under complete random assignment, the number of treatment units, $\mathbf{Z}^{\prime}\mathbf{Z}$, and the number of control units, $\left(\mathbf{1} - \mathbf{Z}\right)^{\prime}\left(\mathbf{1} - \mathbf{Z}\right)$ are fixed constants, which we denote by $n_t$ and $n_c$, respectively.
\begin{align*}
= \mathbb{E}_{\Omega} \left[\frac{\mathbf{Z}^{\prime} \mathbf{y_t}}{n_t}\right] - \mathbb{E}_{\Omega}\left[\frac{(\mathbf{1} - \mathbf{Z})^{\prime} \mathbf{y_c}}{n_c}\right] \\
= \left(\frac{1}{n_t} \right) \mathbb{E}_{\Omega} \left[\mathbf{Z}^{\prime} \mathbf{y_t}\right] - \left(\frac{1}{n_c}\right) \mathbb{E}_{\Omega}\left[(\mathbf{1} - \mathbf{Z})^{\prime} \mathbf{y_c}\right] \\
= \left(\frac{1}{n_t} \right) \mathbb{E}_{\Omega} \left[ \sum \limits_{i = 1}^n Z_i y_{ti}\right] - \left(\frac{1}{n_c}\right) \mathbb{E}_{\Omega}\left[\sum \limits_{i = 1}^n \left(1 - Z_i\right) y_{ci} \right] \\
= \left(\frac{1}{n_t} \right) \mathbb{E}_{\Omega} \left[ Z_1 y_{t1} + \dots + Z_n y_{tn}\right] - \left(\frac{1}{n_c}\right) \mathbb{E}_{\Omega}\left[\left(1 - Z_1\right) y_{c1} + \dots + \left(1 - Z_n\right) y_{cn} \right] \\
= \left(\frac{1}{n_t} \right) \mathbb{E}_{\Omega} \left[ Z_1 y_{t1}\right] + \dots + \mathbb{E}_{\Omega} \left[Z_n y_{tn}\right] - \left(\frac{1}{n_c}\right) \mathbb{E}_{\Omega}\left[\left(1 - Z_1\right) y_{c1}\right] + \dots + \mathbb{E}_{\Omega} \left[\left(1 - Z_n\right) y_{cn} \right] \\
= \left(\frac{1}{n_t} \right) y_{t1}\mathbb{E}_{\Omega} \left[ Z_1 \right] + \dots + y_{tn}\mathbb{E}_{\Omega} \left[Z_n\right] - \left(\frac{1}{n_c}\right) y_{c1}\mathbb{E}_{\Omega}\left[\left(1 - Z_1\right) \right] + \dots + y_{cn}\mathbb{E}_{\Omega} \left[\left(1 - Z_n\right) \right] \\
\end{align*}
In a uniform randomized experiment, $Z_1, \dots , Z_n$ are independent and identically distributed random variables, which implies that the expected value of each independent Bernouli random variable is identical. Hence, we can represent each $\mathbb{E}_{\Omega} \left[ Z_1 \right], \dots , \mathbb{E}_{\Omega} \left[ Z_n \right]$ by $\mathbb{E}_{\Omega} \left[ Z_i \right]$.

\begin{align*}
= \left(\frac{1}{n_t} \right) y_{t1}\mathbb{E}_{\Omega} \left[ Z_i \right] + \dots + y_{tn}\mathbb{E}_{\Omega} \left[Z_i\right] - \left(\frac{1}{n_c}\right) y_{c1}\mathbb{E}_{\Omega}\left[\left(1 - Z_i\right) \right] + \dots + y_{cn}\mathbb{E}_{\Omega} \left[\left(1 - Z_i\right) \right] \\
= \left(\frac{1}{n_t} \right) \mathbb{E}_{\Omega} \left[ Z_i \right] \left(y_{t1} + \dots + y_{tn} \right) - \left(\frac{1}{n_c}\right) \mathbb{E}_{\Omega}\left[\left(1 - Z_i\right) \right] \left(y_{c1} + \dots + y_{cn} \right) \\
\end{align*}
By Lemma \ref{lem: exp val Z}, $\mathbb{E}_{\Omega}\left[Z_i \right] = \left(\frac{n_t}{n}\right)$, which implies that $\mathbb{E}_{\Omega}\left[\left(1 - Z_i\right) \right] = 1 - \left(\frac{n_t}{n}\right) = \left(\frac{n_c}{n}\right)$. Hence, we can substitute $\left(\frac{n_t}{n}\right)$ for $\mathbb{E}_{\Omega}\left[Z_i \right]$ and $\left(\frac{n_c}{n}\right)$ for $\mathbb{E}_{\Omega}\left[1 - Z_i \right]$, respectively.
\begin{align*}
= \left(\frac{1}{n_t} \right) \left(\frac{n_t}{n}\right) \left(y_{t1} + \dots + y_{tn} \right) - \left(\frac{1}{n_c}\right) \left(\frac{n_c}{n}\right) \left(y_{c1} + \dots + y_{cn} \right) \\
= \left(\frac{1}{n} \right) \left(y_{t1} + \dots + y_{tn} \right) - \left(\frac{1}{n}\right) \left(y_{c1} + \dots + y_{cn} \right) \\
= \frac{\left(y_{t1} + \dots + y_{tn} \right)}{n} - \frac{\left(y_{c1} + \dots + y_{cn} \right)}{n} \\
= \overline{y_{t}} - \overline{y_{c}} \\
= \overline{\tau}.
\end{align*}
\end{proof}

\end{document}