# 1.2. Mean Estimation in the Binary Choice Problem


Another motivating example can be found in the estimation of mean of the binary choice model.

## Mean estimation in the binary classification

Our model is from the (case 2) of the previous article.

$P(Y=1|Z=z) = F_0(z),~ Z\sim Q,~ F_0\in\Lambda, \\ \Lambda := \{F:\mathbb R \to [0,1],~ F \text{ is increasing}\}.$

This time the estimand is the population mean

\begin{aligned} \theta_0=\theta_{F_0} &:=\int F_0(z) dQ(z) = E F_0(Z) \\ &= P(Y=1) = EY. \end{aligned}

By definition, it is clear that $E\big[ Y_i - F_0(Z_i) \big] = 0.$

We will only consider the case where $Q$ is known. In this case, the MLE of $F_0$ becomes

$\hat\theta_n = \theta_{\hat F_n} = \int \hat F_n(z) dQ(z),$

where $\hat F_n$ is the MLE of $F_0.$

For $F \in \Lambda,$ define

$\theta_F = \int F(z) dQ(z),$

then by the classical central limit theorem, we get

$\frac 1 {\sqrt n} \sum_{i=1}^n \big( F(Z_i) - \theta_F \big) \overset d \to \mathcal{N}(0, \text{Var}(F(Z))).$

Again, if we prove some form of functional CLT for all $F\in\Lambda,$ then together with the result form the classical CLT, we get the asymptotic distribution of the MLE.

$$\frac 1 {\sqrt n} \sum_{i=1}^n \big( \hat F_n(Z_i) - \hat\theta_n \big) = \frac 1 {\sqrt n} \sum_{i=1}^n \big( F_0(Z_i) - \theta_0 \big) + o_P(1) \\ \implies \sqrt n (\hat\theta_n - \theta_0) \overset d \to \mathcal{N}\left( 0, \int F_0(z)(1-F_0(z)) dQ(z) \right).$$

We will take the fact $\frac1n \sum_{i=1}^n \hat F_n(Z_i)=\bar Y$ for granted. By this fact, \begin{aligned} \sqrt n (\hat\theta_n - \theta_0) &= \sqrt n (\bar Y - \theta_0) - \sqrt n (\bar Y - \hat\theta_n) \\ &= \sqrt n (\bar Y - \theta_0) - \frac 1 {\sqrt n} \sum_{i=1}^n \big( \hat F_n(Z_i) - \hat\theta_n \big) \\ &= \sqrt n (\bar Y - \theta_0) - \frac 1 {\sqrt n} \sum_{i=1}^n \big( F_0(Z_i) - \theta_0 \big) + o_P(1) \\ &= \frac 1 {\sqrt n} \sum_{i=1}^n \big( Y_i - \cancel{\theta_0} - F_0(Z_i) + \cancel{\theta_0} \big) + o_P(1). \end{aligned} By the classical CLT, \begin{aligned} \frac 1 {\sqrt n} \sum_{i=1}^n \big( Y_i - F_0(Z_i) \big) \overset d \to \mathcal{N}\left( 0, \text{Var}(Y-F_0(Z)) \right). \end{aligned} The variance can be rewritten as \begin{aligned} \text{Var}(Y-F_0(Z)) &= E[Y-F_0(Z)]^2 \\ &= E[Y^2 - 2YF_0(Z) + F_0^2(Z)] \\ &= E[Y^2 - F_0^2(Z)] \\ &= EY - EF_0^2(Z) \\ &= EF_0(Z) - EF_0^2(Z). \end{aligned} Hence the result follows from the Slutsky's theorem.

References

• van de Geer. 2000. Empirical Processes in M-estimation. Cambridge University Press.
• Theory of Statistics II (Fall, 2020) @ Seoul National University, Republic of Korea (instructor: Prof. Jaeyong Lee).