## Example Problems on M-Estimators

ST793 Homework 9 on M-Estimators

# 7.7

The generalized method of moments (GMM) is an important estimation method found mainly in the econometrics literature and closely related to M-estimation. Suppose that we have iid random variables $Y_1, \dots , Y_n$ and a $p$ dimensional unknown parameter $\theta$. The key idea is that there are a set of $q \geq p$ possible estimating equations

$\frac{ 1 }{ n } \sum_{i=1}^{n} \psi_j(Y_i; \theta) = 0. \ j = 1, \dots , q$

motivated by the fact that $E_{\psi_j}(Y_1; \theta_0) = 0$ , $j = 1, \dots , q$, where $\theta_0$ is the true value. These motivating zero expectaions come from the theory in the subject area being studied. But notice that if $q > p$, then we have too many equations. The GMM approach is to minimize the objective function

$T = \Big[ \frac{ 1 }{n }\sum_{i=1}^{n} \psi(Y_{i};\theta) \Big]^T W \Big[ \frac{ 1 }{n }\sum_{i=1}^{n} \psi(Y_{i};\theta) \Big].$

To find $\widehat \theta$ we just take the partial derivative of $T$ with respect to $\theta$ and set it equal to 0:

$S(Y; \theta) = 2 w_1 \Big[ \frac{ 1 }{n }\sum_{i=1}^{n} \psi_1(Y_{i};\theta) \Big] \Big[ \frac{ 1 }{n }\sum_{i=1}^{n} \psi_1'(Y_{i};\theta) \Big] + 2 w_2 \Big[ \frac{ 1 }{n }\sum_{i=1}^{n} \psi_2(Y_{i};\theta) \Big]\Big[ \frac{ 1 }{n }\sum_{i=1}^{n} \psi_2'(Y_{i};\theta) \Big] = 0.$

## a

Prove that $S(Y; \theta_0)$ converges to 0 in probability as $\lim_{n\rightarrow \infty}$ making any moment assumptions that you need. (This suggest to you that the solution of the equation $S(Y; \theta)=0$ above is consistent.)

Let’s start by looking at the $\psi$ terms. By WLLN

$\frac{ 1 }{ n } \sum_{i=1}^{n} \psi_j (Y_i; \theta_0) \stackrel{ \text{p,d}}{\rightarrow} E( \psi_j(Y_1; \theta_0)) = 0.$

We can also apply WLLN for the $\psi’$ terms.

$\frac{ 1 }{ n } \sum_{i=1}^{n} \psi'_j (Y_i; \theta_0) \stackrel{ \text{p,d}}{\rightarrow} E( \psi_j(Y_1; \theta_0)) =A_{jj}$

Assuming that these $A_{jj}$ terms exist and are finite, we can use the continuous mapping theorem or Slutsky’s (since convergence in probability and distribution to a constant are the same) to say

$S(Y; \theta_0) \stackrel{ \text{p}}{\rightarrow} 2w_1 \cdot 0 \cdot A_{11} + 2w_2 \cdot 0 \cdot A_{22} = 0.$

## b

To get asymptotic normality for $\widehat \theta$, a direct approach is to expand $S(Y; \widehat \theta)$ around $\theta_0$ and solve for $\widehat \theta - \theta_0$.

$\widehat \theta - \theta_0 = \Big[ - \frac{ \partial S(Y; \theta_0) }{\partial \theta^T} \Big]^{-1} S(Y; \theta_0) + \Big[ - \frac{ \partial S(Y; \theta_0) }{\partial \theta^T} \Big]^{-1} R_n$

Then one ignores the remainder term and uses Slutsky’s Theorem along with asymptotic normality of $S(Y; \theta_0)$. But how to get the asymptotic normality of $S(Y; \theta_0)$? Find $h(Y_i;\theta_0)$ such that

$S(Y; \theta_0) = \frac{ 1 }{ n } \sum_{i=1}^{n} h(Y_i;\theta_0) + R_n$

No proofs required.

Recall that

$S(Y; \theta) = 2 w_1 \Big[ \frac{ 1 }{n }\sum_{i=1}^{n} \psi_1(Y_{i};\theta) \Big] \Big[ \frac{ 1 }{n }\sum_{i=1}^{n} \psi_1'(Y_{i};\theta) \Big] + 2 w_2 \Big[ \frac{ 1 }{n }\sum_{i=1}^{n} \psi_2(Y_{i};\theta) \Big]\Big[ \frac{ 1 }{n }\sum_{i=1}^{n} \psi_2'(Y_{i};\theta) \Big]$

Then,

\begin{align} S(Y; \theta) & = 2 w_1 \Big[ \frac{ 1 }{n }\sum_{i=1}^{n} \psi_1(Y_{i};\theta) \Big] \Big[ \frac{ 1 }{n }\sum_{i=1}^{n} \psi_1'(Y_{i};\theta) \Big] + 2 w_2 \Big[ \frac{ 1 }{n }\sum_{i=1}^{n} \psi_2(Y_{i};\theta) \Big]\Big[ \frac{ 1 }{n }\sum_{i=1}^{n} \psi_2'(Y_{i};\theta) \Big] \\ & =2 w_1 \frac{ 1 }{ n^2 } \sum_{i=1}^{n} \sum_{j=1}^{n} \psi_1(Y_{i};\theta) \psi_1'(Y_{j};\theta) + 2 w_2 \frac{ 1 }{ n^2 } \sum_{i=1}^{n} \sum_{j=1}^{n}\psi_2(Y_{i};\theta) \psi_2'(Y_{j};\theta) \\ & = \frac{ 1 }{ n }\sum_{i=1}^{n} \frac{ 1 }{ n }\sum_{j=1}^{n} 2 w_1 \psi_1(Y_{i};\theta) \psi_1'(Y_{j};\theta) + 2 w_2 \frac{ 1 }{ n^2 } \psi_2(Y_{i};\theta) \psi_2'(Y_{j};\theta) \end{align}

So we have

$h(Y_i; \theta_0) = \frac{ 1 }{ n }\sum_{j=1}^{n} 2 w_1 \psi_1(Y_{i};\theta) \psi_1'(Y_{j};\theta) + 2 w_2 \frac{ 1 }{ n^2 } \psi_2(Y_{i};\theta) \psi_2'(Y_{j};\theta).$

A full proof would require showing that this has expectation 0 and finite variance.

## c

The equation $S(Y; \theta)= 0$ is not in the form for using M-estimation results (because the product of sums is not a simple sum). Show how to get it in M-estimation form by adding two new parameters, $\theta_2$ and $\theta_3$, and two new equations so that the result is a system of three equations with three $\psi$ functions; call them $\psi_1^*, \psi_2^*,$ and $\psi_3^*$ because $\psi_1^*$ is actually a function of the original $\psi_1$ and $\psi_2$.

Take $\theta_2$ and $\theta_3$ such that

$\sum_{i=1}^{n} (\psi'_1(Y_i; \theta) - \theta_2) = \sum_{i=1}^{n} (\psi'_2(Y; \theta) - \theta_3) = 0.$

Then we get

$\psi^*(Y_i; \theta) = \begin{bmatrix} 2 w_1 \theta_2 \psi_1(Y_i; \theta) + 2 w_2 \theta_3 \psi_2(Y_i; \theta) \\ \psi'_1(Y_i; \theta) - \theta_2 \\ \psi'_2(Y_i; \theta) - \theta_3 \end{bmatrix}.$

# 7.9

Let $Y_1, \dots , Y_n$ be iid from some continuous distribution with nonzero density at the population mean, $f(\mu_0)>0$. For defining a “positive” standard deviation similar to the positive mean deviation from the median, one needs to estimate the proportion of the population that lies above the mean, $1-p_0 = 1 - F(\mu_0)$, using $1 - F_n(\overline Y)$ where $F_n$ is the empirical distribution function. Letting $\psi(Y_i; \mu, p)^T = (Y_i - \mu, I(Y_i \leq \mu) - p)$, find $A(\theta_0)$, $B(\theta_0)$, and $V(\theta_0)$. Note that we have to use the nonsmooth definition of $A(\theta_0)$:

$A(\theta_0) - \frac{ \partial }{\partial \theta} \{ E_F(\psi(Y_i; \mu, p)) \}\rvert_{\mu = \mu_0, p = p_0}.$

The off-diagonal element of $B$ turns out to be $b_{12} = \int_{-\infty}^{\mu_0} y f(y) dy - p_0 \mu_0$. In your calculations, just let this be called $b_{12}$.

We can start by finding $A(\theta_0)$.

\begin{align} A(\theta_0) & = \frac{ -\partial }{\partial \theta} \Big( E \begin{bmatrix} Y_i - \mu \\ I(Y_i \leq \mu) - p \end{bmatrix}\Big) \rvert_{\mu =\mu_0, p = p_0} \\ & = \frac{ -\partial }{\partial \theta} \Big( \begin{bmatrix} \mu_0 - \mu \\ P(Y_i \leq \mu) - p \end{bmatrix}\Big) \rvert_{\mu =\mu_0, p = p_0} \\ & = \frac{ -\partial }{\partial \theta} \Big( \begin{bmatrix} \mu_0 - \mu \\ F(\mu)- p \end{bmatrix}\Big) \rvert_{\mu =\mu_0, p = p_0} \\ & = \begin{bmatrix} 1 & 0 \\ -f(\mu_0) & 1 \end{bmatrix} \end{align}

Now we can find $B(\theta_0)$.

\begin{align} B(\theta_0) & = E(\psi(Y_1; \theta_0) \psi(Y_1; \theta_0)^T) \\ & = E \begin{bmatrix} (Y_i - \mu_0)^2 & (Y_i - \mu_0) (I(Y_i \leq \mu_0) - p_0) \\ (Y_i - \mu_0) (I(Y_i \leq \mu_0) - p_0) & (I(Y_i \leq \mu_0) - p_0)^2 \end{bmatrix} \\ & = \begin{bmatrix} \sigma_0^2 & E\Big[ (Y_i - \mu_0) (I(Y_i \leq \mu_0) - p_0) \Big] \\ E\Big[ (Y_i - \mu_0) (I(Y_i \leq \mu_0) - p_0) \Big] & E \Big[ I(Y_i \leq \mu_0)^2 - 2 p_0 I(Y_i \leq \mu_0) + p_0^2 \Big] \end{bmatrix} \\ \\ E\Big[ (Y_i - \mu_0) (I(Y_i \leq \mu_0) - p_0) \Big] & = E(Y_i I(Y_i \leq \mu_0)) - E( \mu_9 I(Y_i \leq \mu_0)) - p_0 E(Y_1) + \mu_0 p_0 \\ & = \int_{-\infty}^{\mu_0} y f(y) dy - p_0 \mu_0 \\ & = b_{12} ✅ \\ E \Big[ I(Y_i \leq \mu_0)^2 - 2 p_0 I(Y_i \leq \mu_0) + p_0^2 \Big] & = F(\mu_0) - 2 p_0 F(\mu_0) + p_0^2 \\ & = p_0 - 2 p_0^2 + p_0^2 \\ & = p_0 (1 - p_0) \\ \\ B(\theta_0) & = \begin{bmatrix} \sigma_0^2 & b_{12} \\ b_{12} & p_0(1-p_0) \end{bmatrix} \end{align}

Now we can find $V(\theta_0)$.

\begin{align} V(\theta_0) & = \begin{bmatrix} 1 & 0 \\ -f(\mu_0) & 1 \end{bmatrix} \begin{bmatrix} \sigma_0^2 & b_{12} \\ b_{12} & p_0(1-p_0) \end{bmatrix} \begin{bmatrix} 1 & -f(\mu_0) \\ 0 & 1 \end{bmatrix} \\ & = \begin{bmatrix} \sigma_0^2 & b_{12} \\ f(\mu_0) \sigma^2 + b_{12} & f(\mu_0) b_{12} + p_0(1-p_0) \end{bmatrix} \begin{bmatrix} 1 & -f(\mu_0) \\ 0 & 1 \end{bmatrix} \\ & = \begin{bmatrix} \sigma_0^2 & \sigma_0^2 f(\mu_0) + b_{12} \\ f(\mu_0) \sigma_0^2 + b_{12} & \sigma_0^2 f(\mu_0) + 2 b_{12} f(\mu_0) + p_0 ( 1- p_0) \end{bmatrix} \end{align}

# 7.11

Consider the simple no intercept model

$Y_i = \beta x_i + e_i, \ i = 1, \dots, n,$

where the $x_1, \dots x_n$ are fixed constants and $e_1, \dots e_n$ are independent with $E(\psi^*(e_i)) = 0$ and $E(\psi^*(e_i)^2) < \infty$ with values possibly depending on $i$, and $\psi^*$ is a real-valued function (with the subscript “$*$” to differentiate from our usual $\psi(Y_i, x_i, \beta)$).

## a

A classical robust M-estimator $\widehat \beta$ satisfies

$\sum_{i=1}^{n} \psi^*(Y_i - \widehat \beta x_i) x_i = 0.$

Write down the asymptotic variance of $\widehat \beta$ and an estimator of this asymptotic variance. You may assume that the derivative $\psi^* ‘$ of $\psi^*$ exists everywhere and that $E | \psi^* ‘ (e_i) |<\infty$ for all $i$.

First we need to find $A(\beta_0)$ and $B(\beta_0)$.

\begin{align} A(\beta_0) & = - E(\frac{ \partial \psi^*(Y_i - \beta x_i) x_i}{\partial \beta} ) \\ & = -E( \psi^{* '}(Y_i - \beta_0 x_i) (-x_i^2)) \\ & = E(\psi^{*'}(Y_i - \beta_0 x_i)) x_i^2 \\ \\ B(\beta_0) & =E((\psi^*(Y_i - \beta_0 x_i)x_i)^2) \\ & = x_i^2 E(\psi^*(Y_i - \beta_0 x_i)^2) \end{align}

Then,

$V(\beta_0 )= A^{-1}(\beta_0) B(\beta_0) A^{-1}(\beta_0) = \frac{ E(\psi^{*}(Y_i - \beta_0 x_i)^2 )}{ x_i^2 E(\psi^{*'}(Y_i - \beta_0 x_i))^2 }$

So our asymptotic variance of an estimate $\widehat \beta$ is

$\text{AsyVar}( \widehat \beta ) = \frac{ 1 }{ n } \sum_{i=1}^{n} \frac{ E(\psi^{*}(Y_i - \widehat \beta x_i)^2 )}{ x_i^2 E(\psi^{*'}(Y_i - \widehat \beta x_i))^2 }.$

As we have done in the past, we can estimate the numerator and denominator using method of moments.

\begin{align} E\Big( \widehat{\psi^*(Y_i - \widehat \beta x_i)^2} \Big) & = \frac{ 1 }{ n } \sum_{i=1}^{n} \psi^*(Y_i - x_i \widehat \beta)^2 \\ E \Big( \widehat{\psi^{*'}(Y_i - \widehat \beta x_i )} \Big) & = \frac{ 1 }{ n } \sum_{i=1}^{n} \psi^{*'}(Y_i - \widehat \beta x_i) \end{align}

## b

Let’s simplify and assume that $e_1, \dots e_n$ are iid. How does the asymptotic variance expression from a) change? Find an improved asymptotic variance estimator in this situation.

With iid $e_i$’s, we now know that the expectations are the same for every $i$ and we get an asymptotic variance

$\text{AsyVar}( \widehat \beta ) = \frac{ E(\psi^{*}(Y_i - \widehat \beta x_i)^2 )}{ x_i^2 E(\psi^{*'}(Y_i - \widehat \beta x_i))^2 }.$

Notice that now we only need to estimate one expected value rather than many.

# 8.10

Suppose we have data $X_1, \dots , X_n$ that are iid. The sign test for $H_0: \text{median} = 0$ is to count the number of $X$’s that are above 0, say $Y$, and compare $Y$ to a binomial($n, p = 1/2$) distribution. Starting with the defining M-estimator equation for the sample median (see Example 7.4.2, p 312, with $p = 1/2$),

## a

Derive the generalized score statistic $T_{GS}$ for $H_0: \text{median} = 0$ and note that it is the large sample version of the two-sided sign statistic.

We know $\psi = 0.5 - I(Y_i \leq \theta)$. Since $\theta$ is a scalar, we know $V(\theta_0) = B(\theta_0) = p(1-p) = \frac{ 1 }{ 4 }$. Thus,

$T_{GS} = \frac{ 1 }{ n } \Big[ \sum_{i=1}^{n} \psi(Y_i; 0) \Big] (\frac{ 1 }{ 4 })^{-1}\Big[ \sum_{i=1}^{n} \psi(Y_i; 0) \Big] = \frac{ 4 }{ n } \Big[ \sum_{i=1}^{n} 0.5 - I(Y_i \leq 0) \Big] ^2$

## b

Using the expression for the asymptotic variance of the sample median, write down the form of a generalized Wald statistic $T_{GW}$, and explain why it is not as attractive to use here as $T_{GS}$.

From 7.4.2 we know $V(\theta) = \frac{ 1 }{ 4 f(\theta)^2 }$ Thus, our generalized Wald statistics is

$T_{GW} = n ( \widehat \theta - 0) \frac{ 1 }{ 4 f(\widehat \theta)^2} (\widehat \theta - 0) = \frac{ n \widehat \theta^2 }{ 4 f(\widehat \theta) }.$

This is less attractive than the score statistics because it requires that we know the distribution $f$, which we are not necessarily given.