### 商科代写|计量经济学代写Econometrics代考|ECON 7204

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• Statistical Inference 统计推断
• Statistical Computing 统计计算
• (Generalized) Linear Models 广义线性模型
• Statistical Machine Learning 统计机器学习
• Longitudinal Data Analysis 纵向数据分析
• Foundations of Data Science 数据科学基础

## 商科代写|计量经济学代写Econometrics代考|Simulation and Comparisons with Other Estimators

In this section, we compare the LSE with the Simple Score Estimator (SSE), the Efficient Score Estimator (ESE), the Effective Dimension Reduction (EDR) estimate, the spline estimate, the MAVE estimate, and the EFM estimate. We take part in the simulation settings in Balabdaoui et al. (2019a), which means that we take the dimension $d$ equal to 2 . Since the parameter belongs to the boundary of a circle in this case, we only have to determine a one-dimensional parameter. Using this fact, we use the parameterization $\alpha=\left(\alpha_{1}, \alpha_{2}\right)=(\cos (\beta), \sin (\beta))$ and determine the angle $\beta$ by a golden section search for the SSE, ESE, and spline estimate. For EDR, we used the $\mathrm{R}$ package edr: the method is discussed in Hristache et al. (2001). The spline method is described in Kuchibhotla and Patra (2020), and there exists an $R$ package simest for it, but we used our own implementation. For the MAVE method, we used the R package MAVE; for theory, see Xia (2006). For the EFM estimate (see Cui et al. 2011), we used an $\mathrm{R}$ script, due to Xia Cui and kindly provided to us by her and Rohit Patra. All runs of our simulations can be reproduced by running the $R$ scripts in Groeneboom $2018 .$

In simulation model 1 , we take $\alpha_{0}=(1 / \sqrt{2}, 1 / \sqrt{2})^{T}$ and $X=\left(X_{1}, X_{2}\right)^{T}$, where $X_{1}$ and $X_{2}$ are independent Uniform $(0,1)$ variables. The model is now
$$Y=\psi_{0}\left(\alpha_{0}^{T} \boldsymbol{X}\right)+\varepsilon$$
where $\psi_{0}(u)=u^{3}$ and $\varepsilon$ is a standard normal random variable, independent of $\boldsymbol{X}$.
In simulation model 2 , we also take $\alpha_{0}=(1 / \sqrt{2}, 1 / \sqrt{2})^{T}$ and $\boldsymbol{X}=\left(X_{1}, X_{2}\right)^{T}$, where $X_{1}$ and $X_{2}$ are independent Uniform $(0,1)$ variables. This time, however, the model is (Table 1)
$$Y=\operatorname{Bin}\left(10, \exp \left(\boldsymbol{\alpha}{0}^{T} \boldsymbol{X}\right) /\left{1+\exp \left(\boldsymbol{\alpha}{0}^{T} \boldsymbol{X}\right)\right}\right)$$
see also Table 2 in Balabdaoui et al. (2019a). This means $$Y=\psi_{0}\left(\boldsymbol{\alpha}{0}^{T} \boldsymbol{X}\right)+\varepsilon$$ where $$\psi{0}\left(\boldsymbol{\alpha}{0}^{T} \boldsymbol{X}\right)=10 \exp \left(\boldsymbol{\alpha}{0}^{T} \boldsymbol{X}\right) /\left{1+\exp \left(\boldsymbol{\alpha}{0}^{T} \boldsymbol{X}\right)\right}, \quad \varepsilon=N{n}-\psi_{0}\left(\boldsymbol{\alpha}{0}^{T} \boldsymbol{X}\right),$$ and $$N{n}=\operatorname{Bin}\left(10, \frac{\exp \left(\boldsymbol{\alpha}{0}^{T} \boldsymbol{X}\right.}{1+\exp \left(\boldsymbol{\alpha}{0}^{T} \boldsymbol{X}\right)}\right)$$
Note that indeed $\mathbb{E}{\varepsilon \mid \boldsymbol{X})=0$, but that we do not have independence of $\varepsilon$ and $\boldsymbol{X}$, as in the previous example.

## 商科代写|计量经济学代写Econometrics代考|Concluding Remarks

We replaced the “crossing of zero” estimators in Balabdaoui et al. (2019b) with profile least squares estimators. The asymptotic distribution of the estimators was determined and its behavior illustrated by a simulation study, using the same models as in Balabdaoui et al. (2019a).

In the first model, the error is independent of the covariate and homoscedastic and in this case, four of the estimators were efficient. In the other (binomial-logistic) model, the error was dependent on the covariates and not homoscedastic. It was shown that the Simple Score Estimate (SSE) had in fact a smaller asymptotic variance in this model than the other estimators for which the asymptotic variance is known, although the difference is very small and does not really show up in the simulations.
There is no uniformly best estimate in our simulation, but the EDR estimate is clearly inferior to the other estimates, including the LSE, in particular for the lower sample sizes. On the other hand, the LSE is inferior to the other estimators except for the EDR. All simulation results can be reproduced by running the $R$ scripts in Groeneboom (2018).

## 商科代写|计量经济学代写Econometrics代考|Mathematical Context

We assume to be given a sample $\mathscr{D}{n}=\left{\left(X{1}, Y_{1}\right), \ldots,\left(X_{n}, Y_{n}\right)\right}$ of i.i.d. observations, where each pair $\left(X_{i}, Y_{i}\right)$ takes values in $\mathscr{X} \times \mathscr{Y}$. Throughout, $\mathscr{X}$ is a Borel subset of $\mathbb{R}^{d}$, and $\mathscr{Y} \subset \mathbb{R}$ is either a finite set of labels (for classification) or a subset of $\mathbb{R}$ (for regression). The vector space $\mathbb{R}^{d}$ is endowed with the Euclidean norm $|\cdot|$.
Our goal is to construct a predictor $F: \mathscr{X} \rightarrow \mathbb{R}$ that assigns a response to each possible value of an independent random observation distributed as $X_{1}$. In the context of gradient boosting, this general problem is addressed by considering a class $\mathscr{F}$ of functions $f: \mathscr{X} \rightarrow \mathbb{R}$ (called the weak or base learners) and minimizing some empirical risk functional
$$C_{n}(F)=\frac{1}{n} \sum_{i=1}^{n} \psi\left(F\left(X_{i}\right), Y_{i}\right)$$
over the linear combinations of functions in $\mathscr{F}$. The function $\psi: \mathbb{R} \times \mathscr{Y} \rightarrow \mathbb{R}{+}$, called the loss, is convex in its first argument and measures the cost incurred by predicting $F\left(X{i}\right)$ when the answer is $Y_{i}$. For example, in the least squares regression problem, $\psi(x, y)=(y-x)^{2}$ and
$$C_{n}(F)=\frac{1}{n} \sum_{i=1}^{n}\left(Y_{i}-F\left(X_{i}\right)\right)^{2} .$$
However, many other examples are possible, as we will see below. Let $\delta_{z}$ denote the Dirac measure at $z$, and let $\mu_{n}=(1 / n) \sum_{i=1}^{n} \delta_{\left(X_{i}, Y_{j}\right)}$ be the empirical measure associated with the sample $\mathscr{D}{n}$. Clearly, $$C{n}(F)=\mathbb{E} \psi(F(X), Y),$$
where $(X, Y)$ denotes a random pair with distribution $\mu_{n}$ and the symbol $\mathbb{E}$ denotes the expectation with respect to $\mu_{n}$. Naturally, the theoretical (i.e., population) version of $C_{n}$ is
$$C(F)=\mathbb{E} \psi\left(F\left(X_{1}\right), Y_{1}\right),$$
where now the expectation is taken with respect to the distribution of $\left(X_{1}, Y_{1}\right)$. It turns out that most of our subsequent developments are independent of the context,whether empirical or theoretical. Therefore, to unify the notation, we let throughout $(X, Y)$ be a generic pair of random variables with distribution $\mu_{X, Y}$, keeping in mind that $\mu_{X, Y}$ may be the distribution of $\left(X_{1}, Y_{1}\right)$ (theoretical risk), the standard empirical measure $\mu_{n}$ (empirical risk), or any smoothed version of $\mu_{n}$.

## 商科代写|计量经济学代写Econometrics代考|Simulation and Comparisons with Other Estimators

Y=\operatorname{Bin}\left(10, \exp \left(\boldsymbol{\alpha}{0}^{T} \boldsymbol{X}\right) /\left{1+\exp \left(\粗体符号{\alpha}{0}^{T} \boldsymbol{X}\right)\right}\right)Y=\operatorname{Bin}\left(10, \exp \left(\boldsymbol{\alpha}{0}^{T} \boldsymbol{X}\right) /\left{1+\exp \left(\粗体符号{\alpha}{0}^{T} \boldsymbol{X}\right)\right}\right)

\psi{0}\left(\boldsymbol{\alpha}{0}^{T} \boldsymbol{X}\right)=10 \exp \left(\boldsymbol{\alpha}{0}^{T} \ boldsymbol{X}\right) /\left{1+\exp \left(\boldsymbol{\alpha}{0}^{T} \boldsymbol{X}\right)\right}, \quad \varepsilon=N{ n}-\psi_{0}\left(\boldsymbol{\alpha}{0}^{T} \boldsymbol{X}\right),\psi{0}\left(\boldsymbol{\alpha}{0}^{T} \boldsymbol{X}\right)=10 \exp \left(\boldsymbol{\alpha}{0}^{T} \ boldsymbol{X}\right) /\left{1+\exp \left(\boldsymbol{\alpha}{0}^{T} \boldsymbol{X}\right)\right}, \quad \varepsilon=N{ n}-\psi_{0}\left(\boldsymbol{\alpha}{0}^{T} \boldsymbol{X}\right),和

ñn=垃圾桶⁡(10,经验⁡(一个0吨X1+经验⁡(一个0吨X))

## 商科代写|计量经济学代写Econometrics代考|Mathematical Context

Cn(F)=1n∑一世=1nψ(F(X一世),是一世)

Cn(F)=1n∑一世=1n(是一世−F(X一世))2.

Cn(F)=和ψ(F(X),是),

C(F)=和ψ(F(X1),是1),

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