Chapter 6: Problem 14
Consider a location model $$ X_{i}=\theta+e_{i}, \quad i=1, \ldots, n $$ where \(e_{1}, e_{2}, \ldots, e_{n}\) are iid with pdf \(f(z)\). There is a nice geometric interpretation for estimating \(\theta\). Let \(\mathbf{X}=\left(X_{1}, \ldots, X_{n}\right)^{\prime}\) and \(\mathbf{e}=\left(e_{1}, \ldots, e_{n}\right)^{\prime}\) be the vectors of observations and random error, respectively, and let \(\boldsymbol{\mu}=\theta \mathbf{1}\), where \(\mathbf{1}\) is a vector with all components equal to 1 . Let \(V\) be the subspace of vectors of the form \(\mu\); i.e., \(V=\\{\mathbf{v}: \mathbf{v}=a \mathbf{1}\), for some \(a \in R\\}\). Then in vector notation we can write the model as $$ \mathbf{X}=\boldsymbol{\mu}+\mathbf{e}, \quad \boldsymbol{\mu} \in V $$ Then we can summarize the model by saying, "Except for the random error vector e, X would reside in \(V\)." Hence, it makes sense intuitively to estimate \(\boldsymbol{\mu}\) by a vector in \(V\) that is "closest" to \(\mathbf{X}\). That is, given a norm \(\|\cdot\|\) in \(R^{n}\), choose $$ \widehat{\boldsymbol{\mu}}=\operatorname{Argmin}\|\mathbf{X}-\mathbf{v}\|, \quad \mathbf{v} \in V $$ (a) If the error pdf is the Laplace, \((2.2 .4)\), show that the minimization in \((6.3 .27)\) is equivalent to maximizing the likelihood when the norm is the \(l_{1}\) norm given by $$ \|\mathbf{v}\|_{1}=\sum_{i=1}^{n}\left|v_{i}\right| $$ (b) If the error pdf is the \(N(0,1)\), show that the minimization in \((6.3 .27)\) is equivalent to maximizing the likelihood when the norm is given by the square of the \(l_{2}\) norm $$ \|\mathbf{v}\|_{2}^{2}=\sum_{i=1}^{n} v_{i}^{2} $$
Short Answer
Step by step solution
Key Concepts
These are the key concepts you need to understand to accurately answer the question.