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world:max

$$ \newcommand{\arginf}{\mathrm{arginf}} \newcommand{\argmin}{\mathrm{argmin}} \newcommand{\argmax}{\mathrm{argmax}} \newcommand{\asconv}[1]{\stackrel{#1-a.s.}{\rightarrow}} \newcommand{\Aset}{\mathsf{A}} \newcommand{\b}[1]{{\mathbf{#1}}} \newcommand{\ball}[1]{\mathsf{B}(#1)} \newcommand{\bbQ}{{\mathbb Q}} \newcommand{\bproof}{\textbf{Proof :}\quad} \newcommand{\bmuf}[2]{b_{#1,#2}} \newcommand{\card}{\mathrm{card}} \newcommand{\chunk}[3]{{#1}_{#2:#3}} \newcommand{\condtrans}[3]{p_{#1}(#2|#3)} \newcommand{\convprob}[1]{\stackrel{#1-\text{prob}}{\rightarrow}} \newcommand{\Cov}{\mathbb{C}\mathrm{ov}} \newcommand{\cro}[1]{\langle #1 \rangle} \newcommand{\CPE}[2]{\PE\lr{#1| #2}} \renewcommand{\det}{\mathrm{det}} \newcommand{\dimlabel}{\mathsf{m}} \newcommand{\dimU}{\mathsf{q}} \newcommand{\dimX}{\mathsf{d}} \newcommand{\dimY}{\mathsf{p}} \newcommand{\dlim}{\Rightarrow} \newcommand{\e}[1]{{\left\lfloor #1 \right\rfloor}} \newcommand{\eproof}{\quad \Box} \newcommand{\eremark}{</WRAP>} \newcommand{\eqdef}{:=} \newcommand{\eqlaw}{\stackrel{\mathcal{L}}{=}} \newcommand{\eqsp}{\;} \newcommand{\Eset}{ {\mathsf E}} \newcommand{\esssup}{\mathrm{essup}} \newcommand{\fr}[1]{{\left\langle #1 \right\rangle}} \newcommand{\falph}{f} \renewcommand{\geq}{\geqslant} \newcommand{\hchi}{\hat \chi} \newcommand{\Hset}{\mathsf{H}} \newcommand{\Id}{\mathrm{Id}} \newcommand{\img}{\text{Im}} \newcommand{\indi}[1]{\mathbf{1}_{#1}} \newcommand{\indiacc}[1]{\mathbf{1}_{\{#1\}}} \newcommand{\indin}[1]{\mathbf{1}\{#1\}} \newcommand{\itemm}{\quad \quad \blacktriangleright \;} \newcommand{\jointtrans}[3]{p_{#1}(#2,#3)} \newcommand{\ker}{\text{Ker}} \newcommand{\klbck}[2]{\mathrm{K}\lr{#1||#2}} \newcommand{\law}{\mathcal{L}} \newcommand{\labelinit}{\pi} \newcommand{\labelkernel}{Q} \renewcommand{\leq}{\leqslant} \newcommand{\lone}{\mathsf{L}_1} \newcommand{\lrav}[1]{\left|#1 \right|} \newcommand{\lr}[1]{\left(#1 \right)} \newcommand{\lrb}[1]{\left[#1 \right]} \newcommand{\lrc}[1]{\left\{#1 \right\}} \newcommand{\lrcb}[1]{\left\{#1 \right\}} \newcommand{\ltwo}[1]{\PE^{1/2}\lrb{\lrcb{#1}^2}} \newcommand{\Ltwo}{\mathrm{L}^2} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mcbb}{\mathcal B} \newcommand{\mcf}{\mathcal{F}} \newcommand{\meas}[1]{\mathrm{M}_{#1}} \newcommand{\norm}[1]{\left\|#1\right\|} \newcommand{\normmat}[1]{{\left\vert\kern-0.25ex\left\vert\kern-0.25ex\left\vert #1 \right\vert\kern-0.25ex\right\vert\kern-0.25ex\right\vert}} \newcommand{\nset}{\mathbb N} \newcommand{\N}{\mathcal{N}} \newcommand{\one}{\mathsf{1}} \newcommand{\PE}{\mathbb E} \newcommand{\pminfty}{_{-\infty}^\infty} \newcommand{\PP}{\mathbb P} \newcommand{\projorth}[1]{\mathsf{P}^\perp_{#1}} \newcommand{\Psif}{\Psi_f} \newcommand{\pscal}[2]{\langle #1,#2\rangle} \newcommand{\pscal}[2]{\langle #1,#2\rangle} \newcommand{\psconv}{\stackrel{\PP-a.s.}{\rightarrow}} \newcommand{\qset}{\mathbb Q} \newcommand{\revcondtrans}[3]{q_{#1}(#2|#3)} \newcommand{\rmd}{\mathrm d} \newcommand{\rme}{\mathrm e} \newcommand{\rmi}{\mathrm i} \newcommand{\Rset}{\mathbb{R}} \newcommand{\rset}{\mathbb{R}} \newcommand{\rti}{\sigma} \newcommand{\section}[1]{==== #1 ====} \newcommand{\seq}[2]{\lrc{#1\eqsp: \eqsp #2}} \newcommand{\set}[2]{\lrc{#1\eqsp: \eqsp #2}} \newcommand{\sg}{\mathrm{sgn}} \newcommand{\supnorm}[1]{\left\|#1\right\|_{\infty}} \newcommand{\thv}{{\theta_\star}} \newcommand{\tmu}{ {\tilde{\mu}}} \newcommand{\Tset}{ {\mathsf{T}}} \newcommand{\Tsigma}{ {\mathcal{T}}} \newcommand{\ttheta}{{\tilde \theta}} \newcommand{\tv}[1]{\left\|#1\right\|_{\mathrm{TV}}} \newcommand{\unif}{\mathrm{Unif}} \newcommand{\weaklim}[1]{\stackrel{\mathcal{L}_{#1}}{\rightsquigarrow}} \newcommand{\Xset}{{\mathsf X}} \newcommand{\Xsigma}{\mathcal X} \newcommand{\Yset}{{\mathsf Y}} \newcommand{\Ysigma}{\mathcal Y} \newcommand{\Var}{\mathbb{V}\mathrm{ar}} \newcommand{\zset}{\mathbb{Z}} \newcommand{\Zset}{\mathsf{Z}} $$

2017/10/07 23:39 · douc

$\newcommand{\sgn}{\mathrm{sgn}}$

Statement

Let $h (\beta)= (\beta -u)^2+c |\beta|$ where $u>0$. Show that the argmin of $h$ can be written as $$ \beta^\star=u \lr{1-\frac{c}{2|u|}}^+ $$

Proof

The function $h$ is strictly convex and $\lim_{\beta \to \pm \infty}|h (x)|=\infty$. This implies that $h$ admits a unique minimizer $\beta^\star$.

  • Case 1: $\beta^\star\neq 0$, in which case $h' (\beta^\star)=0$. This implies $2 (\beta^\star-u)+c \sgn(\beta^\star)=0$. Therefore $2u=\sgn(\beta^\star)\lr{2|\beta^\star|+c}$, which implies $\sgn (u)=\sgn (\beta^\star)$. Therefore $2 (\beta^\star-u)+c \sgn(u)=0$

from which we deduce $\beta^\star=u \lr{1-\frac{c}{2|u|}}$. Using again $\sgn (u)=sgn (\beta^\star)$, we deduce $1-\frac{c}{2|u|}\geq 0$ and finally, $$ \beta^\star=u \lr{1-\frac{c}{2|u|}}^+ $$

  • Case 2: $\beta^\star= 0$. In this case, for all $\beta \neq 0$, $h (\beta) \geq h (0)=u^2$, which is equivalent to $\beta^2-2 \beta u+c|\beta|\geq 0$. Dividing by $|\beta|$ and letting $\beta\to 0$, we get $-2 u\sgn(\beta)+c \geq 0$ which in turn implies $-2|u|+c\geq 0$. This shows $1-\frac{c}{2|u|}\leq 0$ and we therefore have again:

$$ \beta^\star=0=u \lr{1-\frac{c}{2|u|}}^+ $$

world/max.txt · Last modified: 2022/03/16 07:40 (external edit)