exercise:C0a105156b

[math] \newcommand{\DS}{\displaystyle} \newcommand{\intbeta}{\lfloor \beta \rfloor} \newcommand{\cA}{\mathcal{A}} \newcommand{\cB}{\mathcal{B}} \newcommand{\cC}{\mathcal{C}} \newcommand{\cD}{\mathcal{D}} \newcommand{\cE}{\mathcal{E}} \newcommand{\cF}{\mathcal{F}} \newcommand{\cG}{\mathcal{G}} \newcommand{\cH}{\mathcal{H}} \newcommand{\cI}{\mathcal{I}} \newcommand{\cJ}{\mathcal{J}} \newcommand{\cK}{\mathcal{K}} \newcommand{\cL}{\mathcal{L}} \newcommand{\cM}{\mathcal{M}} \newcommand{\cN}{\mathcal{N}} \newcommand{\cO}{\mathcal{O}} \newcommand{\cP}{\mathcal{P}} \newcommand{\cQ}{\mathcal{Q}} \newcommand{\cS}{\mathcal{S}} \newcommand{\cT}{\mathcal{T}} \newcommand{\cU}{\mathcal{U}} \newcommand{\cV}{\mathcal{V}} \newcommand{\cX}{\mathcal{X}} \newcommand{\cY}{\mathcal{Y}} \newcommand{\cZ}{\mathcal{Z}} \newcommand{\sg}{\mathsf{subG}} \newcommand{\subE}{\mathsf{subE}} \newcommand{\bA}{\mathbf{A}} \newcommand{\bB}{\mathbf{B}} \newcommand{\bC}{\mathbf{C}} \newcommand{\bD}{\mathbf{D}} \newcommand{\bE}{\mathbf{E}} \newcommand{\bF}{\mathbf{F}} \newcommand{\bG}{\mathbf{G}} \newcommand{\bH}{\mathbf{H}} \newcommand{\bI}{\mathbf{I}} \newcommand{\bJ}{\mathbf{J}} \newcommand{\bK}{\mathbf{K}} \newcommand{\bM}{\mathbf{M}} \newcommand{\bN}{\mathbf{N}} \newcommand{\bO}{\mathbf{O}} \newcommand{\bP}{\mathbf{P}} \newcommand{\bp}{\mathbf{p}} \newcommand{\bQ}{\mathbf{Q}} \newcommand{\bS}{\mathbf{S}} \newcommand{\bT}{\mathbf{T}} \newcommand{\bU}{\mathbf{U}} \newcommand{\bV}{\mathbf{V}} \newcommand{\bX}{\mathbf{X}} \newcommand{\bY}{\mathbf{Y}} \newcommand{\bZ}{\mathbf{Z}} \newcommand{\bflambda}{\boldsymbol{\lambda}} \newcommand{\bftheta}{\boldsymbol{\theta}} \newcommand{\bfg}{\boldsymbol{g}} \newcommand{\bfy}{\boldsymbol{y}} \def\thetaphat{\hat{\bftheta}_\bp} \def\bflam{\boldsymbol{\lambda}} \def\Lam{\Lambda} \def\lam{\lambda} \def\bfpi{\boldsymbol{\pi}} \def\bfz{\boldsymbol{z}} \def\bfw{\boldsymbol{w}} \def\bfeta{\boldsymbol{\eta}} \newcommand{\R}{\mathrm{ I}\kern-0.18em\mathrm{ R}} \newcommand{\h}{\mathrm{ I}\kern-0.18em\mathrm{ H}} \newcommand{\K}{\mathrm{ I}\kern-0.18em\mathrm{ K}} \newcommand{\p}{\mathrm{ I}\kern-0.18em\mathrm{ P}} \newcommand{\E}{\mathrm{ I}\kern-0.18em\mathrm{ E}} %\newcommand{\Z}{\mathrm{ Z}\kern-0.26em\mathrm{ Z}} \newcommand{\1}{\mathrm{ 1}\kern-0.24em\mathrm{ I}} \newcommand{\N}{\mathrm{ I}\kern-0.18em\mathrm{ N}} \newcommand{\field}[1]{\mathbb{#1}} \newcommand{\q}{\field{Q}} \newcommand{\Z}{\field{Z}} \newcommand{\X}{\field{X}} \newcommand{\Y}{\field{Y}} \newcommand{\bbS}{\field{S}} \newcommand{\n}{\mathcal{N}} \newcommand{\x}{\mathcal{X}} \newcommand{\pp}{{\sf p}} \newcommand{\hh}{{\sf h}} \newcommand{\ff}{{\sf f}} \newcommand{\Bern}{\mathsf{Ber}} \newcommand{\Bin}{\mathsf{Bin}} \newcommand{\Lap}{\mathsf{Lap}} \newcommand{\tr}{\mathsf{Tr}} \newcommand{\phin}{\varphi_n} \newcommand{\phinb}{\overline \varphi_n(t)} \newcommand{\pn}{\p_{\kern-0.25em n}} \newcommand{\pnm}{\p_{\kern-0.25em n,m}} \newcommand{\psubm}{\p_{\kern-0.25em m}} \newcommand{\psubp}{\p_{\kern-0.25em p}} \newcommand{\cfi}{\cF_{\kern-0.25em \infty}} \newcommand{\e}{\mathrm{e}} \newcommand{\ic}{\mathrm{i}} \newcommand{\Leb}{\mathrm{Leb}_d} \newcommand{\Var}{\mathrm{Var}} \newcommand{\ddelta}{d_{\symdiffsmall}} \newcommand{\dsubh}{d_H} \newcommand{\indep}{\perp\kern-0.95em{\perp}} \newcommand{\supp}{\mathop{\mathrm{supp}}} \newcommand{\rank}{\mathop{\mathrm{rank}}} \newcommand{\vol}{\mathop{\mathrm{vol}}} \newcommand{\conv}{\mathop{\mathrm{conv}}} \newcommand{\card}{\mathop{\mathrm{card}}} \newcommand{\argmin}{\mathop{\mathrm{argmin}}} \newcommand{\argmax}{\mathop{\mathrm{argmax}}} \newcommand{\ud}{\mathrm{d}} \newcommand{\var}{\mathrm{var}} \newcommand{\re}{\mathrm{Re}} \newcommand{\MSE}{\mathsf{MSE}} \newcommand{\im}{\mathrm{Im}} \newcommand{\epr}{\hfill\hbox{\hskip 4pt\vrule width 5pt height 6pt depth 1.5pt}\vspace{0.5cm}\par} \newcommand{\bi}[1]{^{(#1)}} \newcommand{\eps}{\varepsilon} \newcommand{\Deq}{\stackrel{\mathcal{D}}{=}} \newcommand{\ubar}{\underbar} \newcommand{\Kbeta}{K_{\hspace{-0.3mm} \beta}} \newcommand{\crzero}[1]{\overset{r_0}{\underset{#1}{\longleftrightarrow}}} \newcommand{\hint}[1]{\texttt{[Hint:#1]}} \newcommand{\vp}{\vspace{.25cm}} \newcommand{\vm}{\vspace{.5cm}} \newcommand{\vg}{\vspace{1cm}} \newcommand{\vgneg}{\vspace{-1cm}} \newcommand{\vneg}{\vspace{-.5cm}} \newcommand{\vpneg}{\vspace{-.25cm}} \newcommand{\tp}{\ptsize{10}} \newcommand{\douzp}{\ptsize{12}} \newcommand{\np}{\ptsize{9}} \newcommand{\hp}{\ptsize{8}} \newcommand{\red}{\color{red}} \newcommand{\ndpr}[1]{{\textsf{\red[#1]}}} \newcommand\iid{i.i.d\@ifnextchar.{}{.\@\xspace} } \newcommand\MoveEqLeft[1][2]{\kern #1em & \kern -#1em} \newcommand{\leadeq}[2][4]{\MoveEqLeft[#1] #2 \nonumber} \newcommand{\leadeqnum}[2][4]{\MoveEqLeft[#1] #2} \newcommand\independent{\protect\mathpalette{\protect\independenT}{\perp}} \def\independenT#1#2{\mathrel{\rlap{$#1#2$}\mkern2mu{#1#2}}} \newcommand{\MIT}[1]{{\color{MITred} #1}} \newcommand{\dHyp}{\{-1,1\}^d} \newcommand{\thetahard}{\hat \theta^{hrd}} \newcommand{\thetasoft}{\hat \theta^{sft}} \newcommand{\thetabic}{\hat \theta^{bic}} \newcommand{\thetalasso}{\hat \theta^{\cL}} \newcommand{\thetaslope}{\hat \theta^{\cS}} \newcommand{\thetahard}{\hat \theta^{hrd}} \newcommand{\thetasoft}{\hat \theta^{sft}} \newcommand{\thetabic}{\hat \theta^{bic}} \newcommand{\thetalasso}{\hat \theta^{\cL}} \newcommand{\thetaslope}{\hat \theta^{\cS}} \newcommand{\thetals}{\hat \theta^{ls}} \newcommand{\thetalsm}{\tilde \theta^{ls_X}} \newcommand{\thetaridge}{\hat\theta^{\mathrm{ridge}}_\tau} \newcommand{\thetalsK}{\hat \theta^{ls}_K} \newcommand{\muls}{\hat \mu^{ls}} [/math]

Consider the multivariate regression model where [math]\Y[/math] has SVD:

[[math]] \Y=\sum_j\hat \lambda_j \hat u_j \hat v_j^\top\,. [[/math]]

Let [math]M[/math] be defined by

[[math]] \hat M=\sum_j\hat \lambda_j \1(|\hat \lambda_j| \gt 2\tau)\hat u_j \hat v_j^\top\,, \tau \gt 0\,. [[/math]]

1. Show that there exists a choice of [math]\tau[/math] such that
   [[math]] \frac{1}{n}\|\hat M -\X \Theta^*\|_F^2 \lesssim \frac{\sigma^2\rank(\Theta^*)}{n}(d\vee T) [[/math]]
   with probability .99.
2. Show that there exists a matrix [math]n \times n[/math] matrix [math]P[/math] such that [math]P\hat M=\X\hat \Theta[/math] for some estimator [math]\hat \Theta[/math] and
   [[math]] \frac{1}{n}\|\X\hat \Theta -\X \Theta^*\|_F^2 \lesssim \frac{\sigma^2\rank(\Theta^*)}{n}(d\vee T) [[/math]]
   with probability .99.
3. Comment on the above results in light of the results obtain in Section.