te max zk ln max etzk ln from this derive that max zk ln finallyshow that this last inequality implies that max |zk ln( nk ( (dreturning to the objective of this exercisedenote the elements of by |gand let zk `tn (gk `(gk by part (ait is sufficient tobound maxk |zk show that the {zk satisfy the conditions of (cwith ( ) for this you will need to apply part (bto the random variable loss( ( ) `( )where (xyis generic data point now complete the proof of ( consider the problem in exercise above show that |` (ggt `(gg ) sup |` ( `( )` (gg `(gg gg from thisconcludee |` (ggt `(gg ) sup |` ( `( )gg the last bound allows us to assess how close the training loss ` (ggt is to the optimal risk `(gg within class show that for the normal linear model (xbs in )the maximum likelihood estimator of is identical to the method of moments estimator ( let gamma(alshow that the pdf of / is equal to - la ( )- - - (zg(az consider the sequence where (this non-degenerate initial guess and wt (thwt- (th) ( th) we assume that ( this not the constant function (with respect to thand that the maximum likelihood value ( thmax ( thth exists (is boundedlet lt : ( th)wt (thdth show that {lt is strictly increasing and bounded sequence henceconclude that its limit is ( th