Task $1.$ We have learned that a bagged model has a $\frac{{}^2}{m}$ error guarantee based on an assumption that each base model suffers an independent $N(0,{}^2)$ noise. Now, consider a weighted bagged model

$f_{}(x)={\displaystyle \underset{i=1}{\overset{m}{}}}{}_i{f}_i(x),$

where $f_i\text{'}$s are learned in the same way as in the bagged model $($not as in the boosting model$).$

Prove that, under the same noise assumption, this weighted bagged model has a better error guarantee than $\frac{{}^2}{m}$ if it is a convex combination of the base models $($i$.$e$.,{\displaystyle \underset{i=1}{\overset{m}{}}}{}_i=1$ and ${}_{i}0)$ and $|||{|}_2\frac{1}{m},$ where $=[{}_1,$dots,${}_m]$ is an $m-$dimensional vector and $||*|{|}_2$ is $l_2$ norm.

Tip: Analysis should be similar to the bagged model analysis, except at the end we take a further argument based on the Cauchy Schwarz inequality ${\displaystyle \underset{i=1}{\overset{m}{}}}{}_i{p}_i\sqrt[\phantom{2}]{{\displaystyle \underset{i=1}{\overset{m}{}}}{}_i^2{\displaystyle \underset{i=1}{\overset{m}{}}}{p}_i^2}$ for any $p_1,$dots,$p_m.$

Please show by hand