Bootstrap and out-of-bootstrap error calculation

I was trying to solve a question in one of the exercises in my class regarding bootstrap sampling but I am stuck with 0-idea how to move forward. The question is about bootstrapping and calculating the bootstrap and the out-of bootstrap estimates of the generalization error. There are 4 samples which are shown below;

Original Data |Z* | Bootstrap 1 |Bootstrap 2 |Bootstrap 3 | Bootstrap 4
Z₁---------------- Z₁*= ---Z₄--------------- Z₅ ---------------Z₃------------ Z₃
Z₂---------------- Z₂*= ---Z₁--------------- Z₅ ---------------Z₁------------ Z₅
Z₃---------------- Z₃*= ---Z₄--------------- Z₂ ---------------Z₃------------ Z₂
Z₄---------------- Z₄*= ---Z₂--------------- Z₁ ---------------Z₄------------ Z₂
Z₅---------------- Z₅*= ---Z₁--------------- Z₃ ---------------Z₂------------ Z₃

Questions can be found here: http://imgur.com/a/sNEZv

(i) The estimator $\hat{m}^{*(j)}$ is simply computed based on the $j^{th}$ Bootstrap set. The estimator is identical for both, the bootstrap and the out-of-bootstrap methods.
• In the standard bootstrap method, we average over all bootstrap samples, i.e. $\alpha_{ij} = \frac{1}{n} = \frac{1}{5}$.
• For the out-of-bootstrap case, the inner sum is only over those elements that were not used for training, i.e. the elements that are not contained in $j^{th}$ bootstrap data set. Therefore, $\alpha_{ij} = 0$ if $Z_i \in \text{Bootstrap } j$ otherwise $\alpha_{ij}$ is equal $\frac{1}{\# \{\text{distinct elements in Bootstrap j\}}}$.
• We already have our bootstrap samples. For Standard bootstrap case shouldn't the calculation for $\alpha_{ij}$ also depend on the drawings bootstrap sample? – NoResultOnTheWeb Aug 13 '17 at 13:52