Past exam of the mathematics course of the University of Cambridge / 2024 / ii / Paper 2 / 31K / b

Table of contents
- Solution b

Solution

Let

m_{i} = E (Y_{i} ∣ X_{i})

. Conditional on

X_{1 : n}

,

γ_{j} = \frac{1}{N _{j} + 1} \sum_{i} m_{i} 1_{R_{j}} (X_{i}) .

The responses are conditionally independent, so cross-covariances vanish and

E [(γ_{j} - γ_{j})^{2} ∣ X_{1 : n}] = \frac{1}{( N _{j} + 1 ) ^{2}} i \sum Var (Y_{i} ∣ X_{i}) 1_{R_{j}} (X_{i}) .

Because the regions partition predictor space,

T (X) - E (T (X) ∣ X, X_{1 : n}) = \sum_{j} 1_{R_{j}} (X) (γ_{j} - γ_{j}) .

Only one summand is nonzero. Put

q_{j} = P (X \in R_{j})

; the partition is deterministic because

D^{'}

is fixed, and

N_{j} \sim Bin (n, q_{j})

. The variance bound and

N_{j} / (N_{j} + 1)^{2} \leq 1/ (N_{j} + 1)

give

E [{T (X) - E (T (X) ∣ X, X_{1 : n})}^{2}] \leq σ^{2} j \sum q_{j} E \frac{1}{N _{j} + 1} \leq σ^{2} j : q_{j} > 0 \sum \frac{1}{n} \leq \frac{σ ^{2} J}{n} .

This is the conditional variance of a fixed regression-tree partition bound.

 Ancestors (10)

 View article source

 Discussion (0)

There are no discussions about this article yet.

 Articles by others on the same topic (0)

There are currently no matching articles.

  See all articles in the same topic Create my own version