Question

Exercise 3 Given B identically distributed (but not necessarily independent) random variables $Z_1, \dots, Z_B$ with positive pairwise correlation $\rho$, show that $$\text{Var}\left(\frac{1}{B}\sum_{b=1}^B Z_b\right) = \rho\sigma^2 + \frac{1-\rho}{B}\sigma^2$$ (4) If we further assume random variables $Z_1, \dots, Z_B$ are independent. What's the variance $\text{Var}\left(\frac{1}{B}\sum_{b=1}^B Z_b\right)$? Show that the variance is larger if random variables are not independent. Use this result to argue that why bagging does not work as we expect.

          Exercise 3
Given B identically distributed (but not necessarily independent) random variables $Z_1, \dots, Z_B$ with positive
pairwise correlation $\rho$, show that
$$\text{Var}\left(\frac{1}{B}\sum_{b=1}^B Z_b\right) = \rho\sigma^2 + \frac{1-\rho}{B}\sigma^2$$
(4)
If we further assume random variables $Z_1, \dots, Z_B$ are independent. What's the variance $\text{Var}\left(\frac{1}{B}\sum_{b=1}^B Z_b\right)$?
Show that the variance is larger if random variables are not independent. Use this result to argue that why
bagging does not work as we expect.

Exercise 3
Given B identically distributed (but not necessarily independent) random variables Z1, …, ZB with positive
pairwise correlation ρ, show that

Var((1)/(B)∑b=1^B Zb) = ρσ^2 + (1-ρ)/(B)σ^2

(4)
If we further assume random variables Z1, …, ZB are independent. What's the variance Var((1)/(B)∑b=1^B Zb)?
Show that the variance is larger if random variables are not independent. Use this result to argue that why
bagging does not work as we expect.

Added by Deborah P.

Elementary Statistics a Step by Step Approach

Allan G. Bluman 9th Edition

Instant Answer

Solved by Expert Kirsty Gledhill

10/11/2023

Step 1

Step 1: The variance of a sum of random variables is equal to the sum of the variances plus twice the sum of the covariances. Show more…

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Exercise 3 Given B identically distributed (but not necessarily independent) random variables Z1,..., ZB with positive pairwise correlation p, show that B Var (1/B Σ Zb) = po² + (1-p)/B σ² b=1 (4) If we further assume random variables Z1,..., ZB are independent. What's the variance Var (1/B Σ Zb)? Show that the variance is larger if random variables are not independent. Use this result to argue that why bagging does not work as we expect.