Well, that last line means we can safely ignore it (since the other variance terms grow with the square of n rather than linearly. Not particularly rewarding work. But, it does preempt any accusations of hand-waving.
Tuesday, January 23, 2018
Grinding it out
You may recall I said I'd derive the exact formulas for the variance of the block sums. There's nothing particularly hard (or enlightening) about doing this, it's just straightforward symbol manipulation. It's also long and tedious symbol manipulation. For example, the covariance term, which I had previously claimed we can safely ignore:
&space;&=&space;2\left[\left(1-\frac{n}{r}\right)^2\sum_{1\le&space;i<j\le&space;r}\sigma{ij}+&space;\left(1-\frac{n}{r}\right)&space;\sum_{i=1}^r\sum_{j=r+1}^n\sigma_{ij}&space;+&space;\sum_{r\le&space;i<j\le&space;n}\sigma_{ij}\right]\\&space;&=2\sigma_{ij}\left[\left(1-\frac{n}{r}\right)^2\frac{r(r-1)}{2}&space;+&space;\left(1-\frac{n}{r}\right)r(n-r)&space;+&space;\frac{(n-r-1)(n-r)}{2}&space;\right]\\&space;&=-2\sigma_{ij}(n-2r)^2\\&space;&=2\sigma^2\frac{(n-2r)^2}{n-1}\textrm{&space;by&space;an&space;equally&space;tedious&space;derivation&space;of&space;}\sigma_{ij}=\frac{-\sigma^2}{n-1}\\&space;&=&space;\sigma^2\textrm{O}(n)&space;\end{align*} "\begin{align*} \textrm{Cov}(*) &= 2\left[\left(1-\frac{n}{r}\right)^2\sum_{1\le i<j\le r}\sigma{ij}+ \left(1-\frac{n}{r}\right) \sum_{i=1}^r\sum_{j=r+1}^n\sigma_{ij} + \sum_{r\le i<j\le n}\sigma_{ij}\right]\\ &=2\sigma_{ij}\left[\left(1-\frac{n}{r}\right)^2\frac{r(r-1)}{2} + \left(1-\frac{n}{r}\right)r(n-r) + \frac{(n-r-1)(n-r)}{2} \right]\\ &=-2\sigma_{ij}(n-2r)^2\\ &=2\sigma^2\frac{(n-2r)^2}{n-1}\textrm{ by an equally tedious derivation of }\sigma_{ij}=\frac{-\sigma^2}{n-1}\\ &= \sigma^2\textrm{O}(n) \end{align*}")
Well, that last line means we can safely ignore it (since the other variance terms grow with the square of n rather than linearly. Not particularly rewarding work. But, it does preempt any accusations of hand-waving.
Well, that last line means we can safely ignore it (since the other variance terms grow with the square of n rather than linearly. Not particularly rewarding work. But, it does preempt any accusations of hand-waving.
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