Exercise 5.5.5

Answers

${\hat{x}}_{k + 1} = {\hat{x}}_{k} + \frac{1}{k + 1} (b_{k + 1} - {\hat{x}}_{k}) = \frac{k}{k + 1} {\hat{x}}_{k} + \frac{1}{k + 1} b_{k + 1} = \frac{k}{k + 1} \frac{1}{k} \sum_{i = 1}^{k} b_{i} + \frac{1}{k + 1} b_{k + 1} = \frac{1}{k + 1} \sum_{i = 1}^{k} b_{i} + \frac{1}{k + 1} b_{k + 1} = \frac{1}{k + 1} \sum_{i = 1}^{k + 1} b_{i}$
$W_{k} = \frac{σ^{2}}{k}$ , also $V_{k + 1} = σ^{2}$ , so we have $W_{k + 1}^{- 1} = W_{k}^{- 1} + A_{k + 1}^{T} V_{k + 1}^{- 1} A_{k + 1} = \frac{k}{σ^{2}} + \frac{1}{σ^{2}} = \frac{k + 1}{σ^{2}}$ , in the end we have $W^{k + 1} = \frac{σ^{2}}{k + 1}$

Where $A_{k + 1} = I$

niuers

2020-03-20 00:00