---
title: "Theoretical Details"
author: ""
date: ""
output: rmarkdown::html_vignette
vignette: >
  %\VignetteIndexEntry{Theoretical Details}
  %\VignetteEngine{knitr::rmarkdown}
  %\VignetteEncoding{UTF-8}
---
```{r, include=FALSE, eval=FALSE}
render.rmd("rmd")
```


# The conditional expectation is optimal in the mean-squared error sense:

First, define the following quantities:

- The past information:
$$
F_{k-1} = \left\{ y_0, y_1, ... , y_{k-1} \right\}
$$
- The conditional expectation:
$$
x_{k \mid k-1} = \mathbb{E}\left( x_t \mid F_{t-1} \right)
$$
- The state residual
$$
r_k = x_{t_k} - x_{k \mid k-1} 
$$

### The conditional expectation of the residual is zero:
$$
\begin{align}
\mathbb{E}\left(r_k \mid F_{k-1}\right) & = \mathbb{E}(x_{t_k} \mid F_{k-1}) - \mathbb{E}(x_{k \mid k-1} \mid F_{k-1})  \\[5pt]
& = x_{k \mid k-1} - x_{k \mid k-1} \\[5pt]
& = 0 
\end{align}
$$

### The residual is orthogonal (and uncorrelated) to every function of the past information
We use the tower property to condition on the filtration $F_{k-1}$.
$$
\begin{align}
\mathbb{E}\left(r_k \cdot h(F_{k-1}) \right) 
& = E(E(r_k \cdot h(F_{k-1})\mid F_{k-1})) \\
& = E(E(r_k \mid F_{k-1}) \cdot h(F_{k-1})) \\ 
& = E(0 \cdot h(F_{k-1})) \\
& = E(0) = 0
\end{align}
$$
We can also conclude from this, since th expectation is also the covariance since $E(XY) = C(X,Y)$ if $E(X)=0$ or $E(Y)=0$, that
$$
C(r_k, h(F_{k-1})) = 0
$$
That is, the state residual and any function of the past information are uncorrelated.

### The conditional expectation is the best estimator
$$

$$