...N1,…,N2k−1 are s for k≥1 . Then, letting n∈N, \begin{align*} [N{2k} = n] &= [X{N{2k-1}+1} n] \in \mathcal{F}{n} which holds because N2k and N2k+2 are both s. ■Other Outlinks......
11/7/2022
...so Xn is a wrt (Bn). Finally, observe that \begin{align*} X{n} + U{n} &= \sum\limits{i=1}^{n} d{i} + \sum\limits{i=1}^{n} u{i}\\ &=\sum\limits{i=1}^{n} Z{i} - \mathbb{E}(Z{i}|\mathcal{B}{i-1}) + \mathbb{E}(Z{i}|\mathcal{B}{i-1}) - Z_{i-1}\\ &=Z{n} - Z{0}\\ &=Z_{n}, \end{align*} completing the proof. ■Other Outlinks......
...: n \in \mathbb{N}\}$ is a if F is a wrt N. Remarks It is sometimes convenient to extend F to a over Nˉ by defining B∞=σ(Bn:n∈N).Other Outlinks......
...T}$ be a wrt F∗:=(Bt)t∈T on Ω. Then ∀r≤t∈T, E(Xr)≤E(Xt). Proof This follows by and : \begin{align*} &\mathbb{E}(X{t}|\mathcal{B}{r}) \geq X_{r}\\ \Rightarrow&\mathbb{E}(X{t}) = \mathbb{E}(\mathbb{E}(X{t}|\mathcal{B}{r})) \geq \mathbb{E}(X{r}) \end{align*}......
...a wrt F∗:=(Bt)t∈T on Ω. Then ∀r≤t∈T, E(Xr)≥E(Xt). Proof This follows by and and . That is, \begin{align*} &\mathbb{E}(-X{t}) \geq \mathbb{E}(-X{r})\\ \Rightarrow&\mathbb{E}(X{t}) \leq \mathbb{E}(X{r})\\ \end{align*}......
...\leq \lambda | X{t{n}} )$$ . Remarks Intuitively we are saying that the only the most recent observation matters. The past is only relevant in so much as it tells us about the most recent observation. Other Outlinks......