From 58f45fd59816428c99c9a2677557d82a9422a40d Mon Sep 17 00:00:00 2001 From: mmcky Date: Sun, 25 Sep 2022 07:41:33 +1000 Subject: [PATCH 1/2] fix align in equation --- lectures/likelihood_bayes.md | 6 ++++-- 1 file changed, 4 insertions(+), 2 deletions(-) diff --git a/lectures/likelihood_bayes.md b/lectures/likelihood_bayes.md index f0e48637a..55dc46a90 100644 --- a/lectures/likelihood_bayes.md +++ b/lectures/likelihood_bayes.md @@ -465,10 +465,12 @@ $$ Notice that $$ -\eqalign{ E(\pi_t | \pi_{t-1}) & = \int \Bigl[ { \pi_{t-1} f(w) \over \pi_{t-1} f(w) + (1-\pi_{t-1})g(w) } \Bigr] +\begin{aligned} +E(\pi_t | \pi_{t-1}) & = \int \Bigl[ { \pi_{t-1} f(w) \over \pi_{t-1} f(w) + (1-\pi_{t-1})g(w) } \Bigr] \Bigl[ \pi_{t-1} f(w) + (1-\pi_{t-1})g(w) \Bigr] d w \cr & = \pi_{t-1} \int f(w) dw \cr - & = \pi_{t-1}, \cr} + & = \pi_{t-1}, \cr +\end{aligned} $$ so that the process $\pi_t$ is a **martingale**. From 6640a86666ceaebf9133b192ef4042c7e1ff4231 Mon Sep 17 00:00:00 2001 From: mmcky Date: Sun, 25 Sep 2022 08:01:24 +1000 Subject: [PATCH 2/2] fix tag --- lectures/likelihood_bayes.md | 2 +- 1 file changed, 1 insertion(+), 1 deletion(-) diff --git a/lectures/likelihood_bayes.md b/lectures/likelihood_bayes.md index 55dc46a90..03e4ca4e5 100644 --- a/lectures/likelihood_bayes.md +++ b/lectures/likelihood_bayes.md @@ -549,7 +549,7 @@ $$ Applying the above formula to $\pi_\infty$, we obtain $$ -E_{-1} \pi_\infty(\omega) = \pi_{-1} \tag{20} +E_{-1} \pi_\infty(\omega) = \pi_{-1} $$ where the mathematical expectation $E_{-1}$ here is taken with respect to the probability