How to derive this proportional conditional probability

Question

In the book Bayesian Data Analysis 3rd edition by Gelman et al., the solution for task 3.15 on time series data involves the following expression:
$$p(y_t|y_{t-1})p(y_{t+1}|y_t) propto p(y_t|y_{t-1}, y_{t+1})$$
This means that $y_t$ depends only on $y_{t-1}$ and $y_{t+1}$. But how do I derive the right-hand side?

hbadger19042 · Answer

Use the defintion of conditional probability to get this formula:
$$
begin{align}
P(y_{t+1}, y_t, y_{t-1}) &= P(y_{t+1}|y_t, y_{t-1}) cdot P(y_t, y_{t-1})\
&=P(y_{t+1}|y_t, y_{t-1}) cdot P(y_t| y_{t-1})cdot P(y_{t-1})\
end{align}
$$
If you assume Markove process, you have $P(y_{t+1}|y_t, y_{t-1}) = P(y_{t+1}|y_t)$.
Plugging the formula, we get
$$
P(y_{t+1}, y_t, y_{t-1}) =P(y_{t+1}|y_t) cdot P(y_t| y_{t-1})cdot P(y_{t-1})\
$$
You  can expand the original probability $P(y_{t+1}, y_t, y_{t-1})$ in this way also:
$$
begin{align}
P(y_{t+1}, y_t, y_{t-1}) &= P(y_{t}|y_{t+1}, y_{t-1}) cdot P(y_{t+1}, y_{t-1})\
&=P(y_{t}|y_{t+1}, y_{t-1})  cdot P(y_{t+1}| y_{t-1})cdot P(y_{t-1})\
end{align}
$$
Under a Markove process hypothesis, $y_{t+1}$ doesn't depend on $y_{t-1}$. Therefore,
we have $P(y_{t+1}| y_{t-1}) = P(y_{t+1})$.
Equating two equations, we have
$$
P(y_{t+1}|y_t) cdot P(y_t| y_{t-1})cdot P(y_{t-1}) = P(y_{t}|y_{t+1}, y_{t-1})  cdot P(y_{t+1})cdot P(y_{t-1})
$$
Cancelling the term $P(y_{t-1})$, we arrive at this equation:
$$
P(y_{t+1}|y_t) cdot P(y_t| y_{t-1}) = P(y_{t}|y_{t+1}, y_{t-1})  cdot P(y_{t+1})
$$

How to derive this proportional conditional probability

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