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Note: I am dealing with a state space model/hidden Markov model, where process $x_t$ is only dependent on $x_{t-1}$ and $y_t$ is only dependent on $x_t$.

\begin{equation} \begin{split} p(x_{0:t}|y_{0:t})& = \frac{p(y_t|x_{0:t},y_{0:t-1}) p(x_{0:t}|,y_{0:t-1}) }{p(y_t|y_{0:t-1})} ~~~\mbox{(bayes)}\\ & = \frac{p(y_t|x_{0:t},y_{0:t-1}) p(x_{t}|x_{0:t-1},y_{0:t-1}) p(x_{0:t-1}|y_{0:t-1})}{p(y_t|y_{0:t-1})}~~~\mbox{(bayes)} \\ &=\frac{p(y_t|x_{t}) p(x_{t}|x_{t-1}) p(x_{0:t-1}|y_{0:t-1})}{p(y_t|y_{0:t-1})}~~~\mbox{(conditional independence)} \end{split} \end{equation}

Question: What is $p(y_t|y_{0:t-1})$ as an integral? I saw here that they wrote $$p(y_t|y_{0:t-1})=\int p(y_t|x_t)p(x_t|x_{t-1})p(x_{t-1}|y_{0:t-1}) dx_{t-1:t} \,.$$ How do they get to that: can someone write a step-by- step derivation (actively stating what steps they used to simplify)?

I find it unusual how they simplified $p(x_{0:t-1}|y_{0:t-1})$ to $p(x_{t-1}|y_{0:t-1})$ and how they knew to integrate away $dx_{t-1:t}$.

If I were to guess what $p(y_t|y_{0:t-1})$ was I would have written

$$p(y_t|y_{0:t-1})=\int p(y_t|x_{0:t},y_{0:t-1})p(x_{0:t}|,y_{0:t-1}) dx_{0:t}$$

However, I do not know how to continue...

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1 Answer 1

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The simplest way is \begin{align*} p(y_t|y_{0:t-1}) &= \int p(y_t,x_t,x_{t-1}|y_{0:t-1}) dx_{t-1:t} \\ &= \int p(y_t|x_{t-1},x_t, y_{0:t-1}) p(x_t|x_{t-1},y_{0:t-1})p(x_{t-1}|y_{0:t-1}) dx_{t-1:t} \\ &= \int p(y_t|x_t)p(x_t|x_{t-1})p(x_{t-1}|y_{0:t-1}) dx_{t-1:t} && \tag{1} \end{align*} with the third equality being because there's a lot of conditional independence.

Your way is also true. But you're missing the conditional independence again. \begin{align*} \int p(y_t|x_{0:t},y_{0:t-1})p(x_{0:t}|,y_{0:t-1}) dx_{0:t} &= \int p(y_t|x_{t})p(x_{0:t}|,y_{0:t-1}) dx_{0:t} \\ &= \int p(y_t|x_{t})p(x_t|x_{0:t-1},y_{0:t-1})p(x_{0:t-1}|,y_{0:t-1}) dx_{0:t} \\ &= \int p(y_t|x_{t})p(x_t|x_{t-1})p(x_{0:t-1}|,y_{0:t-1}) dx_{0:t} && \tag{2} \end{align*}

Notice that (1) and (2) are the same.

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