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I have

$$\pi(a) =P(a|b) = \frac{P(b|a)P(a)}{P(b)}$$

I would like to now update given some new information, $C=c$

Is it possible to write:

$$P(a|b,c) = \frac{P(b,c|a)P(a)}{P(b,c)} \stackrel{?}{=} \frac{P(c|a)\pi(a)}{P(c)} $$

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You can write:

$$P(a,b,c) = P(a \vert b,c)P(b,c) = P(a \vert b,c)P(c \vert b)P(b)$$

or, also valid:

$$P(a,b,c) = P(c \vert a,b)P(a,b) = P(c \vert a,b)P(b \vert a)P(a)$$

Putting together both expressions:

$$P(a \vert b,c) = \frac{P(c \vert a,b)P(b \vert a)P(a)}{P(c \vert b)P(b)} = \frac{P(c \vert a,b) \pi(a)}{P(c \vert b)} $$

And if this new observation $c$ does not depend on the previous observation $b$ (i.e. $P(b,c) = P(b)P(c)$), you can write:

$$P(a \vert b,c) = \frac{P(c \vert a) \pi(a)}{P(c)} $$

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    $\begingroup$ Thank you, this answers the question. For something like a signal, however, doesn't the next observation usually depend on the previous one? $\endgroup$ – user0 Jun 6 '18 at 16:48
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    $\begingroup$ @user2357111317, actually that will depend on the autocorrelation of that signal $\endgroup$ – Carlos Campos Jun 7 '18 at 9:37
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    $\begingroup$ I especially appreciate your method of solving this by essentially deriving Bayes rule by showing how the joint equals the products in each case - it allowed me to solve a similar, more difficult problem. $\endgroup$ – user0 Jan 7 at 5:19
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It might help to use some more specific notation because the Bayesian update will depend on what model you're using.

As an example, say you had a linear regression model where you regressed $y_i$ on $\mathbf{x}_i$ variables. The coefficients/parameters for this model could be called $\theta$. If you had a batch of $n$ data points, you might write your likelihood as $$ p(\mathbf{y}_{1:n} \mid \mathbf{X}_{1:n}, \theta) = \prod_{i=1}^n p(y_i \mid \mathbf{x}_i, \theta). $$ Your model would be complete as soon as you chose some prior distribution $\pi(\theta)$.

Bayes' rule states $$ \pi(\theta \mid \mathbf{y}_{1:n}, \mathbf{x}_{1:n}, \theta) \propto p(\mathbf{y}_{1:n} \mid \mathbf{X}_{1:n}, \theta)\pi(\theta). $$ If you got another row of data ($y_{n+1}, \mathbf{x}_{n+1}$), then you could update your posterior using $$ \pi(\theta \mid \mathbf{y}_{1:n+1}, \mathbf{x}_{1:n+1}, \theta) = p(y_{n+1} \mid \mathbf{x}_{n+1}, \theta)\pi(\theta \mid \mathbf{y}_{1:n}, \mathbf{x}_{1:n}, \theta). $$ So the old posterior distribution takes the place of the prior distribution when you update sequentially.

As I said before, the Bayesian update will depend on what model you're using. A different model would have different conditional independence structure. However, many other models will resemble the last expression in that the old posterior is used as a prior, and it's being multiplied by some marginal "likelihood."

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