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Say we have a random process $X(t, u)$ parametrized by $t$ and $u$ that generates data $x$. We also have a prior on $u$, $p(u)$.

Am I correct in stating that the expression to find the maximum a posteriori (MAP) estimate of only $t$ would be

$$ t_{MAP} = \underset{t}{\operatorname{argmax}} \int p(x | t, u) p(u) du = \underset{t}{\operatorname{argmax}} p(x|t) $$

That is, you'd marginalize the nuisance parameters out, correct?

Now suppose $U\sim \mathcal{N}[\hat{u}, C_u]$ and $X \mid u, t \sim \mathcal{N}[M(t) u, C_x]$.

Is it correct to say that $p(x | t)$ equals the pdf of $M(t) U$? That is, $$X \mid t \sim \mathcal{N}[M(t)\hat{u}, M(t)C_uM(t)^\top]$$

Stated differently and more generally, does deriving the distribution of $f(U, t)$ (where $f$ is a bijective function) give the same result as the $\int p(x|t, u)p(u)du$?

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  • $\begingroup$ Perhaps you benefit from having a peek at my Bayesian tutorial. You can read most of it online under "Read Sample." amazon.com/dp/B0BTNVFR65 $\endgroup$
    – Romke Bontekoe
    Commented Jul 7 at 13:15
  • $\begingroup$ @RomkeBontekoe oops, fixed now! $\endgroup$
    – DangerousTim
    Commented Jul 7 at 13:15
  • $\begingroup$ What is $t$ here, another random parameter or a time index? $\endgroup$
    – Johan de Aguas
    Commented Jul 7 at 14:00
  • $\begingroup$ @JohandeAguas $t$ is a generic parameter. $\endgroup$
    – DangerousTim
    Commented Jul 7 at 15:42
  • $\begingroup$ Hi: the integral gives p(x | t). I'm not sure about the definition of f(U, t). It's not defined. It looks like possibly the joint dist of U and t ? $\endgroup$
    – mlofton
    Commented Jul 7 at 17:10

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If $t$ is a random parameter, for which Bayesian inference is required, then $t$ should have a prior, $p(t)$, so it can have a posterior and a MAP estimate. In the most general case, $t_{MAP}=\arg\max_t p(t)\int p(x\mid u,t)p(u\mid t)du$.

If $u$ and $t$ are prior-independent, then $t_{MAP}=\arg\max_t p(t)\int p(x\mid u,t)p(u)du$.

If, on top, $p(t)$ is the uniform distribution, then $t_{MAP}=\arg\max_t \int p(x\mid u,t)p(u)du$.

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