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I encountered this expression;

How do we get $p(x|A)$ from $p_x(x[n]|A)$? I thought we just sum all discrete x[n] values, but in another example there was this one which got me confused.

$p(x[n]|\theta) = \frac{1}{\theta} \text{ for }0 \leq x[n] \leq \theta $

$p(x|\theta) = \frac{1}{\theta^N}$

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  • $\begingroup$ What does x[n] represent? $\endgroup$ – David Luke Thiessen Jun 11 at 1:27
  • $\begingroup$ Observations (n = 0,1,....N-1) $\endgroup$ – EserRose Jun 11 at 1:44
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It looks like they're just multiplying the probabilities for $N$ different $x[n]$ values.

$$ \prod_{n=0}^{N-1} \frac{1}{\theta} = \frac{1}{\theta^N} $$

$$ \prod_{n=0}^{N-1} \frac{1}{(2\pi\sigma^2)^{1/2}}\exp\left[\frac{-1}{2\sigma^2}(x[n] - A)^2\right] = \frac{1}{(2\pi\sigma^2)^{N/2}}\exp\left[\sum_{n=0}^{N-1}\frac{-1}{2\sigma^2}(x[n] - A)^2\right] $$

This would make sense if

$$ p(x|A) = \prod_{n=0}^{N-1}p(x[n]|A) $$

To me that seems like a strange definition of $p(x|A)$, but maybe it makes sense in the context you're looking at?

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  • $\begingroup$ Yes, I noticed it was simple multiplication way too late. I'm still not sure why though, but thanks $\endgroup$ – EserRose Jun 11 at 2:01

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