# How does this transformation work?

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.

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

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

• What does x[n] represent? – David Luke Thiessen Jun 11 at 1:27
• Observations (n = 0,1,....N-1) – EserRose Jun 11 at 1:44

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?

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