If I have a piecewise distribution with density $f$ on $[0,1)$ so that $f=f_1$ on $[0,0.5)$ and $f=f_2$ on $[0.5,1)$, with both $f_1$ and $f_2$ densities, is it necessarily a mixture distribution? My reasoning is that we can always introduce a latent variable $z$ taking values in $[0,1)$ uniformly and then choose either $f_1$ or $f_2$ depending on whether $z>0.5$ or not. What is the difference between a piecewise distribution and a mixture distribution here?
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1$\begingroup$ you can write any piecewise distribution as a mixture distribution, yes. Actually, there's no need for it to be a piecewise distribution: a standard normal distribution is a 50-50 mixture between two folded normal distributions, one positive and one negative. $\endgroup$– John MaddenCommented May 12, 2023 at 14:06
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1$\begingroup$ Every distribution, without exception, can be expressed as a mixture in many ways. Just arbitrarily partition $\mathbb R$ into up to a countable union of measurable disjoint subsets $\mathbb R = \mathcal E_1\cup \mathcal E_2 \cup \cdots$, let $p_i = {\Pr}_f(\mathcal E_i)$ for each $i,$ and for all nonzero $p_i$ set $$f_i(x) = \frac{f(x)}{p_i}\mathcal{I}_{\mathcal E_i}(x)$$ to be the truncation of $f$ supported on $\mathcal E_i$.. It is immediate that $f$ is the mixture of these $f_i$ with weights $p_i.$ $\endgroup$– whuber ♦Commented May 12, 2023 at 14:47
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1$\begingroup$ It's more clear now that mixture distributions constitute every distribution, in particular the piecewise defined distributions, thanks. $\endgroup$– giorgi nguyenCommented May 13, 2023 at 6:33
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