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(*Question edited for clarification)

Does the lower bound of the unstandarized beta distribution always have to be bigger than 0?

itool
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  • Thanks for the link! It does have to do with what I posted, but I'm interested to see if the lower bound can be negative and that question doesn't explicitly state it. – itool Jan 18 '17 at 13:48
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    It also does not state any restrictions. The only restriction is that lower bound must be smaller then upper bound, what is pretty obvious. – Tim Jan 18 '17 at 14:15

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Non-standard beta distribution has probability density function

$$ f(x) = \frac{(x-a)^{\alpha-1}(b-x)^{\beta-1}} {\mathrm{B}(\alpha,\beta) (b-a)^{\alpha+\beta-1}} $$

where $\alpha,\beta$ are parameters of beta distribution and $a,b$ are lower and upper bounds such that $a<b$, there is no other restrictions (so they can be negative).

The trick is that if random variable $X$ is bounded in $(a,b)$, then it can be normalized to have $(0,1)$ bounds by applying simple normalization $Z = (X-a)/(b-a)$, so it's density function is $f(Z)/(b-a)$. Similar operation can be applied to any bounded distribution, e.g. this is how we calculate probabilities of uniformly distributed variable with bounds other then $(0,1)$.

Tim
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