Let $(aX + bY)$ be a truncated normal and assume $X,Y$ are both identically distributed (but necessarily NOT independent) what is the distribution of $X$ and $Y$? More importantly can the pdf of $X$ be written just in terms of $X$?
Is this known? Or have I just defined a new distribution?
Suppose in the case that $aX + bY$ is a standard normal singly truncated from the top at $\beta$ i.e. $aX + bY \le \beta$ then taking the limit as $b$ goes to 0 we must have $$aX \sim N(0,1)|aX < \beta$$ so from this I can intuitively deduce that $$aX|bY ~ N(0,1)|aX <\beta - bY$$ and symmetrically $$bY|aX ~ N(0,1)|bY <\beta - aX$$ so the distribution of one variable given the other is know to be just a truncated normal but it seems quite different to write out the pdf the unconditional distribution. Indeed, let $f(x)$ be the pdf of $X$ then
$$f(x) = \int_{y=-\infty}^\infty f(x|y)f(y)\ dy$$ the right side also involves the $f$ so it looks pretty hard to solve?
If $X$ and $Y$ are not independent, I doubt that there's much that can be said. There are just too many ways to obtain $Z$ as the sum of $X$ and $Y$. For example, here are some ways to get an arbitrary distribution as $Z = X + Y$ with $X$ and $Y$ identically distributed. Let $B$ be Bernoulli($1/2$) and independent of $Z$. Let $f$ be an arbitrary Borel function. Let $X = B f(Z) + (1-B)(Z-f(Z))$ and $Y = B (Z - f(Z)) + (1-B) f(Z)$.