Suppose $X_1$, $X_2$, $\ldots$ ,$X_N$ are identically distributed (not necessarily independent).
Then, given $a_1+a_2+\ldots+a_N=1$, and let $S=a_1 X_1 + a_2 X_2 + \ldots + a_N X_N$. Does $S$ follow the same distribution as $X$?
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1We can produce examples in which the distribution is the same, but "almost always" it will not be. – André Nicolas Aug 03 '15 at 14:23
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Let the $X_i$ be independent and identically distributed, with non-zero variance $\sigma^2$. Suppose that the $a_i$ are $\ge 0$, and at least $2$ of them are non-zero. Then $a_1X_1+a_2X_2+\cdots a_NX_N$ has variance $(a_1^2+\cdots+a_N^2)\sigma^2$, which is less than $\sigma^2$.
André Nicolas
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Not necessarily. Fix an arbitrary random variable $X$ whose law is not a Dirac mass at $0$ (say standard normal, for example) and set $X_1=X$ and $X_2=-X$. $S=\frac{1}{2}X_1 + \frac{1}{2}X_2 = 0$ does not have the same law as $X$.
Nocturne
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interesting, are $X_1$ and $X_2$ two draws from the distribution of $X$? or one draw? – Math-fun Aug 03 '15 at 14:57
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@Math-fun It's one draw: $X(\omega)$. $X_1(\omega) = X(\omega)$ and $X_2(\omega)=-X(\omega)$. – Nocturne Aug 03 '15 at 15:14