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A continuous function vanishes on $(-\infty,a]$ and on $[c,\infty),$ its graph is a straight line on the interval $[a,b]$ and another straight line on $[b,c],$ and its integral is $1.$

The mean of the probability distribution whose density this is, is $$ \frac{a+b+c} 3 $$ and the variance is $$ \frac{a^2+b^2+c^2-ab-ac-bc} {18}. $$ Both of these are symmetric functions of $a,b,c,$ despite the fact that the role of $b$ in the first paragraph above is different from those of $a$ and $c.$

Is this symmetry somehow surreptitiously present in the characterization of this distribution in the first paragraph above?

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The distribution function equals $$ p(x)=\frac{|x-a|}{(a-b)(a-c)}+\frac{|x-b|}{(b-a)(b-c)}+\frac{|x-c|}{(c-b)(c-a)}. $$ This is pretty symmetric. If you need a $k$-th moment, it equals $\frac2{(k+1)(k+2)}h_k(a,b,c)$, where $h_k$ is complete homogeneous polynomial (sum of all monomials of degree $k$.)

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  • $\begingroup$ Instantly this function looked symmetric in $a,b,c,$ then I took a few seconds to think maybe I'd better check that since I wasn't sure plus and minus signs wout not be altered, then I applied the permutation $a\mapsto b\mapsto c\mapsto a$ to it and ascertained that it does not change, then did the same with $a\leftrightarrow b, c\leftrightarrow c$, and checked that it still didn't change. Now I'm thinking maybe my question was insufficient for what I was wondering about, but also that my question looks perfectly ok. To be continued$\,\ldots\qquad$ $\endgroup$
    – Michael Hardy
    Commented Mar 31, 2018 at 0:25

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