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I read in Boyd's text that over convex $C$ such a distribution:

$f(x) = {1\over a} I_C(x)$

for $I$ the indicator function for $C$ and $a$ the measure of $C$. And that taking $\log 0 = -\infty$ we have that $\log f$ is $-\log a$ for $x \in C$ and $-\infty$ otherwise. Then the text asserts that because this is concave the distribution is log concave. But isn't $-\log x$ a convex function? (Second derivative is $1/x^2$ which is always positive). I keep thinking it should be log convex, what am I missing?

3 Answers 3

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Consider $- \log f$: This is finite on a convex set and $+\infty$ on the complement, hence the epigraph is convex. Hence $- \log f$ is convex, it follows that $\log f$ is concave.

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The function $\log f(x)$ should be concave with respect to $x$, not with respect to $a$. When $x\in C$, your distribution function is constant, and hence all derivatives are zero here.

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    ohhhhhhhhhhhh good point...2012-09-24
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A function $g$ is concave if, for every $x$ and $y$ and every $t$ in $(0,1)$, $ g(tx+(1-t)y)\geqslant tg(x)+(1-t)g(y)\qquad\qquad (\ast) $ If $g=\log f$ for your function $f$, two cases arise:

  • Either $x$ and $y$ are in $C$, then $tx+(1-t)y$ is in $C$ as well because $C$ is convex, hence $g(x)=g(y)=g(tx+(1-t)y)=-\log a$ and $(\ast)$ holds.
  • Or, $x$ or $y$ is not in $C$, then $g(x)=-\infty$ or $g(y)=-\infty$ hence the RHS of $(\ast)$ is $-\infty$ and $(\ast)$ holds.

Hence $g=\log f$ is concave and $f$ is log-concave.

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    Thanks for the explanation2012-09-24