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I am quoting a line from a text:

The Laplace exponent $\Phi$ is concave and non-negative, the inequality $\Phi(\lambda)\leq k\Phi(\lambda/k)$ for all $\lambda>0$ and $k>1$ follows.

Why does being concave positive means $\Phi(\lambda)\leq k\Phi(\lambda/k)$? Can someone please explain? thanks.

Note the $\Phi$ function is defined from $[0,\infty)$ to $\mathbb{R}$. It is the log of the Laplace transform

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Being concave positive implies this (but it does not mean it) because $$ \Phi\left(\frac{\lambda}k\right)=\Phi\left(\frac{\lambda+0+\cdots+0}k\right)\geqslant\frac{\Phi(\lambda)+\Phi(0)+\cdots+\Phi(0)}k\geqslant\frac{\Phi(\lambda)}k. $$

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  • $\begingroup$ how do i make it work for non integer $k$? replace it with random variables taking value $\lambda$ with probability $1/k$ and $0$ otherwise. apply Jensen? $\endgroup$ – Lost1 Aug 21 '13 at 22:22
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    $\begingroup$ Do the same thing than above but using $\lambda$ $q$ times and $0$ $p-q$ times. This gets the $k=p/q$ case. Then approach $k$ by rationals and use the continuity of $\Phi$ on $(0,\infty)$. $\endgroup$ – Did Aug 21 '13 at 22:30

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