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I know that sum of log-concave is not always log-concave.

Could anyone provides me with an example to prove this?

Like probability distribution fn (pdf) of normal distribution is log-concave; on what condition, the sum of two pdf will not be log-concave?

Thanks!!

:)

sleeve chen
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    You remarked that the Gaussian functions associated with normal distributions are log-concave. Have you tried experimenting with adding two Gaussian functions together, to see if you can create an example where the sum is not log-concave? I particularly like the idea of having the two Gaussians be centered at different points.... – Greg Martin Mar 30 '14 at 00:07
  • Given that log-concavity is preserved under convolution, you will have to look only at sums of dependent random variables. There are various way to capture dependency - for example, try a copula approach. – baudolino Mar 30 '14 at 01:09

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Here is an example. Consider the density of two exponential random variables as \begin{align*} f_1(x) &=\lambda_1 \exp(-\lambda_1x) \\ f_2(x) &=\lambda_2 \exp(-\lambda_2x), \end{align*} where $\lambda_1 \neq \lambda_2$.

Consider the sum of these two log concave functions \begin{align*} g(x)=f_1(x)+f_2(x), \end{align*} where the first and second derivatives of $g(x)$ are given as: \begin{align*} g^{'}(x) &=- \lambda_1 ^2 \exp(-\lambda_1x) - \lambda_2^2 \exp(-\lambda_2x) \\ g^{''}(x) &= \lambda_1 ^3 \exp(-\lambda_1x) + \lambda_2^3 \exp(-\lambda_2x) \end{align*}

One can show that $g^{'}(x)g^{'}(x)\le g^{''}(x)g(x)$, hence $g(x)$ is not log-concave. In fact, it is log-convex.

emper
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    Can you expand a little bit on this "on can show"? It's clearly not true for ALL $\lambda_1\neq\lambda_2$ (take one equal to zero...), so does one actually show it? – Clement C. Feb 01 '19 at 16:24
  • (i.e., it only holds as long as the two parameters are both non zero, besides being distinct) – Clement C. Feb 01 '19 at 16:27