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Suppose $f:\mathbb{R}^n\rightarrow \mathbb{R}$ is positive almost everywhere and integrable. We know that if $n=1$ and if $f$ is unimodal then the integral $F(x)=\int_{[-\infty,x]} f$ is convex for all $x<a$ and concave for all $x>a$, where $a$ is the point of unimodality. Suppose now that one extends the definition of unimodality in the following manner: $\{x\ |\ f(x)\ge a\}$ is a convex set for every $a>0$.

I'm wondering if this result holds as well in the sense that the distribution funciton $$F(x_1,\ldots,x_n)=\int_{[-\infty,x_i]} f(x_1,\ldots,x_n)dx_1 \ldots dx_n$$ is eventually concave (i.e., in some region $\cap \{x_i\ge t_i\}$.

Intuitively, the restriction on the density function should yield this kind of outcome since as soon as one moves away from the mode, the function starts to decrease. Am I getting the right idea? Can anyone refer me to reference they know of. Thanks in advance.

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What do you mean by "positive almost"? – Robert Israel Jul 26 '12 at 20:47
sorry - I meant positive almost everywhere. – Stuck_pls_help Jul 26 '12 at 20:48
You also want to assume $f$ is integrable on $(-\infty, t]^n$ for all $t$. – Robert Israel Jul 26 '12 at 20:50
Yes - Thanks for the correction @RobertIsrael. I've added integrability of the function everywhere on $\mathbb{R^n}$. – Stuck_pls_help Jul 26 '12 at 20:52
up vote 1 down vote accepted

With $n=2$, consider $f(x,y) = \sum_{j=1}^\infty 2^{-j} f_j(x,y)$ where $f_j(x,y) = 1$ for $-j \le x,y \le j$ and $0$ otherwise. Thus each nonempty $\{(x,y): f(x,y) \ge a \}$ is a square. $F(x,y) = \sum_{j=1}^\infty F_j(x,y)$ where $F_j(x,y) = (x+j)(y+j)$ for $-j < x,y < j$, while $F_j(x,y)$ is constant for $x,y > j$. Since the Hessian matrix of $(x+j)(y+j)$ is $\pmatrix{0 & 1\cr 1 & 0\cr}$ which is indefinite, $F$ is not concave or convex on $k < x,y < k+1$ for any $k$.

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Thanks @RobertIsrael – Stuck_pls_help Jul 26 '12 at 21:47

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