with vector $\mathbf{x}=[x_1,x_2,..,x_n]$, is $f_i(\mathbf{x})=x_i \exp(-(x_1+x_2+...x_i))$ concave for vector $\mathbf{x}$ ????

Question

It looks not that complicated but I'm stuck in the middle.

$\mathbf{x}=[x_1,x_2, \cdots ,x_n]$.

1. $g(\mathbf{x})=\exp(-\mathbf{x})$ is a decreasing, and convex function.
2. $h(\mathbf{x})=x_1+x_2+x_3+\dotsm\;$ is a linear, increasing function, convex/concave.

So, $g(h(\mathbf{x}))$ is a convex function. Am I right so far?

and the problem is that $x_i$ is multiplied to $g(h(\mathbf{x}))$. $x_i$'s are positive and between 0 to 1 real value.

How can I prove $f_i(\mathbf{x})=x_i g(h(\mathbf{x}))$ is concave?

Thanks a lot.

Answer 1

You can do the hessian H of $f_i(x)$ and prove that H is negative semidefinite (if x'Hx ≤ 0 for all x).

the Hessian of $f_i$ is positive semi definite in this case which means that $f_i$ is convex if $x_i$ >0.

Answer 2

Note that $f(x) = x e^{-x}$ is strictly convex for $x \ge 2$ (just compute $f''(x)$).