How to show the concavity of this function (if possible)?

Question

Let $y_j\in[0, 1]$ for $j\in\{1,\ldots,n\}$. Let $f_j$ be the function

$$ f_j(y_1,\dots,y_n):=\log\left(1+\dfrac{c_j}{1+\sum_{j'\neq j}^nc_{j'}y_{j'}}\right), $$ where $c_j$ are nonnegative integers and $j\in\{1,\ldots,n\}$.

Is this function concave?

I calculated the Hessian and then $z^\top H z$ but I cannot prove if it is positive or negative. .

Answer 1

Without loss of generality, take the Hessian of $f_1(y_1,\ldots,y_n)$. After calculating the partial derivatives, we found

$$\begin{align} \mathbf{H}_1:=g(y_2,\ldots,y_n)\times\begin{bmatrix} 0 & 0 & \cdots & 0\\ 0 & c_1c_2c_2 & \cdots & c_1c_2c_n\\ 0 & c_1c_3c_2 & \cdots & c_1c_3c_n\\ \vdots & \vdots & \ddots & \vdots\\ 0 & c_1c_nc_2 & \cdots & c_1c_nc_n\\ \end{bmatrix}, \end{align}$$ where $g(y_2,\ldots,y_n):=\dfrac{2+c_1+2\sum\limits_{j=2}^nc_{j}y_{j}}{\left(1+\sum\limits_{j=2}^nc_{j}y_{j}\right)^2\times\left(1+c_1+\sum\limits_{j=2}^nc_{j}y_{j}\right)^2}$.

Claculating $\mathbf{z}^\top\mathbf{H}_1\mathbf{z}$ gives

$$\mathbf{z}^\top\mathbf{H}_1\mathbf{z}=c_1\times g(y_2,\ldots,y_n)\times\left(\sum_{i=2}^nc_iz_i\right)^2\geqslant0.$$

Hence the function $f_1(y_1,y_2,\ldots,y_n)$ is convex and not concave.

Answer 2

Your function is not concave. Just consider $n=2$ and $c_j=1$. Then the first component of $f$ is: $$f_1(y_1,y_2) = \log\left( 1+\frac{1}{1+y_2} \right)$$ A plot for $y_2 \in [0,1]$ reveals that $f$ is convex for this particular $n$ and $c$.