I have a somewhat complicated function of $M+1$ variables, which looks as follows.
$$f (x_0, x_1, x_2, \dots, x_M) = \sum_{i=1}^{N_A} \ln \left[1 - \text{erf}\left(x_0 + \sum_{j=1}^M a_{ij} x_j\right) \right] + \sum_{i=1}^{N_B} \ln \left[1 + \text{erf}\left(x_0 + \sum_{j=1}^M b_{ij} x_j\right) \right].$$
All is real here, but in in principle other than that there are no restrictions on the possible values of the coefficients $a_{ij}$ and $b_{ij}$. $N_A$ and $N_B$ are in general some "larger" numbers (say, at least on the order of 100 or 1000), while $M$ tends to be pretty small, say for example 5 to 10 or so. Not that it should matter, but to add some context, this is actually a likelihood expression for some model.
Now my question is, is this a concave function with a unique maximum? Intuitively, the answer seems to be yes, and "numerical evidence" hints to the same direction, but I have a hard time proving it rigorously. I tried to calculate the Hessian, and eventually a general expression for its eigenvalues (which should all be negative in case my assumptions holds true), but it was just too much.
Any suggestions would be appreciated!
Thanks, Lennex
A sum of concave functions is concave, and $\ln(1+\text{erf}(t))$ and $\ln(1-\text{erf}(t))$ are easily seen to be concave. So your function is concave. Moreover, since $\ln(1+\text{erf}(t))$ and $\ln(1-\text{erf}(t))$ are strictly concave, the only way for a maximum of your function (assuming it exists) to be non-unique would be for all the $x_0 + \sum_j a_{ij} x_j$ and all the $x_0 + \sum_j b_{ij} x_j$ to be equal at two different points, i.e. the matrix $\pmatrix{1 & A\cr 1 & B\cr}$ to have rank $< M+1$, where $A$ and $B$ are the matrices of coefficients $a_{ij}$ and $b_{ij}$, and the $1$'s are column vectors of all $1$'s.