Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I think the following problem is convex (due to the results of some simulations), but I am not sure:

$min_x||e^{(Ax)}-b||^2_2$ s.t. x>0

where $A$ is m x n, $x$ is n x 1, and b is m x 1. $A,x,b$ are all real. The exponent of a vector means taking the exponent of each coordinate (is there a better way to write this?).

My reasoning is as follows:

  • $Ax$ is convex and $e^x$ is convex.

  • Composition of convex functions is convex, so $e^{(Ax)}$ is convex.

  • Subtraction shouldn't change convexity, so $e^{(Ax)}-b$ is convex.

  • $||x||^2_2$ is convex so $||e^{(Ax)}-b||^2_2$ is convex due to convex composition.

  • Restriction of variables to be positive won't change convexity.

Is this correct?

1st EDIT:

As justt indicated, my reasoning does not hold because convexity is not defined for vector-valued functions. I will try other approaches and update on my progress.

2nd EDIT:

For future reference, if anyone is trying to solve a similar problem: while this problem is not convex, it can be viewed as fitting the parameters of a log-linear model ($e^{Ax}$) to data ($b$). If $e^{Ax}$ is normalized into a probability distribution, then fitting the parameters using the MLE (Maximum Likelihood Estimator) approach will give a convex problem.

share|cite|improve this question
I don't agree that $Ax$ is convex and $e^x$ is convex. We can't say that a function is convex unless it is real-valued. You won't be able to use convex composition here. – justt Apr 30 '13 at 22:43
You can define vector-valued convexity by the partial ordering over a closed convex cone of vectors. I don't know if that is helpful here, though. – Ross B. May 1 '13 at 0:01
@justt oh, did you mean the problem is that convexity is not defined for a function that maps a vector to a vector? – Bitwise May 1 '13 at 1:01
It's not true in general that a composition of two convex functions is convex. See this question – Matti Åstrand May 1 '13 at 1:41
@MattiÅstrand good point, although justt pointed out that convexity isn't defined anyway until the last step where I take the squared L2 norm. Also, $e^{f(x)}$ would be convex if $f(x)$ is convex, since $e^x$ is non-decreasing. Not sure about the second composition though. – Bitwise May 1 '13 at 1:50
up vote 5 down vote accepted

Think about the case $m=n=1$: your function is just $(e^{ax}-b)^2=e^{2ax}-2be^{ax}+b^2$. It's pretty easy to find a counterexample where this is not convex.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.