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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.

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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
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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

1 Answer 1

up vote 4 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.

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