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Suppose I have the weighted sum of $n$ Gaussian functions with varying means and standard deviations, $$f(x) = \sum_{i=1}^n a_i\exp\left(-\frac{(x-\mu_i)^2}{2\sigma_i^2}\right).$$ All the weights $a_i$ are positive. I want to find the position of the global maximum of $f$.

Clearly $f$ is non-concave and may have multiple local maxima. However, it feels like each local maximum should be close to the peak $\mu_i$ of one of the Gaussians.

Suppose I perform gradient ascent $n$ times starting from $x = \mu_i$ for $i = 1, 2,\ldots, n$. Am I guaranteed to find all the local maxima, and thus the global maximum? Is there a better strategy to find the global maximum?

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Since $x$ is one dimensional, you can do a simple grid search with fine enough grids, which guarantees to find global optimum. It is not necessary the case that each local maximum is very close to $\mu_i$. In fact, new local maxima can be created between two distant $\mu_i$s if their $\sigma_i$s are large enough.


To see how the grid search will give guarantees on global optimal solution. We first compute the bound of $f'(x)$:

$$|f'(x)| \le \sum_i a_i \left|\frac{x-\mu_i}{\sigma_i^2}\right|\exp\left(-\frac{(x-\mu_i)^2}{2\sigma_i^2}\right)$$

since (by taking derivative)

$$\frac{|x|}{\sigma^2}\exp\left(-\frac{x^2}{2\sigma^2}\right) \le \frac{1}{\sigma}e^{-1/2}$$

we have $$|f'(x)| \le e^{-1/2}\sum_i \frac{a_i}{\sigma_i} \equiv M$$

On the other hand, there exists a sufficiently large $K$ so that $|f(x)| \le \max_i a_i$ for $x \notin [-K,K]$ and cannot be the global optimum solution. We thus divide $[-K, K]$ into $\lceil 2KM/\epsilon \rceil$ bins $[l_j, u_j]$ so that $u_j - l_j \le \epsilon/M$. Thus since $|f'(x)|\le M$, within each bin $j$ there exists $c_j$ so that

$$c_j - \frac{\epsilon}{2}\le f(x) \le c_j + \frac{\epsilon}{2}$$

Let $j^* = \arg\max_j c_j$. Then any $\hat x^* \in [l_{j^*}, u_{j^*}]$ gives $f(\hat x^*)$ that is at most $\epsilon$ worse than the true global optimum.

I admit that from this analysis, it is not possible to provide a finite time algorithm that find the exact global optimum.

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How does a grid search guarantee global optimum? – Lord Soth Jul 25 '13 at 21:36
Because you essentially have searched every location of the space if the function does not change too much within each grid. – Yuandong Jul 25 '13 at 21:41
Unless you use some of the properties of the given optimization problem, you can never guarantee finding the global optimum with grid search. For example, at least you should adjust your grid size with respect to the parameters of the mixture Gaussian, and specify where you start your search and where you end it. – Lord Soth Jul 25 '13 at 21:45
"Close" is of course relative to the scale parameter $\sigma_i$. Anyway, how fine a grid is fine enough? I am looking for theoretical guarantees. – Rahul Jul 25 '13 at 22:28

Here's a partial result making the intuition in my question rigorous.

Let $f(x) = \sum\limits_{i=1}^n g_i(x)$, where $$g_i(x) = a_i\exp\left(-\frac{(x-\mu_i)^2}{2\sigma_i^2}\right)$$ are the $n$ Gaussians being summed together.

Suppose $x$ is a critical point of $f$, that is, $f'(x)=0$. For it to be a local maximum, $f''(x)$ must be negative. But $f''(x)=\sum\limits_{i=1}^ng_i''(x)$, and $$g_i''(x)=\frac{(x-\mu_i)^2-\sigma_i^2}{\sigma_i^4}g(x)$$ is negative only when $x\in[\mu_i-\sigma_i,\mu_i+\sigma_i]$. So any local maximum can only occur within $\sigma_i$ of a $\mu_i$.

This allows us to only search within the potentially smaller union of intervals $\bigcap\limits_{i=1}^n [\mu_i-\sigma_i,\mu_i+\sigma_i]$ rather than all of the approximately $[\min\mu_i,\max\mu_i]$ as I interpret Yuandong's answer. This is helpful if the $\mu_i$ are widely separated. Nevertheless, $f$ is not guaranteed to be concave in any of these intervals, so we cannot be assured to find a global maximum simply by gradient ascent. One can get within $\epsilon$ of the global maximum by a grid search with step size $\epsilon/M$, where $M$ is as defined in Yuandong's answer, but it would be nice to know if any faster convergence is possible.

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