Nonlinearity and nonconvexity

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1. Jun 30, 2015

pamparana

I am taking a course on linear regression online and it talks about the sum of square difference cost function and one of the points it makes is that the cost function is always convex i.e. it has only one optima.

Now, reading a bit more it seems that non-linear functions tend to give rise to non-convex functions and I am trying to develop some intuition behind it. So, suppose I take a random model like:

$$f(x) = w_0 x^2 + w_1 \exp(x) + \epsilon$$

And the cost function I choose is the same i.e. I want to minimise:

$$J(w_0, w_1) = (f(x) - w_0 x^2 + w_1 \exp(x))^2$$

What is the intuition that the squared term and the exponential term would give rise to non-convexities?

2. Jul 1, 2015

fzero

For the example that you've chosen (fixing a missing minus sign),

$$\frac{\partial^2 J}{\partial w_0 \partial w_1} = 2 x^2 e^x \geq 0,$$

is positive semidefinite.

In a more general case, we will have residuals

$$r_i = y_i - \sum_\alpha w_\alpha f_\alpha(x_i)$$

and we are minimising

$$S = \sum_i r_i^2.$$

The sign of the expressions

$$m_{\alpha\beta}= \frac{\partial^2 S}{\partial w_\alpha \partial w_\beta} = 2 \sum_i f_\alpha(x_i) f_\beta(x_i),$$

determines the convexivity of $S$ as a function of the $w_\alpha$. When $\beta = \alpha$, $m_{\alpha\alpha}$ is a sum of squares, so is positive semidefinite. When $\beta\neq \alpha$, it is possible to find $m_{\alpha\beta}<0$, depending on the specific functions $f_\alpha$ and the data $x_i$.