Cost function of neural network is non-convex?

Question

The cost function of neural network is $J(W,b)$, and it is claimed to be non-convex. I don't quite understand why it's that way, since as I see that it's quite similar to the cost function of logistic regression, right?

If it is non-convex, so the 2nd order derivative $\frac{\partial J}{\partial W} < 0$, right?

UPDATE

Thanks to the answers below as well as @gung's comment, I got your point, if there's no hidden layers at all, it's convex, just like logistic regression. But if there's hidden layers, by permuting the nodes in the hidden layers as well as the weights in subsequent connections, we could have multiple solutions of the weights resulting to the same loss.

Now more questions,

1) There're multiple local minima, and some of them should be of the same value, since they're corresponding to some nodes and weights permutations, right?

2) If the nodes and weights won't be permuted at all, then it's convex, right? And the minima will be the global minima. If so, the answer to 1) is, all those local minima will be of the same value, correct?

Answer 1

The cost function of a neural network is in general neither convex nor concave. This means that the matrix of all second partial derivatives (the Hessian) is neither positive semidefinite, nor negative semidefinite. Since the second derivative is a matrix, it's possible that it's neither one or the other.

To make this analogous to one-variable functions, one could say that the cost function is neither shaped like the graph of $x^2$ nor like the graph of $-x^2$. Another example of a non-convex, non-concave function is $\sin(x)$ on $\mathbb{R}$. One of the most striking differences is that $\pm x^2$ has only one extremum, whereas $\sin$ has infinitely many maxima and minima.

How does this relate to our neural network? A cost function $J(W,b)$ has also a number of local maxima and minima, as you can see in this picture, for example.

The fact that $J$ has multiple minima can also be interpreted in a nice way. In each layer, you use multiple nodes which are assigned different parameters to make the cost function small. Except for the values of the parameters, these nodes are the same. So you could exchange the parameters of the first node in one layer with those of the second node in the same layer, and accounting for this change in the subsequent layers. You'd end up with a different set of parameters, but the value of the cost function can't be distinguished by (basically you just moved a node, to another place, but kept all the inputs/outputs the same).

Answer 2

If you permute the neurons in the hidden layer and do the same permutation on the weights of the adjacent layers then the loss doesn't change. Hence if there is a non-zero global minimum as a function of weights, then it can't be unique since the permutation of weights gives another minimum. Hence the function is not convex.

Answer 3

Whether the objective function is convex or not depends on the details of the network. In the case where multiple local minima exist, you ask whether they're all equivalent. In general, the answer is no, but the chance of finding a local minimum with good generalization performance appears to increase with network size.

This paper is of interest:

Choromanska et al. (2015). The Loss Surfaces of Multilayer Networks

http://arxiv.org/pdf/1412.0233v3.pdf

From the introduction:

• For large-size networks, most local minima are equivalent and yield similar performance on a test set.

• The probability of finding a "bad" (high value) local minimum is non-zero for small-size networks and decreases quickly with network size.

• Struggling to find the global minimum on the training set (as opposed to one of the many good local ones) is not useful in practice and may lead to overfitting.

They also cite some papers describing how saddle points are a bigger issue than local minima when training large networks.

Answer 4

Some answers for your updates:

1. Yes, there are in general multiple local minima. (If there was only one, it would be called the global minimum.) The local minima will not necessarily be of the same value. In general, there may be no local minima sharing the same value.

2. No, it's not convex unless it's a one-layer network. In the general multiple-layer case, the parameters of the later layers (the weights and activation parameters) can be highly recursive functions of the parameters in previous layers. Generally, multiplication of decision variables introduced by some recursive structure tends to destroy convexity. Another great example of this is MA(q) models in times series analysis.

Side note: I don't really know what you mean by permuting nodes and weights. If the activation function varies across nodes, for instance, and you permute the nodes, you're essentially optimizing a different neural network. That is, while the minima of this permuted network may be the same minima, this is not the same network so you can't make a statement about the multiplicity of the same minima. For an analogy of this in the least-squares framework, you are for example swapping some rows of $y$ and $X$ and saying that since the minimum of $\|y - X\beta\|$ is the same as before that there are as many minimizers as there are permutations.

Answer 5

You will have one global minimum if problem is convex or quasiconvex.

About convex "building blocks" during building neural networks (Computer Science version)

I think there are several of them which can be mentioned:

1. max(0,x) - convex and increasing

2. log-sum-exp - convex and increasing in each parameter

3. y = Ax is affine and so convex in (A), maybe increasing maybe decreasing. y = Ax is affine and so convex in (x), maybe increasing maybe decreasing.

Unfortunately it is not convex in (A, x) because it looks like indefinite quadratic form.

4. Usual math discrete convolution (by "usual" I mean defined with repeating signal) Y=h*X Looks that it is affine function of h or of variable X. So it's a convex in variable h or in variable X. About both variables - I don't think so because when h and X are scalars convolution will reduce to indefinite quadratic form.

5. max(f,g) - if f and g are convex then max(f,g) is also convex.

If you substitute one function into another and create compositions then to still in the convex room for y=h(g(x),q(x)), but h should be convex and should increase (non-decrease) in each argument....

Why neural netwoks in non-convex:

1. I think the convolution Y=h*X is not nessesary increasing in h. So if you not use any extra assumptions about kernel you will go out from convex optimization immediatly after you apply convolution. So there is no all fine with composition.

2. Also convolution and matrix multiplication is not convex if consider couple parameters as mentioned above. So there is evean a problems with matrix multiplication: it is non-convex operation in parameters (A,x)

3. y = Ax can be quasiconvex in (A,x) but also extra assumptions should be taken into account.

Please let me know if you disagree or have any extra consideration. The question is also very interesting to me.

p.s. max-pooling - which is downsamping with selecting max looks like some modification of elementwise max operations with affine precomposition (to pull need blocks) and it looks convex for me.

About other questions

1. No, logistic regression is not convex or concave, but it is log-concave. This means that after apply logarithm you will have concave function in explanatory variables. So here max log-likelihood trick is great.

2. If there are not only one global minimum. Nothing can be said about relation between local minimums. Or at least you can not use convex optimization and it's extensions for it, because this area of math is deeply based on global underestimator.

Maybe you have confusion about this. Because really people who create such schemas just do "something" and they receive "something". Unfortunately because we don't have perfect mechanism for tackle with non-convex optimization (in general).

But there are even more simple things beside Neural Networks - which can not be solved like non-linear least squares -- https://youtu.be/l1X4tOoIHYo?t=2992 (EE263, L8, 50:10)

Answer 6

The composition of multiple layers is what makes the cross-entropy or least-squares loss function of multi-layer neural networks non-convex with respect to the set of all weights and biases. This non-convexity arises primarily from the multiplication of parameters (weights and biases) across layers rather than from the non-linearity of activation functions or the over-parameterization issues like symmetry (which relates to the arguments around weight permutations).

To understand how the multiplication of parameters results in non-convexity, consider the simple function $f(x,y)=xy$. This function is convex in $x$ when $y$ is held constant and convex in $y$ when $x$ is fixed. However, it is not convex in $x$ and $y$ jointly. The graph of this function exhibits a saddle shape (i.e, it is convex along $x=y$ and concave along $x=-y$), reflecting non-convexity in both variables jointly:

Another example is $f(x,y)=x^2y^2$, which is non-convex since it is zero along the axes and positive elsewhere, despite both $x^2$ and $y^2$ being strictly convex functions individually.

A function $f(x,y)$ is convex if, for all points $(x_1,y_1)$ and $(x_2,y_2)$, and for any $\lambda\in[0,1]$, it holds that: $$f(\lambda x_1+(1-\lambda) x_2, \lambda y_1+(1-\lambda) y_2) \le \lambda f(x_1,y_1)+(1-\lambda)f(x_2,y_2)$$ This inequality ensures that the function’s graph lies below the line segment connecting any two points on the graph, giving it a bowl-shaped appearance. For $f(x,y)=xy$, we have $$f(\lambda x_1+(1-\lambda) x_2, \lambda y_1+(1-\lambda) y_2)=\lambda^2 x_1 y_1 + (1-\lambda)^2 x_2 y_2 +\lambda(1-\lambda)(x_1 y_2 + x_2 y_1),$$ which is not necessarily less than or equal to $$\lambda f(x_1,y_1)+(1-\lambda)f(x_2,y_2)=\lambda x_1 y_1 + (1-\lambda) x_2 y_2.$$ Thus, multiplicative interactions between variables can introduce non-convexity across all variables by generating saddle points in the cost function.

Answer 7

By definition, a function $f(x)$ is convex over a convex set $S$ if for all $x, y \in S$ and $t \in [0, 1]$, $tf(x) + (1-t)f(y) \geq f(tx + (1-t)y)$. Think of this as a straight line connecting two points of $y = x^2$ always being above the curve itself.

In the general case, $f$ can be shown to be convex if its Hessian is positive definite. Therefore, most of the cost functions used for training neural networks are convex with respect to the net's final output and expected value. This includes MSE, CCE. There are cherry-picked non-convex loss functions that could be used as well, such as the Rosenbrock function, $f(x, y) = 100(x^2-y)^2 + (x-1)^2$. However, I have not seen non-convex functions be used in literature or in practice unless the author is trying to show the goodness of their new update scheme. That an some $L \leq 1$ regularisation schemes.

As for convexity with respect to the intermediary layer weights, unless the output of these intermediaries is non-convex, convexity is still found. Linear layers, convolutions, and activation functions like ReLU are convex, so the loss is also convex with respect to these layers. Generally you just check the convexity of activation functions.

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The argument about how to permute the weights and get the same loss shows that the loss isn't convex isn't true, and when it is, it's not useful.

Consider again $f(x, y)=x^2$ as a loss function. This is convex. Say the net currently outputs $x=2$ yielding a loss of $4$. But, if the weights are changed so that the net now outputs $x = -2$, the loss is also $4$. This is a convex function that (assuming suitable net expressivity) has a way to get the same loss.

Technically the argument also hinged on the fact that the loss you currently have is a global one - but this is an odd assumption as there's no way of knowing you've attained a global loss unless the function is cherry-picked or convex. There is also no suitable permutation of nodes to use, so there's no proof that the argument can be carried out.