'Comparing MSE loss and cross-entropy loss in terms of convergence

For a very simple classification problem where I have a target vector [0,0,0,....0] and a prediction vector [0,0.1,0.2,....1] would cross-entropy loss converge better/faster or would MSE loss? When I plot them it seems to me that MSE loss has a lower error margin. Why would that be?

Or for example when I have the target as [1,1,1,1....1] I get the following:

Solution 1:^[1]

You sound a little confused...

Comparing the values of MSE & cross-entropy loss and saying that one is lower than the other is like comparing apples to oranges
MSE is for regression problems, while cross-entropy loss is for classification ones; these contexts are mutually exclusive, hence comparing the numerical values of their corresponding loss measures makes no sense
When your prediction vector is like [0,0.1,0.2,....1] (i.e. with non-integer components), as you say, the problem is a regression (and not a classification) one; in classification settings, we usually use one-hot encoded target vectors, where only one component is 1 and the rest are 0
A target vector of [1,1,1,1....1] could be the case either in a regression setting, or in a multi-label multi-class classification, i.e. where the output may belong to more than one class simultaneously

On top of these, your plot choice, with the percentage (?) of predictions in the horizontal axis, is puzzling - I have never seen such plots in ML diagnostics, and I am not quite sure what exactly they represent or why they can be useful...

If you like a detailed discussion of the cross-entropy loss & accuracy in classification settings, you may have a look at this answer of mine.

Solution 2:^[2]

As complement to the accepted answer, I will answer the following questions

What is the interpretation of MSE loss and cross entropy loss from probability perspective.
Why cross entropy is used for classification and MSE is used for linear regression?

TL;DR Use MSE loss if (random) target variable is from Gaussian distribution and categorical cross entropy loss if (random) target variable is from Multinomial distribution.

MSE(Mean squared error)

One of the assumptions of the linear regression is multi-variant normality. From this it follows that the target variable is normally distributed(more on the assumptions of linear regression can be found here and here).

Gaussian distribution(Normal distribution) with mean $\mu$ and variance $\sigma^2$ is given by
$\mathcal{N}(x|\mu,\sigma^2)=\frac{1}{\sqrt{2\pi\sigma^{2}}}e^{-\frac{(x-\mu)^2}{2\sigma^2}}$
Often in machine learning we deal with distribution with mean 0 and variance 1(Or we transform our data to have mean 0 and variance 1). In this case the normal distribution will be,
$\mathcal{N}(x|\mu=0,\sigma^2=1)=\frac{1}{\sqrt{2\pi}}e^{-\frac{x^2}{2}}$ This is called standard normal distribution.
For normal distribution model with weight parameter $\mathbf{w}$ and precision(inverse variance) parameter $\beta$ , the probability of observing a single target t given input x is expressed by the following equation

$\mathcal{p(t|x,\mathbf{w},\beta)=\mathcal{N}(t|y(x,\mathbf{w}),\beta^{-1})$ , where $y(x,\mathbf{w})$ is mean of the distribution and is calculated by model as
$y(x,\mathbf{w})=\sum_{i=1}^{m}w_ix^i$

Now the probability of target vector $\mathbf{t}$ given input $\mathbf{X}$ can be expressed by

$p(\mathbf{t}|\mathbf{X},\mathbf{w},\beta)=\prod_{n=1}^{N}\mathcal{N}(t_n|y(x_n,\mathbf{w}),\beta^{-1})=$ $\prod_{n=1}^{N}\frac{\beta}{\sqrt{2\pi}}e^{-\beta\frac{(t_n-y(x_n,w))^2}{2}}$
Taking natural logarithm of left and right terms yields

$\ln p(\mathbf{t}|\mathbf{X},\mathbf{w},\beta)=\ln \prod_{n=1}^{N}\frac{\beta}{\sqrt{2\pi}}e^{-\beta\frac{(t_n-y(x_n,w))^2}{2}}$
$=-\frac{\beta}{2}\sum_{n=1}^N\left{y(x_n,w)-t_n\right}^2+\frac{N}{2}\ln\beta-\frac{N}{2}\ln(2\pi)=$ $\ln L(\mathbf{w},\beta}|\mathbf{X},\mathbf{t})$
Where $\ln L(\mathbf{w},\beta}|\mathbf{X},\mathbf{t})$ is log likelihood of normal function. Often training a model involves optimizing the likelihood function with respect to $\mathbf{w}$ . Now maximum likelihood function for parameter $\mathbf{w}$ is given by (constant terms with respect to $\mathbf{w}$ can be omitted),

$\ln L(\mathbf{w},\beta}|\mathbf{X},\mathbf{t})=-\frac{\beta}{2}\sum_{n=1}^N\left{y(x_n,w)-t_n\right}^2$

For training the model omitting the constant $\frac{-\beta}{2}$ doesn't affect the convergence. $\ln L(\mathbf{w},\beta}|\mathbf{X},\mathbf{t})=\sum_{n=1}^N\left{y(x_n,w)-t_n\right}^2$ This is called squared error and taking the mean yields mean squared error.
$\frac{1}{N}\ln L(\mathbf{w},\beta}|\mathbf{X},\mathbf{t})=\frac{1}{N}\sum_{n=1}^N\left{y(x_n,w)-t_n\right}^2$ ,

Cross entropy

Before going into more general cross entropy function, I will explain specific type of cross entropy - binary cross entropy.

Binary Cross entropy

The assumption of binary cross entropy is probability distribution of target variable is drawn from Bernoulli distribution. According to Wikipedia

Bernoulli distribution is the discrete probability distribution of a random variable which takes the value 1 with probability p and the value 0 with probability q=1-p

Probability of Bernoulli distribution random variable is given by
$P(Y=k)=p^k(1-p)^{1-k}$ , where $k\in\left{0,1\right}$ and p is probability of success. This can be simply written as $P(y)=p^y(1-p)^{1-y}$
Taking negative natural logarithm of both sides yields

$-\ln P(y)=-y\ln(p)-(1-y)\ln(1-p)$ , this is called binary cross entropy.

Categorical cross entropy

Generalization of the cross entropy follows the general case when the random variable is multi-variant(is from Multinomial distribution ) with the following probability distribution

$P(\mathbf{Y})=\prod_{n=1}^{N}p_n^{y_n}(1-p_n)^{1-y_n}={p_n}^{\sum_{n=1}^{N}y_n}(1-p_n)^{n-\sum_{n=1}^{N}y_n}}$

Taking negative natural logarithm of both sides yields categorical cross entropy loss.

$-\ln P(y)=-(\sum_{n=1}^{N}y_n\ln(p_n)+(1-y_n)\ln(1-p_n))$ ,

Solution 3:^[3]

I tend to disagree with the previously given answers. The point is that the cross-entropy and MSE loss are the same.

The modern NN learn their parameters using maximum likelihood estimation (MLE) of the parameter space. The maximum likelihood estimator is given by argmax of the product of probability distribution over the parameter space. If we apply a log transformation and scale the MLE by the number of free parameters, we will get an expectation of the empirical distribution defined by the training data.

Furthermore, we can assume different priors, e.g. Gaussian or Bernoulli, which yield either the MSE loss or negative log-likelihood of the sigmoid function.

For further reading: Ian Goodfellow "Deep Learning"

Sources

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Source: Stack Overflow

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Solution 3	polkadot21

'Comparing MSE loss and cross-entropy loss in terms of convergence

Solution 1:[1]

Solution 2:[2]