22
$\begingroup$

Essentially, my question is that in multilayer Perceptrons, perceptrons are used with a sigmoid activation function. So that in the update rule $\hat{y}$ is calculated as

$$\hat{y} = \frac{1}{1+\exp(-\mathbf{w}^T\mathbf{x}_i)}$$

How does this "sigmoid" Perceptron differ from a logistic regression then?

I would say that a single-layer sigmoid perceptron is equivalent to a logistic regression in the sense that both use $\hat{y} = \frac{1}{1+\exp(-\mathbf{w}^T\mathbf{x}_i)}$ in the update rule. Also, both return $\operatorname{sign}(\hat{y} = \frac{1}{1+\exp(-\mathbf{w}^T\mathbf{x}_i)})$ in the prediction. However, in multilayer perceptrons, the sigmoid activation function is used to return a probability, not an on off signal in contrast to logistic regression and a single-layer perceptron.

I think the usage of the term "Perceptron" may be a little bit ambiguous, so let me provide some background based on my current understanding about single-layer perceptrons:

Classic perceptron rule

Firstly, the classic perceptron by F. Rosenblatt where we have a step function:

$$\Delta w_d = \eta(y_{i} - \hat{y_i})x_{id} \quad\quad y_{i}, \hat{y_i} \in \{-1,1\}$$

to update the weights

$$w_k := w_k + \Delta w_k \quad \quad (k \in \{1, ..., d\})$$

So that $\hat{y}$ is calculated as

$$\hat{y} = \operatorname{sign}(\mathbf{w}^T\mathbf{x}_i) = \operatorname{sign}(w_0 + w_1x_{i1} + ... + w_dx_{id})$$


Gradient Descent

Using gradient descent, we optimize (minimize) the cost function

$$J(\mathbf{w}) = \sum_{i} \frac{1}{2}(y_i - \hat{y_i})^2 \quad \quad y_i,\hat{y_i} \in \mathbb{R}$$

where we have "real" numbers, so I see this basically analogous to linear regression with the difference that our classification output is thresholded.

Here, we take a step into the negative direction of the gradient when we update the weights

$$\Delta w_k = - \eta \frac{\partial J}{\partial w_k} = - \eta \sum_i (y_i - \hat{y_i})(- x_{ik}) = \eta \sum_i (y_i - \hat{y_i})x_{ik}$$

But here, we have $\hat{y} = \mathbf{w}^T\mathbf{x}_i$ instead of $\hat{y} = \operatorname{sign}(\mathbf{w}^T\mathbf{x}_i)$

$$w_k := w_k + \Delta w_k \quad \quad (k \in \{1, ..., d\})$$

Also, we calculate the sum of squared errors for a complete pass over the entire training dataset (in the batch learning mode) in contrast to the classic perceptron rule which updates the weights as new training samples arrive (analog to stochastic gradient descent -- online learning).


Sigmoid activation function

Now, here is my question:

In multilayer Perceptrons, perceptrons are used with a sigmoid activation function. So that in the update rule $\hat{y}$ is calculated as

$$\hat{y} = \frac{1}{1+\exp(-\mathbf{w}^T\mathbf{x}_i)}$$

How does this "sigmoid" Perceptron differ from a logistic regression then?

$\endgroup$
1
  • 4
    $\begingroup$ Amazing, this question by itself allowed me to condense my machine learning and neural net basics! $\endgroup$
    – varun
    Nov 1, 2017 at 18:28

3 Answers 3

6
$\begingroup$

Using gradient descent, we optimize (minimize) the cost function

$$J(\mathbf{w}) = \sum_{i} \frac{1}{2}(y_i - \hat{y_i})^2 \quad \quad y_i,\hat{y_i} \in \mathbb{R}$$

If you minimize the mean squared error, then it's different from logistic regression. Logistic regression is normally associated with the cross entropy loss, here is an introduction page from the scikit-learn library.


(I'll assume multilayer perceptrons are the same thing called neural networks.)

If you used the cross entropy loss (with regularization) for a single-layer neural network, then it's going to be the same model (log-linear model) as logistic regression. If you use a multi-layer network instead, it can be thought of as logistic regression with parametric nonlinear basis functions.


However, in multilayer perceptrons, the sigmoid activation function is used to return a probability, not an on off signal in contrast to logistic regression and a single-layer perceptron.

The output of both logistic regression and neural networks with sigmoid activation function can be interpreted as probabilities. As the cross entropy loss is actually the negative log likelihood defined through the Bernoulli distribution.

$\endgroup$
2
$\begingroup$

Because gradient descent updates each parameter in a way that it reduces output error which must be continues function of all parameters. Threshold based activation is not differentiable that is why sigmoid or tanh activation is used.

Here is a single-layer NN

$\frac{dJ(w,b)}{d\omega_{kj}} =\frac{dJ(w,b)}{dz_k}\cdot \frac{dz_k}{d\omega_{kj}}$

$\frac{dJ(w,b)}{dz_k} = (a_k -y_k)(a_k(1-a_k))$

$\frac{dz_k}{d\omega_{kj}} = x_k $

$ J(w,b) = \frac{1}{2} (y_k - a_k)^2 $

$ a_k = sigm(z_k) = sigm(W_{kj}*x_k + b_k) $

if activation function were a basic step function (threshold), derivative of $J$ w.r.t $z_k$ would be non-differentiable.

here is a link that explain it in general.

Edit: Maybe, I misunderstood what you mean by perceptron. If I'm not mistaken, perceptron is threholded weighed sum of inputs. If you change threholding with logistic function it turns into logistic regression. Multi-layer NN with sigmoid (logistic) activation functions is cascaded layers composed of logistic regressions.

$\endgroup$
7
  • 3
    $\begingroup$ This doesn't answer the question. $\endgroup$
    – Neil G
    Feb 19, 2015 at 16:29
  • $\begingroup$ Thanks for writing this nice comment, but this was not what I was asking for. My question was not "why gradient descent" but "what makes a perceptron with a sigmoid activation function different from logistic regression" $\endgroup$
    – user39663
    Feb 19, 2015 at 18:31
  • $\begingroup$ @SebastianRaschka They are the same. What makes you think that they are different? I've drove gradient descent because I saw a mistake in your gradient descent evaluation. You assumed $y = W^T X$ when you were driving it. That is why you found the same derivation for both Perceptron and Gradient update. $\endgroup$ Feb 19, 2015 at 18:49
  • 1
    $\begingroup$ "What makes you think that they are different?" -- the nomenclature, thus I was wondering if there is something else; I am just curious why we have 2 different terms for the same thing. Btw. I don't see any mistake in the gradient descent in my question. $y = w_j^Tx_{ji}$ is correct. And I also didn't find the same derivation between "perceptron rule" and "gradient descent" update. The former is done in an online learning manner (sample by sample), the latter is done in batch, and also we minimize the sum of squared errors instead of using a stepwise function. $\endgroup$
    – user39663
    Feb 19, 2015 at 18:59
  • $\begingroup$ I think what might caused the confusion is that you have distinguish between the "classification" and the "learning" step. The classification step is always thresholded (-1 or 1, or 0 and 1 if you like). However, the update is different, in the classic perceptron, the update is done via $\eta (y - sign(w^Tx_i))x$ whereas in let's say stochastic gradient descent it is $\eta (y - w^Tx_i)x_i$ $\endgroup$
    – user39663
    Feb 19, 2015 at 19:02
2
$\begingroup$

Intuitively, I think of a multilayer perceptron as computing a nonlinear transformation on my input features, and then feeding these transformed variables into a logistic regression.

The multinomial (that is, N > 2 possible labels) case may make this more clear. In traditional logistic regression, for a given data point, you want to compute a "score", $\beta_i X$, for each class, $i$. And the way you convert these to probabilities is just by taking the score for the given class over the sum of scores for all classes, $\frac{\beta_i X}{\sum_j \beta_j X}$. So a class with a large score has a larger share of the combined score and so a higher probability. If forced to predict a single class, you choose the class with the largest probability (which is also the largest score).

I don't know about you, but in my modeling courses and research, I tried all kinds of sensible and stupid transformations of the input features to improve their significance and overall model prediction. Squaring things, taking logs, combining two into a rate, etc. I had no shame, but I had limited patience.

A multilayer perceptron is like a graduate student with way too much time on her hands. Through the gradient descent training and sigmoid activations, it's going to compute arbitrary nonlinear combinations of your original input variables. In the final layer of the perceptron, these variables effectively become the $X$ in the above equation, and your gradient descent also computes an associated final $\beta_i$. The MLP framework is just an abstraction of this.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.