# Tag Info

## New answers tagged perceptron

0

It seems loosely reasonable but there are various things which are potentially unclear. What exactly is a prediction, and is it deterministic or stochastic? First, if you are predicting a continuous value, you can never be "correct" - there will always be at least some very small deviation. This makes me assume that you are talking about making ...

0

It depends on your hypothesis $h$. The author of the original article compares the dot product with a threshold: So for a binary classification problem $h$ can be defined as follows: $$h = \begin{cases} 1 & \text{if f>z}\\ 0 & \text{otherwise} \end{cases}$$ That is, $\hat y_i$ is your prediction and $\hat y_i = h(x_i)$, $y_i$ ...

1

Assume we have a binary classification problem, which we want to solve with a simple single-layer perceptron. For a 2d space, a perceptron will have 2 inputs $x_1$ and $x_2$, and a bias denoted $x_0$ which is always $x_0=1$. It also has corresponding learnable weights $w_0$, $w_1$ and $w_2$. This can be vectorized:  \overline{x} = \begin{bmatrix} 1 \\ x_1 \...

1

The loss function is simply a way to measure how wrong a neural network is, it doesn't affect the output of the neuron. Say we have a neural network with 3 output neurons that attempts to classify images of cats, dogs, and humans. The output it gives is the confidence of the neural network's classification. For example if the output is [0, 0.2, 0.8] (0 being ...

2

Loss function is a function used to measure the loss. It is not used in any component of a neuron. It is used in updating the weights of the neuron i.e., in order to train the neuron. The contribution of a loss function is in the updation of $\bar{W}$. For a given $\bar{X}$ and $\bar{W}$, the neuron gives a post-action value $h$. But the desired output may ...

2

I am specifically asking about the probability that the value is 1 (that is, how sigmoid functions specifically check for this). They don't in general. In the quoted text, there is an explicit constraint that means this can be the case: If it is desirable to predict a probability of a binary class (emphasis mine). This means that the target value $y \in \{... 0 Well, there is already a nice answer to this question. I present a similar contradiction based answer using basic coordinate geometry. Is there a proof to explain why$XOR$cannot be linearly separable? Let us suppose, if possible, that the$XOR\$ function, given by following table, is linearly separable. \begin{array}{|c|c|c|} \hline x& y & x \...

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