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My understanding is that a connection between two neurons has a weight, but a neuron itself does not have a weight.

If connection c connects neurons A to B, then c has a weight w, but A and B don't have a weight. w determines if A has a strong influence on B or a weak influence on B.

But this Wikipedia article says: "The connections between artificial neurons are called ‘edges’. Artificial neurons and edges typically have a weight that adjusts as learning proceeds."

https://en.wikipedia.org/wiki/Artificial_neural_network

The 2nd sentence seems to imply that a neuron has a weight, too.

Does a neuron really have a weight?

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All this talking about the connection and neurons have weight is virtual

The point is that each neuron take multiple weights coming from the previous layer through the connection

therefore each neuron will have multiple weights ( you can call them coming from the connection or from the neurons ) the naming is just virtual to present the relation between the weights in the neurons

each weight is different independent from other weight

enter image description here

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If you look at the 2nd equation under "Propagation function"

$p_{j}(t)=\sum _{i=1}^No_{i}(t)w_{ij}+w_{0j}$

The $w_{0j}$ is the constant (bias), so it's reasonable to write that the $w_{ij}$ weights are associated with the connections, and the $w_{0j}$ is a weight associated with the nueron itself.

Additionally if you assume $o_{0}(t)=1$ then you can rewrite the equation above as

$p_{j}(t)=\sum _{i=0}^No_{i}(t)w_{ij}$

(i.e. note the sum is from 0 to N, not 1 to N)

When you do this it no longer really makes sense to make a distinction between a "weight" and a "bias".

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  • $\begingroup$ Hi David, you wrote: "Maybe it's been edited, I cannot find the sentence you refer to". That sentence is still there. Go to en.wikipedia.org/wiki/Artificial_neural_network The 3rd paragraph. The 3rd sentence. "Artificial neurons and edges typically have a weight that adjusts as learning proceeds." $\endgroup$ – Curious Aug 6 '19 at 2:46
  • $\begingroup$ Yes you're correct, I've edited my answer as reading it again I think I see where they're coming from. $\endgroup$ – David Waterworth Aug 6 '19 at 3:28
  • $\begingroup$ Hi David. Does the artificial neuron networking community universally refer to w0j as "bias"? Have you ever heard anyone referring to w0j as "weight" (besides this Wikipedia article)? $\endgroup$ – Curious Aug 7 '19 at 3:01
  • $\begingroup$ I wouldn't say universally, it's more usual to call w_0j the bias when discussing the low level implementation, but equally when you're talking about the trainable parameters of a network at a higher I believe most practitioners would tend to use the term "weights" to refer to all the parameters. $\endgroup$ – David Waterworth Aug 7 '19 at 3:07
  • $\begingroup$ I added a bit more to the answer $\endgroup$ – David Waterworth Aug 7 '19 at 3:15
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Neurons does have a value, which we multiple with the weights to get the activation value for a given neuron.

We generally don't call it as weight of neuron but it means the same.

I'm updating my answer. Let's say we have a input x1,x2 . We want two hidden neuron h1 and h2.

We multiply x1 and x2 with w1 and w2 to get h1:

h1=w1*x1+w2*x2

Similar for h2.

h2=w3*x1+w4*x2

Now h1 and h2 are neurons having some values(that i called as their weight). We use those neurons for next layer value calculation using another set of weights. Hope that clear things out.

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  • $\begingroup$ Hi ashukid, you wrote: "Neurons does have a value, which we multiple with the weights to get the activation value for a given neuron." What we multiply with the weights are the input values, am I correct? Suppose a neuron has two incoming connections, with the weights w1 and w2. Suppose the input values from those two connections are v1 and v2. Then we multiply the input values with the weights, and add them together, to get the weighted sum. weighted sum = v1 x w1 + v2 x w2 Which value here is the "weight of the neuron"? $\endgroup$ – Curious Aug 6 '19 at 2:47

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