I have not seen a neuron that uses both a bias and a threshold. Why is this?

  • $\begingroup$ I'm unsure about what this question is asking. Could you clarify and provide some context? $\endgroup$ Commented Apr 6, 2019 at 4:41
  • $\begingroup$ i have seen that if we set bias in activation function then we don't have to set threshold value, one of them will be use in activation function. is this necessary in every case or there is any specific reason for that? $\endgroup$
    – hina munir
    Commented Apr 6, 2019 at 6:26
  • $\begingroup$ I did my best interpreting and answering your question. Please let me know if I understood you correctly. $\endgroup$ Commented Apr 6, 2019 at 7:23

1 Answer 1


I assume you're talking about a perceptron threshold function. One definition of it with an explicit threshold is $$f(\textbf{x})= \begin{cases} 1& \text{if } \textbf{w}\cdot\textbf{x} > t\\ 0& \text{otherwise} \end{cases}.$$

Another form with a bias is $$f(\textbf{x})= \begin{cases} 1& \text{if } \textbf{w}\cdot\textbf{x} + b > 0\\ 0& \text{otherwise} \end{cases}.$$

But these forms are of course equivalent if you set $b=-t$.

There's nothing stopping you from using a perceptron definition with both a bias and a threshold: $$f(\textbf{x})= \begin{cases} 1& \text{if } \textbf{w}\cdot\textbf{x} +b > t\\ 0& \text{otherwise} \end{cases}.$$

But this is also equivalent to the other two forms. We can rewrite this as $$f(\textbf{x})= \begin{cases} 1& \text{if } \textbf{w}\cdot\textbf{x} > t'\\ 0& \text{otherwise} \end{cases}$$ where $t'=t-b$ is the new threshold. Or, we could rewrite it as $$f(\textbf{x})= \begin{cases} 1& \text{if } \textbf{w}\cdot\textbf{x} + a> 0\\ 0& \text{otherwise} \end{cases}$$ where $a=b-t$ is the new bias.

You never see definitions of this function with both a threshold and a bias because it has simpler forms.


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