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I am reviewing a paper titled " Bayesian Sampling Approach to Decision Fusion" by Biao Chen and Pramod K Varshney. This paper uses an indicator function that I am not being able understand. The indicator function is used as follows

$I_{\mathbf{U}=g(\mathbf{X})}=1\hspace{5mm}if \hspace{5mm}\mathbf{U}=g(\mathbf{X}) \\ =0 \hspace{5mm} otherwise\\$

$\\where,\hspace{5mm} U_i=g(X_i)=0, \hspace{5mm} if \hspace{5mm} X_i \leq \tau \\ =1,\hspace{5mm} if \hspace{5mm} X_i \geq \tau$

where $X_i$ is observation at individual sensors, $U_i$ is the decision made at individual sensors and $\mathbf{X}$ is the vector of observation. How do you interpret indicator function in this case?The paper discussing about formulation is mentioned here

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  • $\begingroup$ You haven't said what $g$ and $\tau$ are, so there isn't much that can be done. Certainly the set contains $\bigcap_{i=1}^n \{ \omega \in \Omega : X_i(\omega) \leq \tau(\omega) \}$, but it is not clear what else it contains. $\endgroup$
    – Ian
    Jun 30, 2014 at 20:08
  • $\begingroup$ $\tau$ is a threshold and $g(X_i)$ is a deterministic mapping from observation $X_i$ to decision $U_i$ , ie. if $X_i\leq \tau$,$U_i=0$ and if $X_i\geq\tau$, $U_i=1$. So $g(X_i)$ is basically a thresholding function. Does that clarify ? $\endgroup$
    – Spandyie
    Jun 30, 2014 at 21:15
  • $\begingroup$ Not really; I do not see how $g(X)$ would not be the same as $U$ in all circumstances the way you have described everything. $\endgroup$
    – Ian
    Jun 30, 2014 at 21:22
  • $\begingroup$ According to the way it is defined in paper, I guess $g(X)$ is always equals to $U$. I have also attached the link to the paper for more reference. $\endgroup$
    – Spandyie
    Jun 30, 2014 at 21:25

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