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In the function below the boldfaced $\mathbf{x}$ is the vector of "all the other $x$'s besides $x_i$, evaluated at $x_i^*$." Does the following notation convey that? If not, or if there is a better way to notate this, I appreciate the advice.

$$f^* =f(x_i^* , \mathbf{x}) \:\:\:\:; \:\:\:\:\mathbf{x}=\mathbf{x}(x_i^*)$$

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I'm not sure I understand what you're trying to do. Are the other $x_j$ functions of $x_i$? –  Jonathan Christensen Dec 19 '12 at 18:51
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In statistics, we may use $\mathbf{x}_{(i)}$ to denote the vector with its $i$th entry removed. –  Nicolás Kim Dec 19 '12 at 18:51
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Do you mean there is some (fixed) index $i$ that you want to avoid, but want to range over all the others from say $1$ to $n$? If so, then say something like, "Let $i$ be fixed. Then $\mathbf{x}=\mathbf{x}(x_j)$, for all $j\not=i$, $j=1,\dots,n$." –  JohnD Dec 19 '12 at 19:05
    
Yes. But now how do I say that the elements of $\mathbf{x}$ are functions of, or evaluated at, $x_i^*$? Maybe this: $$\mathbf{x}^*=\mathbf{x}(x_j(x_i^*))$$ –  ben Dec 19 '12 at 19:11
    
Does $\mathbf{x}(x_j)$ mean that the $j$th component of $\mathbf{x}$ is $x_j$? Because this is easily confused with a function evaluation. –  JohnD Dec 19 '12 at 19:17

2 Answers 2

up vote 1 down vote accepted

Maybe one of these is what you want:

  1. Let $\mathbf{x}=[x_j]$, $j=1,\dots n$, and $j^*\in\{1,\dots,n\}$ be fixed. Set $\mathbf{x}^*=[x_j],\ j\not=j^*$.
  2. Let $\mathbf{x}=[x_j]$, $j=1,\dots n$, and $j^*\in\{1,\dots,n\}$ be fixed. Set $\mathbf{x}^*=[x_j(x_{j^*})],\ j\not=j^*$.
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It seems that for some reason you want to expose the $i$th coordinate variable. One way to do this is defining $${\bf x}_{(i)}':=(x_1,\ldots,x_{i-1},x_{i+1},x_n)$$ and introducing the special notation $(x_i,{\bf x}_{(i)}')$ for arbitrary vectors ${\bf x}\in{\mathbb R}^n$.

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