This example is from Narahari's textbook on Game theory.
I am not sure I understand the last part: the set of $n$ simultaneous equations has the unique solution: $$x_i^*=1/(1+n).$$
Basically, I don't see how that follows and how we actually derive the solution.
In reverse, the above means that $\sum_{j\neq i}x_j=\frac{n-1}{n+1}$ which again doesn't make a lot of sense to me.
I'd appreciate any help. Thank you.
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We want to solve $$2x_i^* + \sum_{j \ne i}x_j^* = 1$$
This can be written as a matrix system:
$$(ee^T+I)x=e$$
where $e$ is the all one vector.
By matrix determinant lemma, we have
$$\det(ee^T+I)=(1+e^Te)\det(I)=n+1\ne0$$ Hence the solution is unique.
Let's verify that $x=\frac{e}{n+1}$ is the solution.
$$(ee^T+I)\frac{e}{n+1}=\frac{1}{n+1}(e(e^Te)+e)=\frac{1}{n+1}(en+e)=e$$
Hence, it is the solution.
Here is a direct approach that avoids using vectors and matrices.
The best reply condition for agent $i$ is $$x_i = \frac{1-\sum_{j\ne i} x_j}{2} \qquad i = 1\, \ldots, n$$ Rewrite this as $$(*) \qquad x_i + \sum_{j=1}^n x_j = 1 \qquad i = 1\, \ldots, n$$ Please note that the sum extends over $j=1, \ldots, n$.
Add up the $n$ equations for $i = 1\, \ldots, n$ to get $$\sum_{j=1}^n x_j + n \left( \sum_{j=1}^n x_j \right) = n$$ Collecting, you find $$\sum_{j=1}^n x_j = \frac{n}{n+1}$$ Replace this in (*) to find $$x_i + \frac{n}{n+1} = 1$$ and the result follows.