OK, so apologies for the easy question, but I'm new to this! This is somewhere between elementary algebra, and beginner's game theory. The question comes from a paper I read here (see p. 193): http://home.uchicago.edu/~sashwort/valence.pdf
The following is a utility function for an individual comparing two alternatives (call them L and R). The individual, $i$, prefers L to R when:
$V_L - (x^* - x_L)^2 > V_R - (x^* - x_R)^2$
So far so good. The difficulty I'm having is figuring out how we can get from here to a cutoff rule, such that $i$ will prefer L if and only if:
$x^* < \hat{x}(x_L,x_R,v_L,v_R)$
The paper says that this can be accomplished via "straightforward algebra" to reach:
$\hat{x}(x_L,x_R,v_L,v_R) = \frac{1}{2}(x_R + x_L) + \frac{V_L - V_R}{2(X_R-X_L)}$
Sadly, for me, this algebra ain't so straightforward. If anyone could walk me through the steps to reach this point (or point out how I should approach this) that'd be great. Of course, in the SO tradition, anything more general that can help make this question more applicable to others is also very welcome.
Thanks!
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EDIT: posted this q this morning, and have had some views but no nibbles... anyone got any suggestions? Thanks so much!
Just to avoid cumbersome effects, use $x_*$ instead of $x^*$.
Then we have (step by step)
$V_R - (x_* - x_R)^2 < V_L - (x_* - x_L)^2 $,
$V_R - x_{*}^2 - x_{R}^2 + 2x_* x_R < V_L - x_{*}^2 - x_{L}^2 + 2x_* x_L $,
$V_R - x_{R}^2 + 2x_* x_R < V_L - x_{L}^2 + 2x_* x_L $,
$2x_* x_R - 2x_* x_L + x_{L}^2 - x_{R}^2 < V_L - V_R $,
$2x_* ( x_R - x_L) < (V_L - V_R) + (x_{R}^2 - x_{L}^2)$,
$x_* < \frac{(V_L - V_R)}{2 ( x_R - x_L)} + \frac{1}{2}(x_{R} + x_{L})$.
As somebody suggested, drop the algebraic topology tag. ;)
I hope it helps!