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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!

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  • $\begingroup$ 1. Expand both sides. 2. Cancel the $(x^*)^2$ that appears on both sides. 3. Solve for $x^*$. 4. Simplify, remembering that $(x_L^2-x_R^2)=(x_L+x_R)(x_L-x_R)$. 5. Drop the "algebraic-geometry" tag! :) $\endgroup$
    – WillO
    Commented Jul 22, 2014 at 16:49
  • $\begingroup$ Thanks so much - that's really great! $\endgroup$ Commented Jul 22, 2014 at 17:20

1 Answer 1

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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!

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  • $\begingroup$ Many thanks for this! It's great. Just one thing: in the last step, when you divide $(V_R - V_R)$ by $2(X_R - X_L)$, why is the other term $(X_R^2 - X_L^2)$ not also divided by $(X_R - X_L)$? I understand the $\frac{1}{2}$ part but don't understand where the other bit goes! Apologies for confusion. Many thanks. $\endgroup$ Commented Jul 22, 2014 at 17:23
  • $\begingroup$ Set $x_R = a$ and $x_L =b$. Then you have $\frac{a^2 - b^2}{2(a-b)}$. But this is nothing more than $\frac{(a - b)(a+b)}{2(a-b)}$, and you simplify to get $\frac{(a+b)}{2}$. $\endgroup$
    – Kolmin
    Commented Jul 22, 2014 at 17:36
  • $\begingroup$ Ah I see. In which case, I think the last term in the last line ought to be $\frac{1}{2}(x_R + x_L)$ i.e., without the square term on $x_R$ and $x_L$? $\endgroup$ Commented Jul 22, 2014 at 22:35
  • $\begingroup$ Indeed, I corrected the typo. $\endgroup$
    – Kolmin
    Commented Jul 23, 2014 at 6:23
  • $\begingroup$ Great, many thanks again. $\endgroup$ Commented Jul 23, 2014 at 8:35

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