<div dir="ltr"><div><div><div><div><div><div>It is more likely that two candidates will have the same median score (an MJ tie situation) than having the same XA score.<br><br></div>Part of the reason is that the XA scores depend continuously on the distribution of ratings, while the median can be a discontinuous function of the distribution.<br><br></div>From another point of view, the graph of y = x is more likely to be perpendicular to the graph of the distribution function F(x) = Probability that on a random ballot candidate X will have a rating of at least x. An orthogonal intersection minimizes error due to random perturbations.<br><br></div>The graph of F stair steps down from some point on the y axis between (0, 0) and (0, 1) to some point on the vertical segment connecting (1, 0) to (1, 1). If the distribution is uniform, then the graph of F is the diagonal line segment connecting (0, 1) to (1, 0), perpendicular to the line y = x.<br><br></div>The median point (used in MJ and other Bucklin variants) is the intersection of the graph of F with the vertical line given by x = 1/2, cutting the square with diagonal corners at (0, 0) and (1, 1) in half.<br><br>The midrange Approval value is the intersection of the graph of F with the horizontal line y = 1/2.<br><br></div>The XA value is the intersection of the graph of F and line y = x, which bisects the right angle formed by x = 1/2 and y = 1/2 at the intersection (1/2, 1/2).<br><br></div><div>So XA can be thought of as a method half way between midrange Approval and score based Bucklin.<br><br></div><div>More later ...<br><br></div><div>Forest<br></div><div><br></div><br><div><div><div><div><div><div><div><div><div><div><div class="gmail_extra"><br><div class="gmail_quote"><blockquote style="margin:0px 0px 0px 0.8ex;border-left:1px solid rgb(204,204,204);padding-left:1ex" class="gmail_quote">
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From: Andy Jennings <<a href="mailto:elections@jenningsstory.com">elections@jenningsstory.com</a>><br>
To: Michael Ossipoff <<a href="mailto:email9648742@gmail.com">email9648742@gmail.com</a>><br>
Cc: "<a href="mailto:election-methods@electorama.com">election-methods@electorama.<wbr>com</a>"<br>
<<a href="mailto:election-methods@electorama.com">election-methods@electorama.<wbr>com</a>><br>
Subject: Re: [EM] XA<br>
<br>
On Mon, Oct 31, 2016 at 7:13 PM, Michael Ossipoff <<a href="mailto:email9648742@gmail.com">email9648742@gmail.com</a>><br>
wrote:<br>
<br>
> What makes XA do that more effectively than MJ? What's the main advantage<br>
> that distinguishes how XA does that from how MJ does it, or the results,<br>
> from the voters' strategic standpoint?<br>
<br>
<br>
Michael,<br>
<br>
As Rob said, the median is not terribly robust if the distribution of votes<br>
is two-peaked:<br>
<a target="_blank" rel="noreferrer" href="http://www.rangevoting.org/MedianVrange.html#twopeak">http://www.rangevoting.org/<wbr>MedianVrange.html#twopeak</a><br>
And I'm afraid many of our contentious political elections are two-peaked,<br>
at least in the current environment.<br>
<br>
With MJ, I like the fact that if the medians for all candidates will fall<br>
between B and D, then I can use the range outside that for honest<br>
expression. Yet in the back of my head, I know that if everyone tries to<br>
"use the range outside that for honest expression", then the medians won't<br>
be in that range anymore and it seems like a slippery slope to everyone<br>
using only the two extreme grades.<br>
<br>
XA solves this problem by making the more extreme grades more difficult to<br>
achieve. As Rob said, in the case where everyone grades at the extremes,<br>
the XA will match the mean.<br>
<br>
On the other hand, I admit that:<br>
1) with the median, 50% would have to give the top grade for a candidate to<br>
receive that grade. And 50% would have to give the bottom grade for a<br>
candidate to receive that grade. I consider both of these very unlikely.<br>
2) MJ is not just "the median", it has a tie-breaking scheme which<br>
mitigates this somewhat.<br>
<br>
~ Andy<br>
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