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	<title>Comments on: Score with scoring rules</title>
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	<link>http://www.decisionsciencenews.com/2009/07/21/score-with-scoring-rules/</link>
	<description>A website about decision research in Marketing, Psychology, Economics, Behavioral Economics, Finance, Medicine, Law, Management, Public Policy, Statistics, Computer Science.</description>
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		<title>By: Michael Smithson</title>
		<link>http://www.decisionsciencenews.com/2009/07/21/score-with-scoring-rules/comment-page-1/#comment-2365</link>
		<dc:creator>Michael Smithson</dc:creator>
		<pubDate>Thu, 22 Oct 2009 22:56:03 +0000</pubDate>
		<guid isPermaLink="false">http://www.decisionsciencenews.com/?p=963#comment-2365</guid>
		<description>Hmm, the clause &quot;a negative score to a probability of 1/n&quot; should have read &quot;a negative score to a probability less than 1/n.&quot;</description>
		<content:encoded><![CDATA[<p>Hmm, the clause &#8220;a negative score to a probability of 1/n&#8221; should have read &#8220;a negative score to a probability less than 1/n.&#8221;</p>
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	</item>
	<item>
		<title>By: Michael Smithson</title>
		<link>http://www.decisionsciencenews.com/2009/07/21/score-with-scoring-rules/comment-page-1/#comment-2364</link>
		<dc:creator>Michael Smithson</dc:creator>
		<pubDate>Thu, 22 Oct 2009 22:53:39 +0000</pubDate>
		<guid isPermaLink="false">http://www.decisionsciencenews.com/?p=963#comment-2364</guid>
		<description>None of the 3 scoring rules take into account the number of alternatives.  A rule that does would assign a score of 0 to a probability of 1/n, where n is the number of alternatives, a negative score to a probability  1/n.  Here&#039;s an example:
f(r_i) = min[(r_i - 1/n)/(1-1/n), (r_i-1/n)/(1/n)],
where r_i is the probability assigned to the ith alternative. We have f(0) = -1, f(1) = 1, and f(1/n) = 0. And of course, the terms inside the min could be raised to a power other than 1.</description>
		<content:encoded><![CDATA[<p>None of the 3 scoring rules take into account the number of alternatives.  A rule that does would assign a score of 0 to a probability of 1/n, where n is the number of alternatives, a negative score to a probability  1/n.  Here&#8217;s an example:<br />
f(r_i) = min[(r_i - 1/n)/(1-1/n), (r_i-1/n)/(1/n)],<br />
where r_i is the probability assigned to the ith alternative. We have f(0) = -1, f(1) = 1, and f(1/n) = 0. And of course, the terms inside the min could be raised to a power other than 1.</p>
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	<item>
		<title>By: dan</title>
		<link>http://www.decisionsciencenews.com/2009/07/21/score-with-scoring-rules/comment-page-1/#comment-1922</link>
		<dc:creator>dan</dc:creator>
		<pubDate>Wed, 22 Jul 2009 12:46:06 +0000</pubDate>
		<guid isPermaLink="false">http://www.decisionsciencenews.com/?p=963#comment-1922</guid>
		<description>I&#039;m going to try something new here. I love R, but sometimes my code isn&#039;t as beautiful as it should be. From now on, I&#039;m going to paste the code behind my simulations so that if anyone sees room to improve it, they can just chime in.

#This first part is based Stephanie Kovalchik&#039;s beta.prior code http://skoval.bol.ucla.edu/beta.prior.R 

 beta.given.region.mode &lt;- function(lowerq,upperq,region,mode,min=2,max=10000,drawplot=FALSE)
 {
     mode.check &lt;- function(lowerq,upperq,mode) {
     if(!(mode &gt; lowerq &amp; mode&lt;upperq )) {
     stop(&quot;Mode must be between lower and upperq&quot;)
     }
     }
     
     region.check=function(region) {
     if(!(region &gt;= 0 &amp; region&lt;=1 )) {
     stop(&quot;Region must be between 0 and 1&quot;)
     }
     }

     abcheck &lt;- function(a,b) {
     if(a&lt;1 &#124;&#124; b&lt;1 ) {
     stop(&quot;a or b was less than 1&quot;)
     }
     }

    mode.check(lowerq,upperq,mode)
    region.check(region)

    #This function will be zeroed when the difference in the pbetas equals the region size
    f &lt;- function(n) {
    a &lt;- 1 + mode*(-2+n)
    b &lt;- -1-mode*(-2+n)+n
    pbeta(upperq,a,b)-pbeta(lowerq,a,b)-region
    }
    
    #This uniroot thing calls f with every n from 2 to 1000 until it finds the n that causes the inner region to hit a certain number.
    n &lt;- uniroot(f,interval=c(min,max))$root
    a &lt;- 1 + mode*(-2+n)
    b &lt;- -1-mode*(-2+n)+n

    abcheck(a,b)
    theta &lt;- c(a,b)
    cat(a,b,&quot;\n&quot;)
    return(theta)
 }

####This is the part that needs work, especially the super-lame pause function!

pause_x = function (x) {
sofar=0
while (sofar&lt;x) {
sofar=sofar+system.time(for (i in 1:100000) {})[3]
}
}

for (i in seq(.29,.71,.01))
{
theta=beta.given.region.mode(lowerq=.2,upperq=.8,region=.99,mode=i,min=2,max=10000,drawplot=FALSE)

numslots=20
breaks=seq(0,1,1/numslots)
cumVals=pbeta(breaks,theta[1],theta[2])
barHeights=cumVals[1:numslots+1]-cumVals[1:numslots]
par(mfrow=c(4,1),mai=c(.3,.3,.2,.1))
barplot(barHeights,ylim=c(0,.5),main=&quot;User&#039;s Distribution of Belief&quot;,col=&quot;blue&quot;)
denom=barHeights%*%barHeights
barplot(barHeights/denom^.5,ylim=c(0,1),main=&quot;Payoff with Spherical Scoring Rule&quot;,col=&quot;green&quot;)
barplot(2*barHeights-denom,ylim=c(-.5,.6),main=&quot;Payoff with Quadratic Scoring Rule&quot;,col=&quot;green&quot;)
barplot(log(barHeights),ylim=c(-12,1),main=&quot;Payoff with Log Scoring Rule&quot;,xpd=FALSE,col=&quot;green&quot;)
pause_x(.1)
}</description>
		<content:encoded><![CDATA[<p>I&#8217;m going to try something new here. I love R, but sometimes my code isn&#8217;t as beautiful as it should be. From now on, I&#8217;m going to paste the code behind my simulations so that if anyone sees room to improve it, they can just chime in.</p>
<p>#This first part is based Stephanie Kovalchik&#8217;s beta.prior code <a href="http://skoval.bol.ucla.edu/beta.prior.R" rel="nofollow">http://skoval.bol.ucla.edu/beta.prior.R</a> </p>
<p> beta.given.region.mode < - function(lowerq,upperq,region,mode,min=2,max=10000,drawplot=FALSE)<br />
 {<br />
     mode.check <- function(lowerq,upperq,mode) {<br />
     if(!(mode > lowerq &#038; mode<upperq )) {<br />
     stop("Mode must be between lower and upperq")<br />
     }<br />
     }</p>
<p>     region.check=function(region) {<br />
     if(!(region >= 0 &#038; region<=1 )) {<br />
     stop(&#8220;Region must be between 0 and 1&#8243;)<br />
     }<br />
     }</p>
<p>     abcheck <- function(a,b) {<br />
     if(a&lt;1 || b&lt;1 ) {<br />
     stop(&#8220;a or b was less than 1&#8243;)<br />
     }<br />
     }</p>
<p>    mode.check(lowerq,upperq,mode)<br />
    region.check(region)</p>
<p>    #This function will be zeroed when the difference in the pbetas equals the region size<br />
    f <- function(n) {<br />
    a <- 1 + mode*(-2+n)<br />
    b <- -1-mode*(-2+n)+n<br />
    pbeta(upperq,a,b)-pbeta(lowerq,a,b)-region<br />
    }</p>
<p>    #This uniroot thing calls f with every n from 2 to 1000 until it finds the n that causes the inner region to hit a certain number.<br />
    n <- uniroot(f,interval=c(min,max))$root<br />
    a <- 1 + mode*(-2+n)<br />
    b <- -1-mode*(-2+n)+n</p>
<p>    abcheck(a,b)<br />
    theta <- c(a,b)<br />
    cat(a,b,&#8221;\n&#8221;)<br />
    return(theta)<br />
 }</p>
<p>####This is the part that needs work, especially the super-lame pause function!</p>
<p>pause_x = function (x) {<br />
sofar=0<br />
while (sofar<x) {<br />
sofar=sofar+system.time(for (i in 1:100000) {})[3]<br />
}<br />
}</p>
<p>for (i in seq(.29,.71,.01))<br />
{<br />
theta=beta.given.region.mode(lowerq=.2,upperq=.8,region=.99,mode=i,min=2,max=10000,drawplot=FALSE)</p>
<p>numslots=20<br />
breaks=seq(0,1,1/numslots)<br />
cumVals=pbeta(breaks,theta[1],theta[2])<br />
barHeights=cumVals[1:numslots+1]-cumVals[1:numslots]<br />
par(mfrow=c(4,1),mai=c(.3,.3,.2,.1))<br />
barplot(barHeights,ylim=c(0,.5),main=&#8221;User&#8217;s Distribution of Belief&#8221;,col=&#8221;blue&#8221;)<br />
denom=barHeights%*%barHeights<br />
barplot(barHeights/denom^.5,ylim=c(0,1),main=&#8221;Payoff with Spherical Scoring Rule&#8221;,col=&#8221;green&#8221;)<br />
barplot(2*barHeights-denom,ylim=c(-.5,.6),main=&#8221;Payoff with Quadratic Scoring Rule&#8221;,col=&#8221;green&#8221;)<br />
barplot(log(barHeights),ylim=c(-12,1),main=&#8221;Payoff with Log Scoring Rule&#8221;,xpd=FALSE,col=&#8221;green&#8221;)<br />
pause_x(.1)<br />
}</upperq></p>
]]></content:encoded>
	</item>
	<item>
		<title>By: dan</title>
		<link>http://www.decisionsciencenews.com/2009/07/21/score-with-scoring-rules/comment-page-1/#comment-1921</link>
		<dc:creator>dan</dc:creator>
		<pubDate>Wed, 22 Jul 2009 12:29:51 +0000</pubDate>
		<guid isPermaLink="false">http://www.decisionsciencenews.com/?p=963#comment-1921</guid>
		<description>Very interesting post, Alan. 

One thing I think about is the following. Suppose research suggests (which it kind of does) that when you ask people for a 95% confidence interval, you get a very reasonable 50% confidence interval. If this is a pretty consistent finding a researcher might finally admit, ok, though it&#039;s not the statistical definition, for whatever reason 95% confidence interval is &quot;human&quot; for 50% confidence interval. In this case, when eliciting a 95% confidence interval, perhaps the right thing to do in consulting settings is treat it as a 50% CI. I reckon that in the simulation above, if I treated the endpoints of the 95% CI as the endpoints of a 50% CI, there&#039;d be very few cases in which 0% probability would be assigned to a cell.

On the other hand, to quote you, this is &quot;teaching&quot; them to not learn the true meaning of a 95% CI, which can&#039;t be a good thing.</description>
		<content:encoded><![CDATA[<p>Very interesting post, Alan. </p>
<p>One thing I think about is the following. Suppose research suggests (which it kind of does) that when you ask people for a 95% confidence interval, you get a very reasonable 50% confidence interval. If this is a pretty consistent finding a researcher might finally admit, ok, though it&#8217;s not the statistical definition, for whatever reason 95% confidence interval is &#8220;human&#8221; for 50% confidence interval. In this case, when eliciting a 95% confidence interval, perhaps the right thing to do in consulting settings is treat it as a 50% CI. I reckon that in the simulation above, if I treated the endpoints of the 95% CI as the endpoints of a 50% CI, there&#8217;d be very few cases in which 0% probability would be assigned to a cell.</p>
<p>On the other hand, to quote you, this is &#8220;teaching&#8221; them to not learn the true meaning of a 95% CI, which can&#8217;t be a good thing.</p>
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	<item>
		<title>By: Alan Schwartz</title>
		<link>http://www.decisionsciencenews.com/2009/07/21/score-with-scoring-rules/comment-page-1/#comment-1918</link>
		<dc:creator>Alan Schwartz</dc:creator>
		<pubDate>Tue, 21 Jul 2009 21:52:52 +0000</pubDate>
		<guid isPermaLink="false">http://www.decisionsciencenews.com/?p=963#comment-1918</guid>
		<description>It&#039;s obviously not mathematically perfect, but you could say that in this case, the study is over and they receive the lowest possible payout or highest possible loss. That&#039;s as close to budget ruin as you can come.

Of course, what you&#039;re teaching them is to never assign a 0 probability to anything; the cost of assigning a tiny probability &gt; 0, even if they truly believe the probability is 0, is like a minuscule insurance payment, and if people are at all risk averse, you&#039;ll have convinced them to misstate their true beliefs, albeit in a tiny way - unless, of course, you believe that what constitutes a &quot;true belief&quot; is your willingness to risk everything for it - a logically consistent position, but a psychological implausible one.</description>
		<content:encoded><![CDATA[<p>It&#8217;s obviously not mathematically perfect, but you could say that in this case, the study is over and they receive the lowest possible payout or highest possible loss. That&#8217;s as close to budget ruin as you can come.</p>
<p>Of course, what you&#8217;re teaching them is to never assign a 0 probability to anything; the cost of assigning a tiny probability &gt; 0, even if they truly believe the probability is 0, is like a minuscule insurance payment, and if people are at all risk averse, you&#8217;ll have convinced them to misstate their true beliefs, albeit in a tiny way &#8211; unless, of course, you believe that what constitutes a &#8220;true belief&#8221; is your willingness to risk everything for it &#8211; a logically consistent position, but a psychological implausible one.</p>
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