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    <title>MaplePrimes - answers and comments on Question, superimposed normalized distribution function / PDF algebra</title>
    <link>http://www.mapleprimes.com/questions/35544-Superimposed-Normalized-Distribution</link>
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    <pubDate>Tue, 09 Jun 2026 14:15:07 GMT</pubDate>
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    <description>The latest answers and comments added to the Question, superimposed normalized distribution function / PDF algebra</description>
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      <title>MaplePrimes - answers and comments on Question, superimposed normalized distribution function / PDF algebra</title>
      <link>http://www.mapleprimes.com/questions/35544-Superimposed-Normalized-Distribution</link>
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    <item>
      <title>Subgroup distribution</title>
      <link>http://www.mapleprimes.com/questions/35544-Superimposed-Normalized-Distribution?ref=Feed:MaplePrimes:superimposed normalized distribution function / PDF algebra:Comments#answer44234</link>
      <itunes:summary>&lt;p&gt;Suppose the probability density for trait A in group B (i.e. the conditional density given that the individual is in group B) is f_B(x),&amp;nbsp; the probability density in the complement of group B is f_C(x), and the probability of being in group B is p, Then the overall probability density for trait A is f_all(x) = p*f_B(x) + (1-p)*f_C(x).&amp;nbsp; Therefore f_C(x) = (f_all(x) - p*f_B(x))/(1-p).&amp;nbsp;&lt;/p&gt;
&lt;p&gt;Caution: if two of these densities are normal, the third will _not_ be normal, except in trivial cases.&lt;/p&gt;
&lt;p&gt;&amp;nbsp;&lt;/p&gt;</itunes:summary>
      <description>&lt;p&gt;Suppose the probability density for trait A in group B (i.e. the conditional density given that the individual is in group B) is f_B(x),&amp;nbsp; the probability density in the complement of group B is f_C(x), and the probability of being in group B is p, Then the overall probability density for trait A is f_all(x) = p*f_B(x) + (1-p)*f_C(x).&amp;nbsp; Therefore f_C(x) = (f_all(x) - p*f_B(x))/(1-p).&amp;nbsp;&lt;/p&gt;
&lt;p&gt;Caution: if two of these densities are normal, the third will _not_ be normal, except in trivial cases.&lt;/p&gt;
&lt;p&gt;&amp;nbsp;&lt;/p&gt;</description>
      <guid>44234</guid>
      <pubDate>Thu, 11 Mar 2010 06:40:00 Z</pubDate>
      <itunes:author>Robert Israel</itunes:author>
      <author>Robert Israel</author>
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    <item>
      <title>Thank you very much for this</title>
      <link>http://www.mapleprimes.com/questions/35544-Superimposed-Normalized-Distribution?ref=Feed:MaplePrimes:superimposed normalized distribution function / PDF algebra:Comments#answer44235</link>
      <itunes:summary>&lt;p&gt;Thank you very much for this reply. Your theoretical considerations are very helpful. But, my little program below produces an unrealistic and unpredictable result, as the resulting probability function becomes negative. The reason is that depending on the difference of these two distribution functions there is a restriction on probability p. Do you know something about these restrictions and how they can be calculated?&lt;/p&gt;
&lt;p&gt;K:=200;&lt;br /&gt;
M:=RandomVariable(Normal(80,15));&lt;br /&gt;
N:=RandomVariable(Normal(100,15));&lt;br /&gt;
p:=0.3;&lt;br /&gt;
f_all:=t-&amp;gt;PDF(N,t);&lt;br /&gt;
f_B:=t-&amp;gt;PDF(M,t);&lt;br /&gt;
f_C:=t-&amp;gt;((f_all(t)-p*f_B(t))/(1-p));&lt;br /&gt;
Apx:=plot(f_all(x),x=0..K);&lt;br /&gt;
Bpx:=plot(f_B(x),x=0..K);&lt;br /&gt;
Cpx:=plot(f_C(x),x=0..K,color=blue);&lt;br /&gt;
plots[display](Apx,Bpx,Cpx);&lt;br /&gt;
&lt;br /&gt;
Thanks Peter&lt;/p&gt;</itunes:summary>
      <description>&lt;p&gt;Thank you very much for this reply. Your theoretical considerations are very helpful. But, my little program below produces an unrealistic and unpredictable result, as the resulting probability function becomes negative. The reason is that depending on the difference of these two distribution functions there is a restriction on probability p. Do you know something about these restrictions and how they can be calculated?&lt;/p&gt;
&lt;p&gt;K:=200;&lt;br /&gt;
M:=RandomVariable(Normal(80,15));&lt;br /&gt;
N:=RandomVariable(Normal(100,15));&lt;br /&gt;
p:=0.3;&lt;br /&gt;
f_all:=t-&amp;gt;PDF(N,t);&lt;br /&gt;
f_B:=t-&amp;gt;PDF(M,t);&lt;br /&gt;
f_C:=t-&amp;gt;((f_all(t)-p*f_B(t))/(1-p));&lt;br /&gt;
Apx:=plot(f_all(x),x=0..K);&lt;br /&gt;
Bpx:=plot(f_B(x),x=0..K);&lt;br /&gt;
Cpx:=plot(f_C(x),x=0..K,color=blue);&lt;br /&gt;
plots[display](Apx,Bpx,Cpx);&lt;br /&gt;
&lt;br /&gt;
Thanks Peter&lt;/p&gt;</description>
      <guid>44235</guid>
      <pubDate>Thu, 11 Mar 2010 16:46:24 Z</pubDate>
      <itunes:author>Peter Mond</itunes:author>
      <author>Peter Mond</author>
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    <item>
      <title>Thank you very much for your help</title>
      <link>http://www.mapleprimes.com/questions/35544-Superimposed-Normalized-Distribution?ref=Feed:MaplePrimes:superimposed normalized distribution function / PDF algebra:Comments#answer44237</link>
      <itunes:summary>&lt;p&gt;All questions solved. Your cantribution was of great help.&lt;/p&gt;
&lt;p&gt;Peter&lt;/p&gt;</itunes:summary>
      <description>&lt;p&gt;All questions solved. Your cantribution was of great help.&lt;/p&gt;
&lt;p&gt;Peter&lt;/p&gt;</description>
      <guid>44237</guid>
      <pubDate>Fri, 12 Mar 2010 22:38:05 Z</pubDate>
      <itunes:author>Peter Mond</itunes:author>
      <author>Peter Mond</author>
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    <item>
      <title>Restriction</title>
      <link>http://www.mapleprimes.com/questions/35544-Superimposed-Normalized-Distribution?ref=Feed:MaplePrimes:superimposed normalized distribution function / PDF algebra:Comments#comment44236</link>
      <itunes:summary>&lt;p&gt;The requirement that f_C(t) &amp;gt;= 0 means that f_all(t) &amp;gt;= p*f_B(t) for all t, i.e. that p is bounded by the infimum of f_all(t)/f_B(t).&amp;nbsp; If, as in your case, f_all and f_B are normal with the same variance sigma^2 but different means mu_all and mu_B, that infimum is actually 0, because f_all(t)/f_B(t) = exp(((t - mu_B)^2 - (t - mu_all)^2)/(2*sigma^2))&lt;br /&gt;
= const * exp(t*(mu_all - mu_B)/sigma^2) -&amp;gt; 0 as t -&amp;gt; +infinity or -infinity, depending on the sign of mu_all - mu_B.&amp;nbsp; &lt;/p&gt;
&lt;p&gt;&amp;nbsp;&lt;/p&gt;
&lt;p&gt;&amp;nbsp;&lt;/p&gt;</itunes:summary>
      <description>&lt;p&gt;The requirement that f_C(t) &amp;gt;= 0 means that f_all(t) &amp;gt;= p*f_B(t) for all t, i.e. that p is bounded by the infimum of f_all(t)/f_B(t).&amp;nbsp; If, as in your case, f_all and f_B are normal with the same variance sigma^2 but different means mu_all and mu_B, that infimum is actually 0, because f_all(t)/f_B(t) = exp(((t - mu_B)^2 - (t - mu_all)^2)/(2*sigma^2))&lt;br /&gt;
= const * exp(t*(mu_all - mu_B)/sigma^2) -&amp;gt; 0 as t -&amp;gt; +infinity or -infinity, depending on the sign of mu_all - mu_B.&amp;nbsp; &lt;/p&gt;
&lt;p&gt;&amp;nbsp;&lt;/p&gt;
&lt;p&gt;&amp;nbsp;&lt;/p&gt;</description>
      <guid>44236</guid>
      <pubDate>Thu, 11 Mar 2010 22:00:23 Z</pubDate>
      <itunes:author>Robert Israel</itunes:author>
      <author>Robert Israel</author>
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