Variance of the truncated normal distribution (truncated from below) increases in $sigma$?











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I'm wondering whether the variance of the truncated normal distribution increases in $sigma$ (which seems to hold numerically), where the untruncated normal distribution is $N(mu,sigma^2)$ and the truncated normal distribution is truncated from zero.



I found that there were some discussions on the relationship between the mean of the truncated normal distribution and $mu$ (Is the mean of the truncated normal distribution monotone in $mu$?), and the relationship between the mean of the truncated normal distribution and $sigma$ (Effect of variance on truncated normal) but couldn't find any discussion on the relationship between the variance (or standard deviation) of the truncated normal distribution and $sigma$.



The variance of the truncated normal distribution (truncated from below) is:



$Var(X|X>0)=sigma^2 left[1+frac{left(-frac{mu}{sigma}right) phileft(-frac{mu}{sigma}right)}{1-Phileft(-frac{mu}{sigma}right)} -left( frac{phileft(-frac{mu}{sigma}right)}{1-Phileft(-frac{mu}{sigma}right)} right)^2right]$



$Phi,phi$ are cdf and pdf of the standard normal distribution.



Is there any proved claim that $Var(X|X>0)$ increases in $sigma$? Or can we prove it? Any information or insight would greatly help.






statistics derivatives normal-distribution







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    up vote
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    down vote

    favorite












    I'm wondering whether the variance of the truncated normal distribution increases in $sigma$ (which seems to hold numerically), where the untruncated normal distribution is $N(mu,sigma^2)$ and the truncated normal distribution is truncated from zero.



    I found that there were some discussions on the relationship between the mean of the truncated normal distribution and $mu$ (Is the mean of the truncated normal distribution monotone in $mu$?), and the relationship between the mean of the truncated normal distribution and $sigma$ (Effect of variance on truncated normal) but couldn't find any discussion on the relationship between the variance (or standard deviation) of the truncated normal distribution and $sigma$.



    The variance of the truncated normal distribution (truncated from below) is:



    $Var(X|X>0)=sigma^2 left[1+frac{left(-frac{mu}{sigma}right) phileft(-frac{mu}{sigma}right)}{1-Phileft(-frac{mu}{sigma}right)} -left( frac{phileft(-frac{mu}{sigma}right)}{1-Phileft(-frac{mu}{sigma}right)} right)^2right]$



    $Phi,phi$ are cdf and pdf of the standard normal distribution.



    Is there any proved claim that $Var(X|X>0)$ increases in $sigma$? Or can we prove it? Any information or insight would greatly help.






    statistics derivatives normal-distribution







    share|cite|improve this question


























      up vote
      0
      down vote

      favorite









      up vote
      0
      down vote

      favorite











      I'm wondering whether the variance of the truncated normal distribution increases in $sigma$ (which seems to hold numerically), where the untruncated normal distribution is $N(mu,sigma^2)$ and the truncated normal distribution is truncated from zero.



      I found that there were some discussions on the relationship between the mean of the truncated normal distribution and $mu$ (Is the mean of the truncated normal distribution monotone in $mu$?), and the relationship between the mean of the truncated normal distribution and $sigma$ (Effect of variance on truncated normal) but couldn't find any discussion on the relationship between the variance (or standard deviation) of the truncated normal distribution and $sigma$.



      The variance of the truncated normal distribution (truncated from below) is:



      $Var(X|X>0)=sigma^2 left[1+frac{left(-frac{mu}{sigma}right) phileft(-frac{mu}{sigma}right)}{1-Phileft(-frac{mu}{sigma}right)} -left( frac{phileft(-frac{mu}{sigma}right)}{1-Phileft(-frac{mu}{sigma}right)} right)^2right]$



      $Phi,phi$ are cdf and pdf of the standard normal distribution.



      Is there any proved claim that $Var(X|X>0)$ increases in $sigma$? Or can we prove it? Any information or insight would greatly help.






      statistics derivatives normal-distribution







      share|cite|improve this question















      I'm wondering whether the variance of the truncated normal distribution increases in $sigma$ (which seems to hold numerically), where the untruncated normal distribution is $N(mu,sigma^2)$ and the truncated normal distribution is truncated from zero.



      I found that there were some discussions on the relationship between the mean of the truncated normal distribution and $mu$ (Is the mean of the truncated normal distribution monotone in $mu$?), and the relationship between the mean of the truncated normal distribution and $sigma$ (Effect of variance on truncated normal) but couldn't find any discussion on the relationship between the variance (or standard deviation) of the truncated normal distribution and $sigma$.



      The variance of the truncated normal distribution (truncated from below) is:



      $Var(X|X>0)=sigma^2 left[1+frac{left(-frac{mu}{sigma}right) phileft(-frac{mu}{sigma}right)}{1-Phileft(-frac{mu}{sigma}right)} -left( frac{phileft(-frac{mu}{sigma}right)}{1-Phileft(-frac{mu}{sigma}right)} right)^2right]$



      $Phi,phi$ are cdf and pdf of the standard normal distribution.



      Is there any proved claim that $Var(X|X>0)$ increases in $sigma$? Or can we prove it? Any information or insight would greatly help.






      statistics derivatives normal-distribution




      statistics derivatives normal-distribution






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      share|cite|improve this question













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      edited Nov 23 at 7:53

























      asked Nov 23 at 6:16









      sndwec

      12




      12






















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          A straightforward (computer) symbolic integration shows that the variance of the truncated distribution for arbitrary $mu$ gives:



          $$sigma left(-frac{2 sigma e^{-frac{mu ^2}{sigma ^2}}}{pi
          left(text{erf}left(frac{mu }{sqrt{2} sigma
          }right)+1right)^2}+frac{sqrt{frac{2}{pi }} mu e^{-frac{mu ^2}{2 sigma
          ^2}}}{text{erfc}left(frac{mu }{sqrt{2} sigma }right)-2}+sigma right).$$



          Here's a graph (for fixed $mu = 1$):



          enter image description here



          Here's the derivative of the variance with respect to $sigma$:



          $$-frac{4 sqrt{2} mu e^{-frac{3 mu ^2}{2 sigma ^2}}}{pi ^{3/2}
          left(text{erf}left(frac{mu }{sqrt{2} sigma }right)+1right)^3}-frac{2
          e^{-frac{mu ^2}{sigma ^2}} left(3 mu ^2+2 sigma ^2right)}{pi sigma
          left(text{erf}left(frac{mu }{sqrt{2} sigma
          }right)+1right)^2}+frac{sqrt{frac{2}{pi }} mu e^{-frac{mu ^2}{2 sigma ^2}}
          left(mu ^2+sigma ^2right)}{sigma ^2 left(text{erfc}left(frac{mu }{sqrt{2}
          sigma }right)-2right)}+2 sigma$$



          Here's a graph of the variance with respect to $sigma$ and $mu$: always monotonic:



          enter image description here






          share|cite|improve this answer



















          • 2




            You do not give any proofs whether the variance of the truncated normal distribution (truncated from below at zero) increases in $sigma$... The variance formula has been already given in this question.
            – sndwec
            Nov 23 at 7:57











          Your Answer





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          A straightforward (computer) symbolic integration shows that the variance of the truncated distribution for arbitrary $mu$ gives:



          $$sigma left(-frac{2 sigma e^{-frac{mu ^2}{sigma ^2}}}{pi
          left(text{erf}left(frac{mu }{sqrt{2} sigma
          }right)+1right)^2}+frac{sqrt{frac{2}{pi }} mu e^{-frac{mu ^2}{2 sigma
          ^2}}}{text{erfc}left(frac{mu }{sqrt{2} sigma }right)-2}+sigma right).$$



          Here's a graph (for fixed $mu = 1$):



          enter image description here



          Here's the derivative of the variance with respect to $sigma$:



          $$-frac{4 sqrt{2} mu e^{-frac{3 mu ^2}{2 sigma ^2}}}{pi ^{3/2}
          left(text{erf}left(frac{mu }{sqrt{2} sigma }right)+1right)^3}-frac{2
          e^{-frac{mu ^2}{sigma ^2}} left(3 mu ^2+2 sigma ^2right)}{pi sigma
          left(text{erf}left(frac{mu }{sqrt{2} sigma
          }right)+1right)^2}+frac{sqrt{frac{2}{pi }} mu e^{-frac{mu ^2}{2 sigma ^2}}
          left(mu ^2+sigma ^2right)}{sigma ^2 left(text{erfc}left(frac{mu }{sqrt{2}
          sigma }right)-2right)}+2 sigma$$



          Here's a graph of the variance with respect to $sigma$ and $mu$: always monotonic:



          enter image description here






          share|cite|improve this answer



















          • 2




            You do not give any proofs whether the variance of the truncated normal distribution (truncated from below at zero) increases in $sigma$... The variance formula has been already given in this question.
            – sndwec
            Nov 23 at 7:57















          up vote
          0
          down vote













          A straightforward (computer) symbolic integration shows that the variance of the truncated distribution for arbitrary $mu$ gives:



          $$sigma left(-frac{2 sigma e^{-frac{mu ^2}{sigma ^2}}}{pi
          left(text{erf}left(frac{mu }{sqrt{2} sigma
          }right)+1right)^2}+frac{sqrt{frac{2}{pi }} mu e^{-frac{mu ^2}{2 sigma
          ^2}}}{text{erfc}left(frac{mu }{sqrt{2} sigma }right)-2}+sigma right).$$



          Here's a graph (for fixed $mu = 1$):



          enter image description here



          Here's the derivative of the variance with respect to $sigma$:



          $$-frac{4 sqrt{2} mu e^{-frac{3 mu ^2}{2 sigma ^2}}}{pi ^{3/2}
          left(text{erf}left(frac{mu }{sqrt{2} sigma }right)+1right)^3}-frac{2
          e^{-frac{mu ^2}{sigma ^2}} left(3 mu ^2+2 sigma ^2right)}{pi sigma
          left(text{erf}left(frac{mu }{sqrt{2} sigma
          }right)+1right)^2}+frac{sqrt{frac{2}{pi }} mu e^{-frac{mu ^2}{2 sigma ^2}}
          left(mu ^2+sigma ^2right)}{sigma ^2 left(text{erfc}left(frac{mu }{sqrt{2}
          sigma }right)-2right)}+2 sigma$$



          Here's a graph of the variance with respect to $sigma$ and $mu$: always monotonic:



          enter image description here






          share|cite|improve this answer



















          • 2




            You do not give any proofs whether the variance of the truncated normal distribution (truncated from below at zero) increases in $sigma$... The variance formula has been already given in this question.
            – sndwec
            Nov 23 at 7:57













          up vote
          0
          down vote










          up vote
          0
          down vote









          A straightforward (computer) symbolic integration shows that the variance of the truncated distribution for arbitrary $mu$ gives:



          $$sigma left(-frac{2 sigma e^{-frac{mu ^2}{sigma ^2}}}{pi
          left(text{erf}left(frac{mu }{sqrt{2} sigma
          }right)+1right)^2}+frac{sqrt{frac{2}{pi }} mu e^{-frac{mu ^2}{2 sigma
          ^2}}}{text{erfc}left(frac{mu }{sqrt{2} sigma }right)-2}+sigma right).$$



          Here's a graph (for fixed $mu = 1$):



          enter image description here



          Here's the derivative of the variance with respect to $sigma$:



          $$-frac{4 sqrt{2} mu e^{-frac{3 mu ^2}{2 sigma ^2}}}{pi ^{3/2}
          left(text{erf}left(frac{mu }{sqrt{2} sigma }right)+1right)^3}-frac{2
          e^{-frac{mu ^2}{sigma ^2}} left(3 mu ^2+2 sigma ^2right)}{pi sigma
          left(text{erf}left(frac{mu }{sqrt{2} sigma
          }right)+1right)^2}+frac{sqrt{frac{2}{pi }} mu e^{-frac{mu ^2}{2 sigma ^2}}
          left(mu ^2+sigma ^2right)}{sigma ^2 left(text{erfc}left(frac{mu }{sqrt{2}
          sigma }right)-2right)}+2 sigma$$



          Here's a graph of the variance with respect to $sigma$ and $mu$: always monotonic:



          enter image description here






          share|cite|improve this answer














          A straightforward (computer) symbolic integration shows that the variance of the truncated distribution for arbitrary $mu$ gives:



          $$sigma left(-frac{2 sigma e^{-frac{mu ^2}{sigma ^2}}}{pi
          left(text{erf}left(frac{mu }{sqrt{2} sigma
          }right)+1right)^2}+frac{sqrt{frac{2}{pi }} mu e^{-frac{mu ^2}{2 sigma
          ^2}}}{text{erfc}left(frac{mu }{sqrt{2} sigma }right)-2}+sigma right).$$



          Here's a graph (for fixed $mu = 1$):



          enter image description here



          Here's the derivative of the variance with respect to $sigma$:



          $$-frac{4 sqrt{2} mu e^{-frac{3 mu ^2}{2 sigma ^2}}}{pi ^{3/2}
          left(text{erf}left(frac{mu }{sqrt{2} sigma }right)+1right)^3}-frac{2
          e^{-frac{mu ^2}{sigma ^2}} left(3 mu ^2+2 sigma ^2right)}{pi sigma
          left(text{erf}left(frac{mu }{sqrt{2} sigma
          }right)+1right)^2}+frac{sqrt{frac{2}{pi }} mu e^{-frac{mu ^2}{2 sigma ^2}}
          left(mu ^2+sigma ^2right)}{sigma ^2 left(text{erfc}left(frac{mu }{sqrt{2}
          sigma }right)-2right)}+2 sigma$$



          Here's a graph of the variance with respect to $sigma$ and $mu$: always monotonic:



          enter image description here







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Nov 23 at 8:44

























          answered Nov 23 at 6:41









          David G. Stork

          9,32721232




          9,32721232








          • 2




            You do not give any proofs whether the variance of the truncated normal distribution (truncated from below at zero) increases in $sigma$... The variance formula has been already given in this question.
            – sndwec
            Nov 23 at 7:57














          • 2




            You do not give any proofs whether the variance of the truncated normal distribution (truncated from below at zero) increases in $sigma$... The variance formula has been already given in this question.
            – sndwec
            Nov 23 at 7:57








          2




          2




          You do not give any proofs whether the variance of the truncated normal distribution (truncated from below at zero) increases in $sigma$... The variance formula has been already given in this question.
          – sndwec
          Nov 23 at 7:57




          You do not give any proofs whether the variance of the truncated normal distribution (truncated from below at zero) increases in $sigma$... The variance formula has been already given in this question.
          – sndwec
          Nov 23 at 7:57


















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