How does entropy depend on location and scale?Entropy of Cauchy (Lorentz) DistributionDeriving Negentropy. Getting stuck[Revised]Proving the expected bolddensity of being the Nth order statistics is decreasing in sample sizeRelationship between least-squares regression and information theoryAsymptotic relation for a class of probability distribution functionsLet $X_(1)leq X_(2)$ be the order statistics. Evaluate $operatornameVar(X_(j))$, $operatornameCov(X_(1),X_(2))$Why we multiply the density instead of taking the integral?Is continuous mutual information the correct analogue of the discrete version?What does the integral of a function times a function of a random variable represent, conceptually?If $Xsim operatornamelognormal$ then $Y:=(X-dmid xgeq d)$ has approximately a Generalized Pareto distributionSequence of shifted exponential distributions has uniform conditionals?

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How does entropy depend on location and scale?


Entropy of Cauchy (Lorentz) DistributionDeriving Negentropy. Getting stuck[Revised]Proving the expected bolddensity of being the Nth order statistics is decreasing in sample sizeRelationship between least-squares regression and information theoryAsymptotic relation for a class of probability distribution functionsLet $X_(1)leq X_(2)$ be the order statistics. Evaluate $operatornameVar(X_(j))$, $operatornameCov(X_(1),X_(2))$Why we multiply the density instead of taking the integral?Is continuous mutual information the correct analogue of the discrete version?What does the integral of a function times a function of a random variable represent, conceptually?If $Xsim operatornamelognormal$ then $Y:=(X-dmid xgeq d)$ has approximately a Generalized Pareto distributionSequence of shifted exponential distributions has uniform conditionals?













4












$begingroup$


The entropy of a continuous distribution with density function $f$ is defined to be the negative of the expectation of $log(f),$ and therefore equals



$$H_f = -int_-infty^infty log(f(x)) f(x)mathrmdx.$$



We also say that any random variable $X$ whose distribution has density $f$ has entropy $H_f.$ (This integral is well-defined even when $f$ has zeros, because $log(f(x))f(x)$ can be taken to equal zero at such values.)



When $X$ and $Y$ are random variables for which $Y = X+mu$ ($mu$ is a constant), $Y$ is said to be a version of $X$ shifted by $mu.$ Similarly, when $Y = Xsigma$ ($sigma$ is a positive constant), $Y$ is said to be a version of $X$ scaled by $sigma.$ Combining a scale with a shift gives $Y=Xsigma + mu.$



These relations occur frequently. For instance, changing the units of measurement of $X$ shifts and scales it.



How is the entropy of $Y = Xsigma + mu$ related to that of $X?$










share|cite|improve this question









$endgroup$
















    4












    $begingroup$


    The entropy of a continuous distribution with density function $f$ is defined to be the negative of the expectation of $log(f),$ and therefore equals



    $$H_f = -int_-infty^infty log(f(x)) f(x)mathrmdx.$$



    We also say that any random variable $X$ whose distribution has density $f$ has entropy $H_f.$ (This integral is well-defined even when $f$ has zeros, because $log(f(x))f(x)$ can be taken to equal zero at such values.)



    When $X$ and $Y$ are random variables for which $Y = X+mu$ ($mu$ is a constant), $Y$ is said to be a version of $X$ shifted by $mu.$ Similarly, when $Y = Xsigma$ ($sigma$ is a positive constant), $Y$ is said to be a version of $X$ scaled by $sigma.$ Combining a scale with a shift gives $Y=Xsigma + mu.$



    These relations occur frequently. For instance, changing the units of measurement of $X$ shifts and scales it.



    How is the entropy of $Y = Xsigma + mu$ related to that of $X?$










    share|cite|improve this question









    $endgroup$














      4












      4








      4


      1



      $begingroup$


      The entropy of a continuous distribution with density function $f$ is defined to be the negative of the expectation of $log(f),$ and therefore equals



      $$H_f = -int_-infty^infty log(f(x)) f(x)mathrmdx.$$



      We also say that any random variable $X$ whose distribution has density $f$ has entropy $H_f.$ (This integral is well-defined even when $f$ has zeros, because $log(f(x))f(x)$ can be taken to equal zero at such values.)



      When $X$ and $Y$ are random variables for which $Y = X+mu$ ($mu$ is a constant), $Y$ is said to be a version of $X$ shifted by $mu.$ Similarly, when $Y = Xsigma$ ($sigma$ is a positive constant), $Y$ is said to be a version of $X$ scaled by $sigma.$ Combining a scale with a shift gives $Y=Xsigma + mu.$



      These relations occur frequently. For instance, changing the units of measurement of $X$ shifts and scales it.



      How is the entropy of $Y = Xsigma + mu$ related to that of $X?$










      share|cite|improve this question









      $endgroup$




      The entropy of a continuous distribution with density function $f$ is defined to be the negative of the expectation of $log(f),$ and therefore equals



      $$H_f = -int_-infty^infty log(f(x)) f(x)mathrmdx.$$



      We also say that any random variable $X$ whose distribution has density $f$ has entropy $H_f.$ (This integral is well-defined even when $f$ has zeros, because $log(f(x))f(x)$ can be taken to equal zero at such values.)



      When $X$ and $Y$ are random variables for which $Y = X+mu$ ($mu$ is a constant), $Y$ is said to be a version of $X$ shifted by $mu.$ Similarly, when $Y = Xsigma$ ($sigma$ is a positive constant), $Y$ is said to be a version of $X$ scaled by $sigma.$ Combining a scale with a shift gives $Y=Xsigma + mu.$



      These relations occur frequently. For instance, changing the units of measurement of $X$ shifts and scales it.



      How is the entropy of $Y = Xsigma + mu$ related to that of $X?$







      distributions data-transformation entropy






      share|cite|improve this question













      share|cite|improve this question











      share|cite|improve this question




      share|cite|improve this question










      asked 9 hours ago









      whuberwhuber

      212k34 gold badges465 silver badges847 bronze badges




      212k34 gold badges465 silver badges847 bronze badges




















          1 Answer
          1






          active

          oldest

          votes


















          4












          $begingroup$

          Since the probability element of $X$ is $f(x)mathrmdx,$ the change of variable $y = xsigma + mu$ is equivalent to $x = (y-mu)/sigma,$ whence from



          $$f(x)mathrmdx = fleft(fracy-musigmaright)mathrmdleft(fracy-musigmaright) = frac1sigma fleft(fracy-musigmaright) mathrmdy$$



          it follows that the density of $Y$ is



          $$f_Y(y) = frac1sigmafleft(fracy-musigmaright).$$



          Consequently the entropy of $Y$ is



          $$H(Y) = -int_-infty^infty logleft(frac1sigmafleft(fracy-musigmaright)right) frac1sigmafleft(fracy-musigmaright) mathrmdy$$



          which, upon changing the variable back to $x = (y-mu)/sigma,$ produces



          $$eqalign
          H(Y) &= -int_-infty^infty logleft(frac1sigmafleft(xright)right) fleft(xright) mathrmdx \
          &= -int_-infty^infty left(logleft(frac1sigmaright) + logleft(fleft(xright)right)right) fleft(xright) mathrmdx \
          &= logleft(sigmaright) int_-infty^infty f(x) mathrmdx -int_-infty^infty logleft(fleft(xright)right) fleft(xright) mathrmdx \
          &= log(sigma) + H_f.
          $$



          These calculations used basic properties of the logarithm, the linearity of integration, and that fact that $f(x)mathrmdx$ integrates to unity (the Law of Total Probability).



          The conclusion is




          The entropy of $Y = Xsigma + mu$ is the entropy of $X$ plus $log(sigma).$




          In words, shifting a random variable does not change its entropy (we may think of the entropy as depending on the values of the probability density, but not on where those values occur), while scaling a variable (which, for $sigma ge 1$ "stretches" or "smears" it out) increases its entropy by $log(sigma).$ This supports the intuition that high-entropy distributions are "more spread out" than low-entropy distributions.




          As a consequence of this result, we are free to choose convenient values of $mu$ and $sigma$ when computing the entropy of any distribution. For example, the entropy of a Normal$(mu,sigma)$ distribution can be found by setting $mu=0$ and $sigma=1.$ The logarithm of the density in this case is



          $$log(f(x)) = -frac12log(2pi) - x^2/2,$$



          whence



          $$H = -E[-frac12log(2pi) - X^2/2] = frac12log(2pi) + frac12.$$



          Consequently the entropy of a Normal$(mu,sigma)$ distribution is obtained simply by adding $logsigma$ to this result, giving



          $$H = frac12log(2pi) + frac12 + log(sigma) = frac12log(2pi,e,sigma^2)$$



          as reported by Wikipedia.






          share|cite|improve this answer









          $endgroup$















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            1 Answer
            1






            active

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            active

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            active

            oldest

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            4












            $begingroup$

            Since the probability element of $X$ is $f(x)mathrmdx,$ the change of variable $y = xsigma + mu$ is equivalent to $x = (y-mu)/sigma,$ whence from



            $$f(x)mathrmdx = fleft(fracy-musigmaright)mathrmdleft(fracy-musigmaright) = frac1sigma fleft(fracy-musigmaright) mathrmdy$$



            it follows that the density of $Y$ is



            $$f_Y(y) = frac1sigmafleft(fracy-musigmaright).$$



            Consequently the entropy of $Y$ is



            $$H(Y) = -int_-infty^infty logleft(frac1sigmafleft(fracy-musigmaright)right) frac1sigmafleft(fracy-musigmaright) mathrmdy$$



            which, upon changing the variable back to $x = (y-mu)/sigma,$ produces



            $$eqalign
            H(Y) &= -int_-infty^infty logleft(frac1sigmafleft(xright)right) fleft(xright) mathrmdx \
            &= -int_-infty^infty left(logleft(frac1sigmaright) + logleft(fleft(xright)right)right) fleft(xright) mathrmdx \
            &= logleft(sigmaright) int_-infty^infty f(x) mathrmdx -int_-infty^infty logleft(fleft(xright)right) fleft(xright) mathrmdx \
            &= log(sigma) + H_f.
            $$



            These calculations used basic properties of the logarithm, the linearity of integration, and that fact that $f(x)mathrmdx$ integrates to unity (the Law of Total Probability).



            The conclusion is




            The entropy of $Y = Xsigma + mu$ is the entropy of $X$ plus $log(sigma).$




            In words, shifting a random variable does not change its entropy (we may think of the entropy as depending on the values of the probability density, but not on where those values occur), while scaling a variable (which, for $sigma ge 1$ "stretches" or "smears" it out) increases its entropy by $log(sigma).$ This supports the intuition that high-entropy distributions are "more spread out" than low-entropy distributions.




            As a consequence of this result, we are free to choose convenient values of $mu$ and $sigma$ when computing the entropy of any distribution. For example, the entropy of a Normal$(mu,sigma)$ distribution can be found by setting $mu=0$ and $sigma=1.$ The logarithm of the density in this case is



            $$log(f(x)) = -frac12log(2pi) - x^2/2,$$



            whence



            $$H = -E[-frac12log(2pi) - X^2/2] = frac12log(2pi) + frac12.$$



            Consequently the entropy of a Normal$(mu,sigma)$ distribution is obtained simply by adding $logsigma$ to this result, giving



            $$H = frac12log(2pi) + frac12 + log(sigma) = frac12log(2pi,e,sigma^2)$$



            as reported by Wikipedia.






            share|cite|improve this answer









            $endgroup$

















              4












              $begingroup$

              Since the probability element of $X$ is $f(x)mathrmdx,$ the change of variable $y = xsigma + mu$ is equivalent to $x = (y-mu)/sigma,$ whence from



              $$f(x)mathrmdx = fleft(fracy-musigmaright)mathrmdleft(fracy-musigmaright) = frac1sigma fleft(fracy-musigmaright) mathrmdy$$



              it follows that the density of $Y$ is



              $$f_Y(y) = frac1sigmafleft(fracy-musigmaright).$$



              Consequently the entropy of $Y$ is



              $$H(Y) = -int_-infty^infty logleft(frac1sigmafleft(fracy-musigmaright)right) frac1sigmafleft(fracy-musigmaright) mathrmdy$$



              which, upon changing the variable back to $x = (y-mu)/sigma,$ produces



              $$eqalign
              H(Y) &= -int_-infty^infty logleft(frac1sigmafleft(xright)right) fleft(xright) mathrmdx \
              &= -int_-infty^infty left(logleft(frac1sigmaright) + logleft(fleft(xright)right)right) fleft(xright) mathrmdx \
              &= logleft(sigmaright) int_-infty^infty f(x) mathrmdx -int_-infty^infty logleft(fleft(xright)right) fleft(xright) mathrmdx \
              &= log(sigma) + H_f.
              $$



              These calculations used basic properties of the logarithm, the linearity of integration, and that fact that $f(x)mathrmdx$ integrates to unity (the Law of Total Probability).



              The conclusion is




              The entropy of $Y = Xsigma + mu$ is the entropy of $X$ plus $log(sigma).$




              In words, shifting a random variable does not change its entropy (we may think of the entropy as depending on the values of the probability density, but not on where those values occur), while scaling a variable (which, for $sigma ge 1$ "stretches" or "smears" it out) increases its entropy by $log(sigma).$ This supports the intuition that high-entropy distributions are "more spread out" than low-entropy distributions.




              As a consequence of this result, we are free to choose convenient values of $mu$ and $sigma$ when computing the entropy of any distribution. For example, the entropy of a Normal$(mu,sigma)$ distribution can be found by setting $mu=0$ and $sigma=1.$ The logarithm of the density in this case is



              $$log(f(x)) = -frac12log(2pi) - x^2/2,$$



              whence



              $$H = -E[-frac12log(2pi) - X^2/2] = frac12log(2pi) + frac12.$$



              Consequently the entropy of a Normal$(mu,sigma)$ distribution is obtained simply by adding $logsigma$ to this result, giving



              $$H = frac12log(2pi) + frac12 + log(sigma) = frac12log(2pi,e,sigma^2)$$



              as reported by Wikipedia.






              share|cite|improve this answer









              $endgroup$















                4












                4








                4





                $begingroup$

                Since the probability element of $X$ is $f(x)mathrmdx,$ the change of variable $y = xsigma + mu$ is equivalent to $x = (y-mu)/sigma,$ whence from



                $$f(x)mathrmdx = fleft(fracy-musigmaright)mathrmdleft(fracy-musigmaright) = frac1sigma fleft(fracy-musigmaright) mathrmdy$$



                it follows that the density of $Y$ is



                $$f_Y(y) = frac1sigmafleft(fracy-musigmaright).$$



                Consequently the entropy of $Y$ is



                $$H(Y) = -int_-infty^infty logleft(frac1sigmafleft(fracy-musigmaright)right) frac1sigmafleft(fracy-musigmaright) mathrmdy$$



                which, upon changing the variable back to $x = (y-mu)/sigma,$ produces



                $$eqalign
                H(Y) &= -int_-infty^infty logleft(frac1sigmafleft(xright)right) fleft(xright) mathrmdx \
                &= -int_-infty^infty left(logleft(frac1sigmaright) + logleft(fleft(xright)right)right) fleft(xright) mathrmdx \
                &= logleft(sigmaright) int_-infty^infty f(x) mathrmdx -int_-infty^infty logleft(fleft(xright)right) fleft(xright) mathrmdx \
                &= log(sigma) + H_f.
                $$



                These calculations used basic properties of the logarithm, the linearity of integration, and that fact that $f(x)mathrmdx$ integrates to unity (the Law of Total Probability).



                The conclusion is




                The entropy of $Y = Xsigma + mu$ is the entropy of $X$ plus $log(sigma).$




                In words, shifting a random variable does not change its entropy (we may think of the entropy as depending on the values of the probability density, but not on where those values occur), while scaling a variable (which, for $sigma ge 1$ "stretches" or "smears" it out) increases its entropy by $log(sigma).$ This supports the intuition that high-entropy distributions are "more spread out" than low-entropy distributions.




                As a consequence of this result, we are free to choose convenient values of $mu$ and $sigma$ when computing the entropy of any distribution. For example, the entropy of a Normal$(mu,sigma)$ distribution can be found by setting $mu=0$ and $sigma=1.$ The logarithm of the density in this case is



                $$log(f(x)) = -frac12log(2pi) - x^2/2,$$



                whence



                $$H = -E[-frac12log(2pi) - X^2/2] = frac12log(2pi) + frac12.$$



                Consequently the entropy of a Normal$(mu,sigma)$ distribution is obtained simply by adding $logsigma$ to this result, giving



                $$H = frac12log(2pi) + frac12 + log(sigma) = frac12log(2pi,e,sigma^2)$$



                as reported by Wikipedia.






                share|cite|improve this answer









                $endgroup$



                Since the probability element of $X$ is $f(x)mathrmdx,$ the change of variable $y = xsigma + mu$ is equivalent to $x = (y-mu)/sigma,$ whence from



                $$f(x)mathrmdx = fleft(fracy-musigmaright)mathrmdleft(fracy-musigmaright) = frac1sigma fleft(fracy-musigmaright) mathrmdy$$



                it follows that the density of $Y$ is



                $$f_Y(y) = frac1sigmafleft(fracy-musigmaright).$$



                Consequently the entropy of $Y$ is



                $$H(Y) = -int_-infty^infty logleft(frac1sigmafleft(fracy-musigmaright)right) frac1sigmafleft(fracy-musigmaright) mathrmdy$$



                which, upon changing the variable back to $x = (y-mu)/sigma,$ produces



                $$eqalign
                H(Y) &= -int_-infty^infty logleft(frac1sigmafleft(xright)right) fleft(xright) mathrmdx \
                &= -int_-infty^infty left(logleft(frac1sigmaright) + logleft(fleft(xright)right)right) fleft(xright) mathrmdx \
                &= logleft(sigmaright) int_-infty^infty f(x) mathrmdx -int_-infty^infty logleft(fleft(xright)right) fleft(xright) mathrmdx \
                &= log(sigma) + H_f.
                $$



                These calculations used basic properties of the logarithm, the linearity of integration, and that fact that $f(x)mathrmdx$ integrates to unity (the Law of Total Probability).



                The conclusion is




                The entropy of $Y = Xsigma + mu$ is the entropy of $X$ plus $log(sigma).$




                In words, shifting a random variable does not change its entropy (we may think of the entropy as depending on the values of the probability density, but not on where those values occur), while scaling a variable (which, for $sigma ge 1$ "stretches" or "smears" it out) increases its entropy by $log(sigma).$ This supports the intuition that high-entropy distributions are "more spread out" than low-entropy distributions.




                As a consequence of this result, we are free to choose convenient values of $mu$ and $sigma$ when computing the entropy of any distribution. For example, the entropy of a Normal$(mu,sigma)$ distribution can be found by setting $mu=0$ and $sigma=1.$ The logarithm of the density in this case is



                $$log(f(x)) = -frac12log(2pi) - x^2/2,$$



                whence



                $$H = -E[-frac12log(2pi) - X^2/2] = frac12log(2pi) + frac12.$$



                Consequently the entropy of a Normal$(mu,sigma)$ distribution is obtained simply by adding $logsigma$ to this result, giving



                $$H = frac12log(2pi) + frac12 + log(sigma) = frac12log(2pi,e,sigma^2)$$



                as reported by Wikipedia.







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered 9 hours ago









                whuberwhuber

                212k34 gold badges465 silver badges847 bronze badges




                212k34 gold badges465 silver badges847 bronze badges



























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