Monotonic operations and integralsDouble IntegralsWhen are sums and integrals “identical” in form?Difficult Indefinite IntegralImproper integrals and right-hand Riemann sumsinterpreting triple integralsTranslating integrals by scalar premultiplicationRegularity of Cramer-Rao inequality for unbiased estimatorsCramer-Rao Lower Bound when the support for $x$ depends on $theta$?Why does the parameter estimator not depend on the parameter?Find $int^pi/2_0 operatornamearccot(1-x+x^2),dx $

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Monotonic operations and integrals


Double IntegralsWhen are sums and integrals “identical” in form?Difficult Indefinite IntegralImproper integrals and right-hand Riemann sumsinterpreting triple integralsTranslating integrals by scalar premultiplicationRegularity of Cramer-Rao inequality for unbiased estimatorsCramer-Rao Lower Bound when the support for $x$ depends on $theta$?Why does the parameter estimator not depend on the parameter?Find $int^pi/2_0 operatornamearccot(1-x+x^2),dx $













2












$begingroup$


If I have a monotonic function, say ln, can I bring it inside an integral?



in other words, is $$lnleft[ int f(x), dxright] = int ln(f(x)), dx.$$



My limits of integration don't depend on $x,$ so I think I can.



Could I move expectation inside the integral?



(I am working on something with the Cramer-Rao inequality, which requires that I take the natural log and then the expectation of a function.)






calculus integration statistics expected-value







share|cite|improve this question











$endgroup$











  • $begingroup$
    You can say, I think, that $lnint fge intln f,$ since $ln$ is a concave function. See Jensen's Inequality.
    $endgroup$
    – Adrian Keister
    8 hours ago
















2












$begingroup$


If I have a monotonic function, say ln, can I bring it inside an integral?



in other words, is $$lnleft[ int f(x), dxright] = int ln(f(x)), dx.$$



My limits of integration don't depend on $x,$ so I think I can.



Could I move expectation inside the integral?



(I am working on something with the Cramer-Rao inequality, which requires that I take the natural log and then the expectation of a function.)






calculus integration statistics expected-value







share|cite|improve this question











$endgroup$











  • $begingroup$
    You can say, I think, that $lnint fge intln f,$ since $ln$ is a concave function. See Jensen's Inequality.
    $endgroup$
    – Adrian Keister
    8 hours ago














2












2








2


1



$begingroup$


If I have a monotonic function, say ln, can I bring it inside an integral?



in other words, is $$lnleft[ int f(x), dxright] = int ln(f(x)), dx.$$



My limits of integration don't depend on $x,$ so I think I can.



Could I move expectation inside the integral?



(I am working on something with the Cramer-Rao inequality, which requires that I take the natural log and then the expectation of a function.)






calculus integration statistics expected-value







share|cite|improve this question











$endgroup$




If I have a monotonic function, say ln, can I bring it inside an integral?



in other words, is $$lnleft[ int f(x), dxright] = int ln(f(x)), dx.$$



My limits of integration don't depend on $x,$ so I think I can.



Could I move expectation inside the integral?



(I am working on something with the Cramer-Rao inequality, which requires that I take the natural log and then the expectation of a function.)






calculus integration statistics expected-value




calculus integration statistics expected-value






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited 8 hours ago









Adrian Keister

5,56872133




5,56872133










asked 8 hours ago









JessJess

135111




135111











  • $begingroup$
    You can say, I think, that $lnint fge intln f,$ since $ln$ is a concave function. See Jensen's Inequality.
    $endgroup$
    – Adrian Keister
    8 hours ago

















  • $begingroup$
    You can say, I think, that $lnint fge intln f,$ since $ln$ is a concave function. See Jensen's Inequality.
    $endgroup$
    – Adrian Keister
    8 hours ago
















$begingroup$
You can say, I think, that $lnint fge intln f,$ since $ln$ is a concave function. See Jensen's Inequality.
$endgroup$
– Adrian Keister
8 hours ago





$begingroup$
You can say, I think, that $lnint fge intln f,$ since $ln$ is a concave function. See Jensen's Inequality.
$endgroup$
– Adrian Keister
8 hours ago











2 Answers
2






active

oldest

votes


















6












$begingroup$

$lnleft(int_0^1 x,dxright)=lnleft(frac12right),$ but $int_0^1 ln x,dx=-1$. So no, it's not that simple.






share|cite|improve this answer











$endgroup$




















    2












    $begingroup$

    No. As a counterexample, take $f(x)=e^x$.






    share|cite|improve this answer









    $endgroup$













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      2 Answers
      2






      active

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      2 Answers
      2






      active

      oldest

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      active

      oldest

      votes






      active

      oldest

      votes









      6












      $begingroup$

      $lnleft(int_0^1 x,dxright)=lnleft(frac12right),$ but $int_0^1 ln x,dx=-1$. So no, it's not that simple.






      share|cite|improve this answer











      $endgroup$

















        6












        $begingroup$

        $lnleft(int_0^1 x,dxright)=lnleft(frac12right),$ but $int_0^1 ln x,dx=-1$. So no, it's not that simple.






        share|cite|improve this answer











        $endgroup$















          6












          6








          6





          $begingroup$

          $lnleft(int_0^1 x,dxright)=lnleft(frac12right),$ but $int_0^1 ln x,dx=-1$. So no, it's not that simple.






          share|cite|improve this answer











          $endgroup$



          $lnleft(int_0^1 x,dxright)=lnleft(frac12right),$ but $int_0^1 ln x,dx=-1$. So no, it's not that simple.







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited 8 hours ago









          Adrian Keister

          5,56872133




          5,56872133










          answered 8 hours ago









          MarkMark

          13.3k1825




          13.3k1825





















              2












              $begingroup$

              No. As a counterexample, take $f(x)=e^x$.






              share|cite|improve this answer









              $endgroup$

















                2












                $begingroup$

                No. As a counterexample, take $f(x)=e^x$.






                share|cite|improve this answer









                $endgroup$















                  2












                  2








                  2





                  $begingroup$

                  No. As a counterexample, take $f(x)=e^x$.






                  share|cite|improve this answer









                  $endgroup$



                  No. As a counterexample, take $f(x)=e^x$.







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered 8 hours ago









                  J. W. TannerJ. W. Tanner

                  7,9941723




                  7,9941723



























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