Why is $left.fracmathrmdmathrmdtmathbbE[|X+t|^p]right|_t=0=0$, where $X$ is standard Gaussian?

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Let $XsimmathcalN(0,1)$ and $p>2$. I would like to prove that
$$left.fracmathrmdmathrmdtmathbbE[|X+t|^p]right|_t=0=0,$$
but I don't see a (simple) way to do it.



Written explicitly, the derivative would be



$$lim_trightarrow 0frac1sqrt2piint_-infty^inftyfrac^ptexp(-x^2/2)~mathrmdx.$$



The denominator inside the integral doesn't generally converge to zero for fixed $x$, so it is not that simple. I also thought of using the bound
$$left| |x+t|^p-|x|^pright|leq p2^p-1|t|(|t|^p-1+|x|^p-1),$$
but I don't see this working either, the limit is strictly positive.



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    Let $XsimmathcalN(0,1)$ and $p>2$. I would like to prove that
    $$left.fracmathrmdmathrmdtmathbbE[|X+t|^p]right|_t=0=0,$$
    but I don't see a (simple) way to do it.



    Written explicitly, the derivative would be



    $$lim_trightarrow 0frac1sqrt2piint_-infty^inftyfrac^ptexp(-x^2/2)~mathrmdx.$$



    The denominator inside the integral doesn't generally converge to zero for fixed $x$, so it is not that simple. I also thought of using the bound
    $$left| |x+t|^p-|x|^pright|leq p2^p-1|t|(|t|^p-1+|x|^p-1),$$
    but I don't see this working either, the limit is strictly positive.



    What can I do?






    integration probability-theory derivatives probability-distributions normal-distribution







    share|cite|improve this question























      up vote
      0
      down vote

      favorite









      up vote
      0
      down vote

      favorite











      Let $XsimmathcalN(0,1)$ and $p>2$. I would like to prove that
      $$left.fracmathrmdmathrmdtmathbbE[|X+t|^p]right|_t=0=0,$$
      but I don't see a (simple) way to do it.



      Written explicitly, the derivative would be



      $$lim_trightarrow 0frac1sqrt2piint_-infty^inftyfrac^ptexp(-x^2/2)~mathrmdx.$$



      The denominator inside the integral doesn't generally converge to zero for fixed $x$, so it is not that simple. I also thought of using the bound
      $$left| |x+t|^p-|x|^pright|leq p2^p-1|t|(|t|^p-1+|x|^p-1),$$
      but I don't see this working either, the limit is strictly positive.



      What can I do?






      integration probability-theory derivatives probability-distributions normal-distribution







      share|cite|improve this question













      Let $XsimmathcalN(0,1)$ and $p>2$. I would like to prove that
      $$left.fracmathrmdmathrmdtmathbbE[|X+t|^p]right|_t=0=0,$$
      but I don't see a (simple) way to do it.



      Written explicitly, the derivative would be



      $$lim_trightarrow 0frac1sqrt2piint_-infty^inftyfrac^ptexp(-x^2/2)~mathrmdx.$$



      The denominator inside the integral doesn't generally converge to zero for fixed $x$, so it is not that simple. I also thought of using the bound
      $$left| |x+t|^p-|x|^pright|leq p2^p-1|t|(|t|^p-1+|x|^p-1),$$
      but I don't see this working either, the limit is strictly positive.



      What can I do?






      integration probability-theory derivatives probability-distributions normal-distribution




      integration probability-theory derivatives probability-distributions normal-distribution






      share|cite|improve this question













      share|cite|improve this question











      share|cite|improve this question




      share|cite|improve this question










      asked Aug 30 at 8:14









      Mau314

      34418




      34418




















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          Hints: $frac d dt |x+t|^p|_t=0 =p|x+t|^p-1 s(x)$ where $s(x)=1$ if $x geq 0$ and $s(x)=-1$ otherwise. Using the inequality you have stated in combination with D CT show that you can differentiate under the expectation. Finally. $Ep|X+t|^p-1 s(X)=0$ because the distribution of $X$ is symmetric.






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          • That's it, thank you.
            – Mau314
            Aug 30 at 8:30










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          Hints: $frac d dt |x+t|^p|_t=0 =p|x+t|^p-1 s(x)$ where $s(x)=1$ if $x geq 0$ and $s(x)=-1$ otherwise. Using the inequality you have stated in combination with D CT show that you can differentiate under the expectation. Finally. $Ep|X+t|^p-1 s(X)=0$ because the distribution of $X$ is symmetric.






          share|cite|improve this answer




















          • That's it, thank you.
            – Mau314
            Aug 30 at 8:30














          up vote
          1
          down vote



          accepted










          Hints: $frac d dt |x+t|^p|_t=0 =p|x+t|^p-1 s(x)$ where $s(x)=1$ if $x geq 0$ and $s(x)=-1$ otherwise. Using the inequality you have stated in combination with D CT show that you can differentiate under the expectation. Finally. $Ep|X+t|^p-1 s(X)=0$ because the distribution of $X$ is symmetric.






          share|cite|improve this answer




















          • That's it, thank you.
            – Mau314
            Aug 30 at 8:30












          up vote
          1
          down vote



          accepted







          up vote
          1
          down vote



          accepted






          Hints: $frac d dt |x+t|^p|_t=0 =p|x+t|^p-1 s(x)$ where $s(x)=1$ if $x geq 0$ and $s(x)=-1$ otherwise. Using the inequality you have stated in combination with D CT show that you can differentiate under the expectation. Finally. $Ep|X+t|^p-1 s(X)=0$ because the distribution of $X$ is symmetric.






          share|cite|improve this answer












          Hints: $frac d dt |x+t|^p|_t=0 =p|x+t|^p-1 s(x)$ where $s(x)=1$ if $x geq 0$ and $s(x)=-1$ otherwise. Using the inequality you have stated in combination with D CT show that you can differentiate under the expectation. Finally. $Ep|X+t|^p-1 s(X)=0$ because the distribution of $X$ is symmetric.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Aug 30 at 8:23









          Kavi Rama Murthy

          25.3k31335




          25.3k31335











          • That's it, thank you.
            – Mau314
            Aug 30 at 8:30
















          • That's it, thank you.
            – Mau314
            Aug 30 at 8:30















          That's it, thank you.
          – Mau314
          Aug 30 at 8:30




          That's it, thank you.
          – Mau314
          Aug 30 at 8:30

















           

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