When does the integral over points on the unit circle get 1 for a probability measure?












0












$begingroup$


I am studying probability theory and want to solve the following exercise.




Exercise.
Let $X$ be a real random variable. Then the following are equal:





  • $mathbb{P}left( X in a + bmathbb{Z}right) = 1$ for some $a, b in mathbb{R}$.

  • There exists a $z in mathbb{R}$ such that $lvert mathbb{hat{P}}left(zright)rvert = 1$.




Here, $mathbb{hat{P}}$ is the characteristic function of the probability measure $mathbb{P}$.



I was able to show "$Rightarrow$" but for the other direction I guess I need this statement:




For any probability measure $mathbb{P}$, if we have



$$
leftlvert int_mathbb{R} e^{itx} dmathbb{P}left(xright) rightrvert = 1 ;text{for a}; t inmathbb{R}setminus{0},
$$



then $exp(itx) = u$ for a $u in mathbb{C}$ with $lvert u rvert = 1$ (almost surely).




This is clear to me from graphical considerations but I have no clue how to show it formally.






real-analysis probability-theory







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  • $begingroup$
    is the first bullet supposed to be "for some $a,b in mathbb{R}$"?
    $endgroup$
    – mathworker21
    Nov 29 '18 at 17:43












  • $begingroup$
    @mathworker21Yes. I'll change that.
    $endgroup$
    – fpmoo
    Nov 29 '18 at 17:45
















0












$begingroup$


I am studying probability theory and want to solve the following exercise.




Exercise.
Let $X$ be a real random variable. Then the following are equal:





  • $mathbb{P}left( X in a + bmathbb{Z}right) = 1$ for some $a, b in mathbb{R}$.

  • There exists a $z in mathbb{R}$ such that $lvert mathbb{hat{P}}left(zright)rvert = 1$.




Here, $mathbb{hat{P}}$ is the characteristic function of the probability measure $mathbb{P}$.



I was able to show "$Rightarrow$" but for the other direction I guess I need this statement:




For any probability measure $mathbb{P}$, if we have



$$
leftlvert int_mathbb{R} e^{itx} dmathbb{P}left(xright) rightrvert = 1 ;text{for a}; t inmathbb{R}setminus{0},
$$



then $exp(itx) = u$ for a $u in mathbb{C}$ with $lvert u rvert = 1$ (almost surely).




This is clear to me from graphical considerations but I have no clue how to show it formally.






real-analysis probability-theory







share|cite|improve this question











$endgroup$












  • $begingroup$
    is the first bullet supposed to be "for some $a,b in mathbb{R}$"?
    $endgroup$
    – mathworker21
    Nov 29 '18 at 17:43












  • $begingroup$
    @mathworker21Yes. I'll change that.
    $endgroup$
    – fpmoo
    Nov 29 '18 at 17:45














0












0








0





$begingroup$


I am studying probability theory and want to solve the following exercise.




Exercise.
Let $X$ be a real random variable. Then the following are equal:





  • $mathbb{P}left( X in a + bmathbb{Z}right) = 1$ for some $a, b in mathbb{R}$.

  • There exists a $z in mathbb{R}$ such that $lvert mathbb{hat{P}}left(zright)rvert = 1$.




Here, $mathbb{hat{P}}$ is the characteristic function of the probability measure $mathbb{P}$.



I was able to show "$Rightarrow$" but for the other direction I guess I need this statement:




For any probability measure $mathbb{P}$, if we have



$$
leftlvert int_mathbb{R} e^{itx} dmathbb{P}left(xright) rightrvert = 1 ;text{for a}; t inmathbb{R}setminus{0},
$$



then $exp(itx) = u$ for a $u in mathbb{C}$ with $lvert u rvert = 1$ (almost surely).




This is clear to me from graphical considerations but I have no clue how to show it formally.






real-analysis probability-theory







share|cite|improve this question











$endgroup$




I am studying probability theory and want to solve the following exercise.




Exercise.
Let $X$ be a real random variable. Then the following are equal:





  • $mathbb{P}left( X in a + bmathbb{Z}right) = 1$ for some $a, b in mathbb{R}$.

  • There exists a $z in mathbb{R}$ such that $lvert mathbb{hat{P}}left(zright)rvert = 1$.




Here, $mathbb{hat{P}}$ is the characteristic function of the probability measure $mathbb{P}$.



I was able to show "$Rightarrow$" but for the other direction I guess I need this statement:




For any probability measure $mathbb{P}$, if we have



$$
leftlvert int_mathbb{R} e^{itx} dmathbb{P}left(xright) rightrvert = 1 ;text{for a}; t inmathbb{R}setminus{0},
$$



then $exp(itx) = u$ for a $u in mathbb{C}$ with $lvert u rvert = 1$ (almost surely).




This is clear to me from graphical considerations but I have no clue how to show it formally.






real-analysis probability-theory




real-analysis probability-theory






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Nov 29 '18 at 17:45







fpmoo

















asked Nov 29 '18 at 17:34









fpmoofpmoo

382113




382113












  • $begingroup$
    is the first bullet supposed to be "for some $a,b in mathbb{R}$"?
    $endgroup$
    – mathworker21
    Nov 29 '18 at 17:43












  • $begingroup$
    @mathworker21Yes. I'll change that.
    $endgroup$
    – fpmoo
    Nov 29 '18 at 17:45


















  • $begingroup$
    is the first bullet supposed to be "for some $a,b in mathbb{R}$"?
    $endgroup$
    – mathworker21
    Nov 29 '18 at 17:43












  • $begingroup$
    @mathworker21Yes. I'll change that.
    $endgroup$
    – fpmoo
    Nov 29 '18 at 17:45
















$begingroup$
is the first bullet supposed to be "for some $a,b in mathbb{R}$"?
$endgroup$
– mathworker21
Nov 29 '18 at 17:43






$begingroup$
is the first bullet supposed to be "for some $a,b in mathbb{R}$"?
$endgroup$
– mathworker21
Nov 29 '18 at 17:43














$begingroup$
@mathworker21Yes. I'll change that.
$endgroup$
– fpmoo
Nov 29 '18 at 17:45




$begingroup$
@mathworker21Yes. I'll change that.
$endgroup$
– fpmoo
Nov 29 '18 at 17:45










2 Answers
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The characteristic function's square modulus is $(mathbb{E}cos tX)^2+(mathbb{E}sin tX)^2=1-operatorname{Var}cos tX-operatorname{Var}sin tX$, requiring $cos tX,,sin tX$ to be constant variables, which is equivalent to $tantfrac{tX}{2}$ being constant.






share|cite|improve this answer









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    0












    $begingroup$

    You wish to show that if $f:mathbb{R} to mathbb{C}$ is s.t. $|f(x)| le 1$ for each $x$ and $|int_mathbb{R} f(x)dmu(x)| = 1$ for some probability measure $mu$, then $f$ is constant $mu$-a.e..



    By multiplying $f$ by a constant, we may assume $int_mathbb{R} f(x)dmu(x) = 1$. Then, $int_mathbb{R} Re[f(x)]dmu(x) = 1$ and $Re[f(x)] le 1$ for each $x$, so we must have $Re[f(x)] = 1$ for $mu$-a.e. $x$. It follows that $f = 1$ $mu$-a.e..






    share|cite|improve this answer









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      2 Answers
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      2 Answers
      2






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      active

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      1












      $begingroup$

      The characteristic function's square modulus is $(mathbb{E}cos tX)^2+(mathbb{E}sin tX)^2=1-operatorname{Var}cos tX-operatorname{Var}sin tX$, requiring $cos tX,,sin tX$ to be constant variables, which is equivalent to $tantfrac{tX}{2}$ being constant.






      share|cite|improve this answer









      $endgroup$


















        1












        $begingroup$

        The characteristic function's square modulus is $(mathbb{E}cos tX)^2+(mathbb{E}sin tX)^2=1-operatorname{Var}cos tX-operatorname{Var}sin tX$, requiring $cos tX,,sin tX$ to be constant variables, which is equivalent to $tantfrac{tX}{2}$ being constant.






        share|cite|improve this answer









        $endgroup$
















          1












          1








          1





          $begingroup$

          The characteristic function's square modulus is $(mathbb{E}cos tX)^2+(mathbb{E}sin tX)^2=1-operatorname{Var}cos tX-operatorname{Var}sin tX$, requiring $cos tX,,sin tX$ to be constant variables, which is equivalent to $tantfrac{tX}{2}$ being constant.






          share|cite|improve this answer









          $endgroup$



          The characteristic function's square modulus is $(mathbb{E}cos tX)^2+(mathbb{E}sin tX)^2=1-operatorname{Var}cos tX-operatorname{Var}sin tX$, requiring $cos tX,,sin tX$ to be constant variables, which is equivalent to $tantfrac{tX}{2}$ being constant.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Nov 29 '18 at 17:58









          J.G.J.G.

          25.7k22539




          25.7k22539























              0












              $begingroup$

              You wish to show that if $f:mathbb{R} to mathbb{C}$ is s.t. $|f(x)| le 1$ for each $x$ and $|int_mathbb{R} f(x)dmu(x)| = 1$ for some probability measure $mu$, then $f$ is constant $mu$-a.e..



              By multiplying $f$ by a constant, we may assume $int_mathbb{R} f(x)dmu(x) = 1$. Then, $int_mathbb{R} Re[f(x)]dmu(x) = 1$ and $Re[f(x)] le 1$ for each $x$, so we must have $Re[f(x)] = 1$ for $mu$-a.e. $x$. It follows that $f = 1$ $mu$-a.e..






              share|cite|improve this answer









              $endgroup$


















                0












                $begingroup$

                You wish to show that if $f:mathbb{R} to mathbb{C}$ is s.t. $|f(x)| le 1$ for each $x$ and $|int_mathbb{R} f(x)dmu(x)| = 1$ for some probability measure $mu$, then $f$ is constant $mu$-a.e..



                By multiplying $f$ by a constant, we may assume $int_mathbb{R} f(x)dmu(x) = 1$. Then, $int_mathbb{R} Re[f(x)]dmu(x) = 1$ and $Re[f(x)] le 1$ for each $x$, so we must have $Re[f(x)] = 1$ for $mu$-a.e. $x$. It follows that $f = 1$ $mu$-a.e..






                share|cite|improve this answer









                $endgroup$
















                  0












                  0








                  0





                  $begingroup$

                  You wish to show that if $f:mathbb{R} to mathbb{C}$ is s.t. $|f(x)| le 1$ for each $x$ and $|int_mathbb{R} f(x)dmu(x)| = 1$ for some probability measure $mu$, then $f$ is constant $mu$-a.e..



                  By multiplying $f$ by a constant, we may assume $int_mathbb{R} f(x)dmu(x) = 1$. Then, $int_mathbb{R} Re[f(x)]dmu(x) = 1$ and $Re[f(x)] le 1$ for each $x$, so we must have $Re[f(x)] = 1$ for $mu$-a.e. $x$. It follows that $f = 1$ $mu$-a.e..






                  share|cite|improve this answer









                  $endgroup$



                  You wish to show that if $f:mathbb{R} to mathbb{C}$ is s.t. $|f(x)| le 1$ for each $x$ and $|int_mathbb{R} f(x)dmu(x)| = 1$ for some probability measure $mu$, then $f$ is constant $mu$-a.e..



                  By multiplying $f$ by a constant, we may assume $int_mathbb{R} f(x)dmu(x) = 1$. Then, $int_mathbb{R} Re[f(x)]dmu(x) = 1$ and $Re[f(x)] le 1$ for each $x$, so we must have $Re[f(x)] = 1$ for $mu$-a.e. $x$. It follows that $f = 1$ $mu$-a.e..







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Nov 29 '18 at 17:52









                  mathworker21mathworker21

                  8,9421928




                  8,9421928






























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