Is $gamma(t)=cos^2(t)$ a characteristic function?












-1












$begingroup$


Is $gamma(t)=cos^2(t)$ a characteristic function?



I don't know how can I show that $gamma(t)=cos^2(t)$ is a positive function.










share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    Do you know what the characteristic function of a finite distribution looks like? Like, say, a coin flip (i.e. even odds Bernoulli distribution)?
    $endgroup$
    – Arthur
    Nov 4 '14 at 22:04












  • $begingroup$
    I know that, in this function we can compute: $X ~~~ (2,1/4), (-2,1/4), (0,1/2)$
    $endgroup$
    – Adams_20
    Nov 4 '14 at 22:07








  • 1




    $begingroup$
    $cos^2 t = dfrac{1+cos 2t}{2}=dfrac{1}{2}e^{0}+...$ and $cos x= dfrac{e^{ix}+e^{-ix}}{2}$
    $endgroup$
    – Cortizol
    Nov 4 '14 at 22:09


















-1












$begingroup$


Is $gamma(t)=cos^2(t)$ a characteristic function?



I don't know how can I show that $gamma(t)=cos^2(t)$ is a positive function.










share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    Do you know what the characteristic function of a finite distribution looks like? Like, say, a coin flip (i.e. even odds Bernoulli distribution)?
    $endgroup$
    – Arthur
    Nov 4 '14 at 22:04












  • $begingroup$
    I know that, in this function we can compute: $X ~~~ (2,1/4), (-2,1/4), (0,1/2)$
    $endgroup$
    – Adams_20
    Nov 4 '14 at 22:07








  • 1




    $begingroup$
    $cos^2 t = dfrac{1+cos 2t}{2}=dfrac{1}{2}e^{0}+...$ and $cos x= dfrac{e^{ix}+e^{-ix}}{2}$
    $endgroup$
    – Cortizol
    Nov 4 '14 at 22:09
















-1












-1








-1





$begingroup$


Is $gamma(t)=cos^2(t)$ a characteristic function?



I don't know how can I show that $gamma(t)=cos^2(t)$ is a positive function.










share|cite|improve this question











$endgroup$




Is $gamma(t)=cos^2(t)$ a characteristic function?



I don't know how can I show that $gamma(t)=cos^2(t)$ is a positive function.







probability probability-theory characteristic-functions






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share|cite|improve this question













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share|cite|improve this question








edited Nov 4 '14 at 22:31









mookid

25.6k52547




25.6k52547










asked Nov 4 '14 at 21:57









Adams_20Adams_20

1




1








  • 1




    $begingroup$
    Do you know what the characteristic function of a finite distribution looks like? Like, say, a coin flip (i.e. even odds Bernoulli distribution)?
    $endgroup$
    – Arthur
    Nov 4 '14 at 22:04












  • $begingroup$
    I know that, in this function we can compute: $X ~~~ (2,1/4), (-2,1/4), (0,1/2)$
    $endgroup$
    – Adams_20
    Nov 4 '14 at 22:07








  • 1




    $begingroup$
    $cos^2 t = dfrac{1+cos 2t}{2}=dfrac{1}{2}e^{0}+...$ and $cos x= dfrac{e^{ix}+e^{-ix}}{2}$
    $endgroup$
    – Cortizol
    Nov 4 '14 at 22:09
















  • 1




    $begingroup$
    Do you know what the characteristic function of a finite distribution looks like? Like, say, a coin flip (i.e. even odds Bernoulli distribution)?
    $endgroup$
    – Arthur
    Nov 4 '14 at 22:04












  • $begingroup$
    I know that, in this function we can compute: $X ~~~ (2,1/4), (-2,1/4), (0,1/2)$
    $endgroup$
    – Adams_20
    Nov 4 '14 at 22:07








  • 1




    $begingroup$
    $cos^2 t = dfrac{1+cos 2t}{2}=dfrac{1}{2}e^{0}+...$ and $cos x= dfrac{e^{ix}+e^{-ix}}{2}$
    $endgroup$
    – Cortizol
    Nov 4 '14 at 22:09










1




1




$begingroup$
Do you know what the characteristic function of a finite distribution looks like? Like, say, a coin flip (i.e. even odds Bernoulli distribution)?
$endgroup$
– Arthur
Nov 4 '14 at 22:04






$begingroup$
Do you know what the characteristic function of a finite distribution looks like? Like, say, a coin flip (i.e. even odds Bernoulli distribution)?
$endgroup$
– Arthur
Nov 4 '14 at 22:04














$begingroup$
I know that, in this function we can compute: $X ~~~ (2,1/4), (-2,1/4), (0,1/2)$
$endgroup$
– Adams_20
Nov 4 '14 at 22:07






$begingroup$
I know that, in this function we can compute: $X ~~~ (2,1/4), (-2,1/4), (0,1/2)$
$endgroup$
– Adams_20
Nov 4 '14 at 22:07






1




1




$begingroup$
$cos^2 t = dfrac{1+cos 2t}{2}=dfrac{1}{2}e^{0}+...$ and $cos x= dfrac{e^{ix}+e^{-ix}}{2}$
$endgroup$
– Cortizol
Nov 4 '14 at 22:09






$begingroup$
$cos^2 t = dfrac{1+cos 2t}{2}=dfrac{1}{2}e^{0}+...$ and $cos x= dfrac{e^{ix}+e^{-ix}}{2}$
$endgroup$
– Cortizol
Nov 4 '14 at 22:09












1 Answer
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$begingroup$

Use and show the following facts:




  • if $X$ and $Y$ are two independent random variables, the characteristic function of $X+Y$ is the product of the characteristic functions of $X$ with that of $Y$;

  • the function $tmapsto cos t$ is a characteristic function.






share|cite|improve this answer









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    0












    $begingroup$

    Use and show the following facts:




    • if $X$ and $Y$ are two independent random variables, the characteristic function of $X+Y$ is the product of the characteristic functions of $X$ with that of $Y$;

    • the function $tmapsto cos t$ is a characteristic function.






    share|cite|improve this answer









    $endgroup$


















      0












      $begingroup$

      Use and show the following facts:




      • if $X$ and $Y$ are two independent random variables, the characteristic function of $X+Y$ is the product of the characteristic functions of $X$ with that of $Y$;

      • the function $tmapsto cos t$ is a characteristic function.






      share|cite|improve this answer









      $endgroup$
















        0












        0








        0





        $begingroup$

        Use and show the following facts:




        • if $X$ and $Y$ are two independent random variables, the characteristic function of $X+Y$ is the product of the characteristic functions of $X$ with that of $Y$;

        • the function $tmapsto cos t$ is a characteristic function.






        share|cite|improve this answer









        $endgroup$



        Use and show the following facts:




        • if $X$ and $Y$ are two independent random variables, the characteristic function of $X+Y$ is the product of the characteristic functions of $X$ with that of $Y$;

        • the function $tmapsto cos t$ is a characteristic function.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Nov 5 '14 at 11:38









        Davide GiraudoDavide Giraudo

        128k17154268




        128k17154268






























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