Cramers decomposition theorem: why are the functions entire and existence of higher moments











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I have some questions on the proof of cramers decomposition theorem.
Cramers original paper:
Suppose we have the integral equation:
begin{equation}tag{$star$}
int_{-infty}^infty F_1(x-t)dF_2(t)=Phileft( xright),
end{equation}



With the first moments $m_{F_1}=m_{F_2}=m=0$ and $sigma_1^2+sigma_2^2=1$. The random variables $X_1$ and $X_2$ with $X_1+X_2sim Phi(X)$ are independent.



First he shows the first and second moments of $X_1$ and $X_2$ exist. Then he shows the integrals
$$int e^{frac{x^2}{2}}dF_i(x)$$
are finite by showing
$$x<0:~F_1(x)leq frac{Phi(x-lambda)}{F_2(-lambda)}color{blue}{<Ae^{-frac{x^2}{2}+lambda x}}, quad x>0: 1-F_1(x)<Be^{-frac{x^2}{2}-nu x}.$$



And similarly for $F_2$.



He goes on to say this also shows the existence of higher moments and that the characteristic functions
$$int e^{ixi x}F_i(dx)$$
are entire.




Question: why does that imply the integrals are finite and if they are why does that show the existence of higher moments and entirety? How does the inequality in blue follow




Attempt:
I see that if we write
$$int e^frac{x^2}{2}dF_1(x)color{red}{<} int e^frac{x^2}{2}Ae^{-frac{x^2}{2}+λx}left(-x+λright) dx$$
and then integrate
$$int_{-infty}^0 A:e^{λx}left(-x+λright)dx=Aleft[left(frac{e^{λx}left(-x+λright)}{λ}+frac{e^{λx}}{λ^2}right)right]_{-infty}^0=A(1+frac{1}{lambda^2})$$
But I dont see why the inequality in red must hold.










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    I have some questions on the proof of cramers decomposition theorem.
    Cramers original paper:
    Suppose we have the integral equation:
    begin{equation}tag{$star$}
    int_{-infty}^infty F_1(x-t)dF_2(t)=Phileft( xright),
    end{equation}



    With the first moments $m_{F_1}=m_{F_2}=m=0$ and $sigma_1^2+sigma_2^2=1$. The random variables $X_1$ and $X_2$ with $X_1+X_2sim Phi(X)$ are independent.



    First he shows the first and second moments of $X_1$ and $X_2$ exist. Then he shows the integrals
    $$int e^{frac{x^2}{2}}dF_i(x)$$
    are finite by showing
    $$x<0:~F_1(x)leq frac{Phi(x-lambda)}{F_2(-lambda)}color{blue}{<Ae^{-frac{x^2}{2}+lambda x}}, quad x>0: 1-F_1(x)<Be^{-frac{x^2}{2}-nu x}.$$



    And similarly for $F_2$.



    He goes on to say this also shows the existence of higher moments and that the characteristic functions
    $$int e^{ixi x}F_i(dx)$$
    are entire.




    Question: why does that imply the integrals are finite and if they are why does that show the existence of higher moments and entirety? How does the inequality in blue follow




    Attempt:
    I see that if we write
    $$int e^frac{x^2}{2}dF_1(x)color{red}{<} int e^frac{x^2}{2}Ae^{-frac{x^2}{2}+λx}left(-x+λright) dx$$
    and then integrate
    $$int_{-infty}^0 A:e^{λx}left(-x+λright)dx=Aleft[left(frac{e^{λx}left(-x+λright)}{λ}+frac{e^{λx}}{λ^2}right)right]_{-infty}^0=A(1+frac{1}{lambda^2})$$
    But I dont see why the inequality in red must hold.










    share|cite|improve this question


























      up vote
      1
      down vote

      favorite









      up vote
      1
      down vote

      favorite











      I have some questions on the proof of cramers decomposition theorem.
      Cramers original paper:
      Suppose we have the integral equation:
      begin{equation}tag{$star$}
      int_{-infty}^infty F_1(x-t)dF_2(t)=Phileft( xright),
      end{equation}



      With the first moments $m_{F_1}=m_{F_2}=m=0$ and $sigma_1^2+sigma_2^2=1$. The random variables $X_1$ and $X_2$ with $X_1+X_2sim Phi(X)$ are independent.



      First he shows the first and second moments of $X_1$ and $X_2$ exist. Then he shows the integrals
      $$int e^{frac{x^2}{2}}dF_i(x)$$
      are finite by showing
      $$x<0:~F_1(x)leq frac{Phi(x-lambda)}{F_2(-lambda)}color{blue}{<Ae^{-frac{x^2}{2}+lambda x}}, quad x>0: 1-F_1(x)<Be^{-frac{x^2}{2}-nu x}.$$



      And similarly for $F_2$.



      He goes on to say this also shows the existence of higher moments and that the characteristic functions
      $$int e^{ixi x}F_i(dx)$$
      are entire.




      Question: why does that imply the integrals are finite and if they are why does that show the existence of higher moments and entirety? How does the inequality in blue follow




      Attempt:
      I see that if we write
      $$int e^frac{x^2}{2}dF_1(x)color{red}{<} int e^frac{x^2}{2}Ae^{-frac{x^2}{2}+λx}left(-x+λright) dx$$
      and then integrate
      $$int_{-infty}^0 A:e^{λx}left(-x+λright)dx=Aleft[left(frac{e^{λx}left(-x+λright)}{λ}+frac{e^{λx}}{λ^2}right)right]_{-infty}^0=A(1+frac{1}{lambda^2})$$
      But I dont see why the inequality in red must hold.










      share|cite|improve this question















      I have some questions on the proof of cramers decomposition theorem.
      Cramers original paper:
      Suppose we have the integral equation:
      begin{equation}tag{$star$}
      int_{-infty}^infty F_1(x-t)dF_2(t)=Phileft( xright),
      end{equation}



      With the first moments $m_{F_1}=m_{F_2}=m=0$ and $sigma_1^2+sigma_2^2=1$. The random variables $X_1$ and $X_2$ with $X_1+X_2sim Phi(X)$ are independent.



      First he shows the first and second moments of $X_1$ and $X_2$ exist. Then he shows the integrals
      $$int e^{frac{x^2}{2}}dF_i(x)$$
      are finite by showing
      $$x<0:~F_1(x)leq frac{Phi(x-lambda)}{F_2(-lambda)}color{blue}{<Ae^{-frac{x^2}{2}+lambda x}}, quad x>0: 1-F_1(x)<Be^{-frac{x^2}{2}-nu x}.$$



      And similarly for $F_2$.



      He goes on to say this also shows the existence of higher moments and that the characteristic functions
      $$int e^{ixi x}F_i(dx)$$
      are entire.




      Question: why does that imply the integrals are finite and if they are why does that show the existence of higher moments and entirety? How does the inequality in blue follow




      Attempt:
      I see that if we write
      $$int e^frac{x^2}{2}dF_1(x)color{red}{<} int e^frac{x^2}{2}Ae^{-frac{x^2}{2}+λx}left(-x+λright) dx$$
      and then integrate
      $$int_{-infty}^0 A:e^{λx}left(-x+λright)dx=Aleft[left(frac{e^{λx}left(-x+λright)}{λ}+frac{e^{λx}}{λ^2}right)right]_{-infty}^0=A(1+frac{1}{lambda^2})$$
      But I dont see why the inequality in red must hold.







      complex-analysis probability-theory






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      edited Dec 1 at 18:49

























      asked Nov 24 at 16:08









      orange

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          You can use the bounds on the probability distribution function $F_i$ together with Fubini to deduce that $int e^{frac{x^2}{2}} F_i(dx)$ converge. I will use the probability notation of expectations here. Notice that the bounds given state that there are positive constants $A$ and $lambda$ such that for any $t > 0$ we have
          $$
          F_i(-t) = P[X_i le -t] le A exp[- frac{t^2}{2} - lambda t]
          quad text{and} quad
          1 - F_i(t) = P[X_i > t] le A exp[- frac{t^2}{2} - lambda t]
          $$

          Here, the actual values of $A$ and $lambda$ are not important. Notice that we can use the same constants for both bounds if we increase the constants a little. Now, for showing that the integral above is finite we write using the Fundamental Theorem of Calculus and Fubini that
          begin{align*}
          EBig[expBig[frac{X_i^2}{2}Big]Big]
          &= EBig[ 1 + int_0^{|X_i|} t e^{t^2/2} dt Big]
          = 1 + EBig[ int_0^{infty} mathbb{1}{|X_i| > t} cdot t e^{t^2/2} dt Big] \
          &= 1 + int_0^{infty} t e^{t^2/2} PBig[ |X_i| > t Big] dt
          le 1 + int_0^{infty} t e^{t^2/2} A exp[- frac{t^2}{2} - lambda t] dt \
          &= 1 + A int_0^{infty} t e^{- lambda t} dt < infty.
          end{align*}

          This proves that some exponential moments exist. Notice that the existence of exponential moments imply the existence of $E[|X_i|^n]$ for any $n geq 1$. This is simply because fixed $n$ we have $exp[x^2/2] geq |x|^n$ for any $x$ with $|x| geq M = M(n)$, some positive constant depending on $n$. Then, we have
          $$
          E[|X_i|^n] le M^n + E[|X_i|^n cdot mathbb{1}{|X_i| geq M}] le M^n + E[exp[X_i^2/2]] < infty.
          $$






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            You can use the bounds on the probability distribution function $F_i$ together with Fubini to deduce that $int e^{frac{x^2}{2}} F_i(dx)$ converge. I will use the probability notation of expectations here. Notice that the bounds given state that there are positive constants $A$ and $lambda$ such that for any $t > 0$ we have
            $$
            F_i(-t) = P[X_i le -t] le A exp[- frac{t^2}{2} - lambda t]
            quad text{and} quad
            1 - F_i(t) = P[X_i > t] le A exp[- frac{t^2}{2} - lambda t]
            $$

            Here, the actual values of $A$ and $lambda$ are not important. Notice that we can use the same constants for both bounds if we increase the constants a little. Now, for showing that the integral above is finite we write using the Fundamental Theorem of Calculus and Fubini that
            begin{align*}
            EBig[expBig[frac{X_i^2}{2}Big]Big]
            &= EBig[ 1 + int_0^{|X_i|} t e^{t^2/2} dt Big]
            = 1 + EBig[ int_0^{infty} mathbb{1}{|X_i| > t} cdot t e^{t^2/2} dt Big] \
            &= 1 + int_0^{infty} t e^{t^2/2} PBig[ |X_i| > t Big] dt
            le 1 + int_0^{infty} t e^{t^2/2} A exp[- frac{t^2}{2} - lambda t] dt \
            &= 1 + A int_0^{infty} t e^{- lambda t} dt < infty.
            end{align*}

            This proves that some exponential moments exist. Notice that the existence of exponential moments imply the existence of $E[|X_i|^n]$ for any $n geq 1$. This is simply because fixed $n$ we have $exp[x^2/2] geq |x|^n$ for any $x$ with $|x| geq M = M(n)$, some positive constant depending on $n$. Then, we have
            $$
            E[|X_i|^n] le M^n + E[|X_i|^n cdot mathbb{1}{|X_i| geq M}] le M^n + E[exp[X_i^2/2]] < infty.
            $$






            share|cite|improve this answer

























              up vote
              1
              down vote



              accepted
              +50










              You can use the bounds on the probability distribution function $F_i$ together with Fubini to deduce that $int e^{frac{x^2}{2}} F_i(dx)$ converge. I will use the probability notation of expectations here. Notice that the bounds given state that there are positive constants $A$ and $lambda$ such that for any $t > 0$ we have
              $$
              F_i(-t) = P[X_i le -t] le A exp[- frac{t^2}{2} - lambda t]
              quad text{and} quad
              1 - F_i(t) = P[X_i > t] le A exp[- frac{t^2}{2} - lambda t]
              $$

              Here, the actual values of $A$ and $lambda$ are not important. Notice that we can use the same constants for both bounds if we increase the constants a little. Now, for showing that the integral above is finite we write using the Fundamental Theorem of Calculus and Fubini that
              begin{align*}
              EBig[expBig[frac{X_i^2}{2}Big]Big]
              &= EBig[ 1 + int_0^{|X_i|} t e^{t^2/2} dt Big]
              = 1 + EBig[ int_0^{infty} mathbb{1}{|X_i| > t} cdot t e^{t^2/2} dt Big] \
              &= 1 + int_0^{infty} t e^{t^2/2} PBig[ |X_i| > t Big] dt
              le 1 + int_0^{infty} t e^{t^2/2} A exp[- frac{t^2}{2} - lambda t] dt \
              &= 1 + A int_0^{infty} t e^{- lambda t} dt < infty.
              end{align*}

              This proves that some exponential moments exist. Notice that the existence of exponential moments imply the existence of $E[|X_i|^n]$ for any $n geq 1$. This is simply because fixed $n$ we have $exp[x^2/2] geq |x|^n$ for any $x$ with $|x| geq M = M(n)$, some positive constant depending on $n$. Then, we have
              $$
              E[|X_i|^n] le M^n + E[|X_i|^n cdot mathbb{1}{|X_i| geq M}] le M^n + E[exp[X_i^2/2]] < infty.
              $$






              share|cite|improve this answer























                up vote
                1
                down vote



                accepted
                +50







                up vote
                1
                down vote



                accepted
                +50




                +50




                You can use the bounds on the probability distribution function $F_i$ together with Fubini to deduce that $int e^{frac{x^2}{2}} F_i(dx)$ converge. I will use the probability notation of expectations here. Notice that the bounds given state that there are positive constants $A$ and $lambda$ such that for any $t > 0$ we have
                $$
                F_i(-t) = P[X_i le -t] le A exp[- frac{t^2}{2} - lambda t]
                quad text{and} quad
                1 - F_i(t) = P[X_i > t] le A exp[- frac{t^2}{2} - lambda t]
                $$

                Here, the actual values of $A$ and $lambda$ are not important. Notice that we can use the same constants for both bounds if we increase the constants a little. Now, for showing that the integral above is finite we write using the Fundamental Theorem of Calculus and Fubini that
                begin{align*}
                EBig[expBig[frac{X_i^2}{2}Big]Big]
                &= EBig[ 1 + int_0^{|X_i|} t e^{t^2/2} dt Big]
                = 1 + EBig[ int_0^{infty} mathbb{1}{|X_i| > t} cdot t e^{t^2/2} dt Big] \
                &= 1 + int_0^{infty} t e^{t^2/2} PBig[ |X_i| > t Big] dt
                le 1 + int_0^{infty} t e^{t^2/2} A exp[- frac{t^2}{2} - lambda t] dt \
                &= 1 + A int_0^{infty} t e^{- lambda t} dt < infty.
                end{align*}

                This proves that some exponential moments exist. Notice that the existence of exponential moments imply the existence of $E[|X_i|^n]$ for any $n geq 1$. This is simply because fixed $n$ we have $exp[x^2/2] geq |x|^n$ for any $x$ with $|x| geq M = M(n)$, some positive constant depending on $n$. Then, we have
                $$
                E[|X_i|^n] le M^n + E[|X_i|^n cdot mathbb{1}{|X_i| geq M}] le M^n + E[exp[X_i^2/2]] < infty.
                $$






                share|cite|improve this answer












                You can use the bounds on the probability distribution function $F_i$ together with Fubini to deduce that $int e^{frac{x^2}{2}} F_i(dx)$ converge. I will use the probability notation of expectations here. Notice that the bounds given state that there are positive constants $A$ and $lambda$ such that for any $t > 0$ we have
                $$
                F_i(-t) = P[X_i le -t] le A exp[- frac{t^2}{2} - lambda t]
                quad text{and} quad
                1 - F_i(t) = P[X_i > t] le A exp[- frac{t^2}{2} - lambda t]
                $$

                Here, the actual values of $A$ and $lambda$ are not important. Notice that we can use the same constants for both bounds if we increase the constants a little. Now, for showing that the integral above is finite we write using the Fundamental Theorem of Calculus and Fubini that
                begin{align*}
                EBig[expBig[frac{X_i^2}{2}Big]Big]
                &= EBig[ 1 + int_0^{|X_i|} t e^{t^2/2} dt Big]
                = 1 + EBig[ int_0^{infty} mathbb{1}{|X_i| > t} cdot t e^{t^2/2} dt Big] \
                &= 1 + int_0^{infty} t e^{t^2/2} PBig[ |X_i| > t Big] dt
                le 1 + int_0^{infty} t e^{t^2/2} A exp[- frac{t^2}{2} - lambda t] dt \
                &= 1 + A int_0^{infty} t e^{- lambda t} dt < infty.
                end{align*}

                This proves that some exponential moments exist. Notice that the existence of exponential moments imply the existence of $E[|X_i|^n]$ for any $n geq 1$. This is simply because fixed $n$ we have $exp[x^2/2] geq |x|^n$ for any $x$ with $|x| geq M = M(n)$, some positive constant depending on $n$. Then, we have
                $$
                E[|X_i|^n] le M^n + E[|X_i|^n cdot mathbb{1}{|X_i| geq M}] le M^n + E[exp[X_i^2/2]] < infty.
                $$







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                answered Dec 2 at 0:16









                Daniel

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