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

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

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 $$X_1+X_2\sim \Phi(X)$$ 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}{0: 1-F_1(x)

And similarly for $$F_2$$.

He goes on to say this also shows the existence of higher moments and that the characteristic functions $$\int e^{i\xi x}F_i(dx)$$ 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 $$\int e^\frac{x^2}{2}dF_1(x)\color{red}{<} \int e^\frac{x^2}{2}Ae^{-\frac{x^2}{2}+λx}\left(-x+λ\right) dx$$ and then integrate $$\int_{-\infty}^0 A\:e^{λx}\left(-x+λ\right)dx=A\left[\left(\frac{e^{λx}\left(-x+λ\right)}{λ}+\frac{e^{λx}}{λ^2}\right)\right]_{-\infty}^0=A(1+\frac{1}{\lambda^2})$$ But I dont see why the inequality in red must hold.

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*} E\Big[\exp\Big[\frac{X_i^2}{2}\Big]\Big] &= E\Big[ 1 + \int_0^{|X_i|} t e^{t^2/2} \ dt \Big] = 1 + E\Big[ \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} P\Big[ |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 $$E[|X_i|^n] \le M^n + E[|X_i|^n \cdot \mathbb{1}\{|X_i| \geq M\}] \le M^n + E[\exp[X_i^2/2]] < \infty.$$