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In Cramer 1936 the proof of the Cramer decomposition theorem contains proving the following integrals are finite $$\int e^{x^2/2} dF_{1}(x), \quad \int e^{x^2/2} dF_{2}(x)$$ to later use in finding a bound for the characteristic functions.

We are solving the integral equation $$\int F_1(x-t)F_2(dt)=\Phi(x),\quad \Phi(x)=\frac{1}{\sqrt{2\pi}}\int_{-\infty}^x e^{-t^2/2} dt.$$

The proof goes as follows:

Since the first moment of $F_2$ is $0$ there exists a $\lambda >0$ so that $F_2(-\lambda)>0$. For all $x<0$: \begin{align}\Phi(x-\lambda)&=\int F_1(x-\lambda-t)F_2(dt) \\ &\geq\int_{-\infty}^{-\lambda} F_1(x-\lambda-t)F_2(dt) \\ &\rm \color{red}{\geq F_1(x)F_2(-\lambda)} \end{align} And therefore $$ F_1(x)\leq \frac{\Phi(x-\lambda)}{F_2(-\lambda)}<Ae^{-\frac{x^2}{2}+\lambda x}$$ Where $A$ is independent of $x$. Analogue for $x>0$: $$1-F_1(x)<Be^{-\frac{x^2}{2}-\mu x}$$ Where $B$ and $\mu$ are positive numbers independent of $x$.

I don't understand how the inequality in red follows. How did this show the integral is finite?

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To me, he is simply using monotonicity and positivity of distribution functions and the definition itself of distribution function: indeed the argument $x-\lambda-t$ attains it minimum on $[-\lambda,-\infty)$ at $-\lambda$ and thus by monotonicity of the distribution $$F_1(x-\lambda-t)\geq F_1(x-\lambda-(-\lambda))=F_1(x)\geq 0.$$ Now we simply have $$\int_{-\infty}^{-\lambda}F_1(x-\lambda-t)\ F_2(dt)\geq F_1(x)\int_{-\infty}^{-\lambda}F_2(dt)\overset{def}{=}F_1(x)\cdot F_2(-\lambda),$$ where I used the definition of distribution in the last step.

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    $\begingroup$ Thank you. Simple enough :). $\endgroup$ – badatmath Oct 21 '18 at 10:28
  • $\begingroup$ glad to be of help :) $\endgroup$ – b00n heT Oct 21 '18 at 10:28

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