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?


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.

  • 1
    $\begingroup$ 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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