# Proving the convergence of $\int e^{x^2/2} dF_{1,2}(x)$ in Cramers proof of the

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 $$F_1(x)\leq \frac{\Phi(x-\lambda)}{F_2(-\lambda)} Where $$A$$ is independent of $$x$$. Analogue for $$x>0$$: $$1-F_1(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?

## 1 Answer

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.

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