Yesterday I stumbled upon this question:

For a distribution function $F(x)$ and constant $a$, integral of $F(x + a) - F(x)$ is $a$.

It is basically proved that if $F(x)$ is a cumulative distribution function, then $$\int_{\mathbb R} F(x+a) - F(x) \ dx = a$$

I understood the proof that was given, but it got me thinking:

Let's suppose $F \in C^\infty$. Then I can write $$F(x+a) - F(x) = f(x) a + \sum_{k=2}^\infty \frac{F^{(k)}(x)}{k!}a^k$$

However, since $\int_\mathbb R f(x) = 1$, we have

$$\int_{\mathbb R} F(x+a)-F(x) \ dx= a + \int_\mathbb R \sum_{k=2}^\infty \frac{F^{(k)}(x)}{k!}a^k \ dx$$

Is it true then that $$\int_\mathbb R \sum_{k=2}^\infty \frac{F^{(k)}(x)}{k!}a^k \ dx= 0$$ for all $a$ if $F \in C^\infty$ or am I misinterpreting something?

Thank you in advance!

  • $\begingroup$ $F\in C^\infty$ doesn't suffice for the Taylor series to represent $F$ (the Taylor series generally doesn't converge). $\endgroup$ – Daniel Fischer Oct 5 '14 at 12:01

Indeed, assuming the density $f$ is analytic and such that every derivative is integrable and that one can exchange summation and integral, you proved that, for every $k\geqslant1$, $$\int_\mathbb Rf^{(k)}(x)\mathrm dx=0.$$ Note that $C^\infty$ is not enough to guarantee the function is equal to its Taylor series and to exchange summation and integral.

  • $\begingroup$ You are right, sorry $\endgroup$ – Yulia V Oct 5 '14 at 12:04
  • $\begingroup$ Thank you very much! I think it is clearer now. I have two observation though: 1) Given that we can exchange summation and integral, dont we first need to prove that $\int_\mathbb R f^{(k)}(x)dx \ge 0$ to conclude that is actually $\int_\mathbb R f^{(k)}(x)dx = 0$? 2) Anyhow, isn't it true that if $\int_\mathbb R f(x) dx$ converges, then $\int_\mathbb R f'(x) dx = 0$ (provided that $f'$ exists continuos) ? This should make the statement in the question trivial $\endgroup$ – Ant Oct 5 '14 at 21:54
  • $\begingroup$ It is rather direct (and classical) to build densities $f$ that are sums of "spikes" such that $f'$ is smooth but not integrable. $\endgroup$ – Did Oct 5 '14 at 22:03
  • $\begingroup$ Uhm, I'll have to look into that. Thank you Did for all these great answers! $\endgroup$ – Ant Oct 6 '14 at 20:00

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