1
$\begingroup$

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!

$\endgroup$
  • $\begingroup$ $F\in C^\infty$ 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
3
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.