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If the characteristics function of a random variable is differentiable $2n$ times then it has finite moment up to even order $2n$. We know the converse is correct, but how can we prove this statement?

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    $\begingroup$ There are several possibilites to prove this. Just google it or have a look at (almost any) book on probability theory... $\endgroup$
    – saz
    Jan 9, 2015 at 4:34
  • $\begingroup$ @saz I think u mean the proof of the converse of this statement is everywhere? $\endgroup$
    – Math Boy
    Jan 9, 2015 at 4:58

1 Answer 1

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It suffices to consider the case $n=1$ (for $n \geq 2$ use induction). Denote by $\phi$ the characteristic function and $\mu$ the distribution of the random variable $X$. Then

$$\begin{align*} \phi(2h)-2\phi(0)+\phi(-2h) &= \int (e^{-\imath \, 2h x} -2 + e^{\imath \, 2hx}) \, \mu(dx) \\ &= 2 \int (\cos(2hx)-1) \mu(dx). \tag{1} \end{align*}$$ Since \begin{equation*} \frac{1-\cos(2y)}{4y^2} \xrightarrow[]{y \to 0} \frac{1}{2} \end{equation*} we obtain by applying Fatou's lemma \begin{align*} \int x^2 \frac{1}{2} \, d\mu(x) &= \int x^2 \lim_{h \to 0} \frac{1-\cos(2hx)}{4(hx)^2} \, \mu(dx) \\ &\leq \liminf_{h \to 0} \frac{1}{4h^2} \int (1-\cos(2hx)) \, \mu(dx) \\ &\stackrel{(1)}{=} - \frac{1}{2} \liminf_{h \to 0} \frac{1}{4 h^2} \big(\phi(2h)-2\phi(0)+\phi(-2h)) \\ &= - \frac{1}{2} \phi''(0)<\infty. \end{align*}

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  • $\begingroup$ I got it, and can I use the formula $\phi^{(n)}(t)= \int (ix)^{n}e^{itx} \, d\mu(x) $ directly, which gives $\phi^{(n)}(0)= \int (ix)^{n} \, d\mu(x)$ is finite? $\endgroup$
    – Math Boy
    Jan 9, 2015 at 5:16
  • $\begingroup$ @user147893 The proof of the formula $$\phi^{(n)}(t) = \int (\imath x)^n e^{\imath \, t x} \, d\mu(x)$$ requires that $\int x^n \, d\mu<\infty$. Therefore, we cannot use this identity to prove the existence of moments. $\endgroup$
    – saz
    Jan 9, 2015 at 5:22
  • $\begingroup$ @ saz, one more question, why our statement is not true when it is $2n+1$ instead of $2n$? $\endgroup$
    – Math Boy
    Jan 9, 2015 at 6:20
  • $\begingroup$ @user147893 The proof makes use of the fact that $1-\cos(y) \approx \frac{y^2}{2}$. In particular, we cannot expect $1-\cos(y) \approx c |y|$ for some constant $c$. And at least if one wants to use the same strategy as in the proof above to prove the existence of the first moment, one would need something of this form. $\endgroup$
    – saz
    Jan 9, 2015 at 11:54

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