= Characteristic-function derivatives from finite absolute moments
{title2=$\varphi_X^{(k)}(0)=i^k\mathbb E[X^k]$}
If $m\geq1$ is an <integer> and $\mathbb E[|X|^m]<\infty$, the <characteristic function> has continuous <derivatives> through order $m$, with
$$
\varphi_X^{(k)}(t)=\mathbb E[(iX)^ke^{itX}],\qquad
\varphi_X^{(k)}(0)=i^k\mathbb E[X^k],\qquad 1\leq k\leq m.
$$
Thus finite <absolute moments> justify recovering <moments> from <derivatives>.<Probability: Theory and Examples (fifth edition) by Rick Durrett>[§3.3.3, Theorem 3.3.18, printed p. 134]{ref}
For a short justification of <differentiation under the integral sign>, the difference quotient for the $(k-1)$st <derivative> has integrand $(iX)^{k-1}e^{itX}(e^{ihX}-1)/h$. Its absolute value is at most $|X|^k$, since $|e^{iu}-1|\leq|u|$ for real $u$. Lower <absolute moments> are finite because $|X|^k\leq1+|X|^m$. Apply the <dominated convergence theorem> successively, then use the same integrable bound to obtain <continuity> of each <derivative>.
For the <Bernoulli distribution> with parameter $1/3$, direct <differentiation> gives $\varphi_B'(0)=i/3$ and $\varphi_B''(0)=-1/3$, agreeing with $\mathbb E[B]=\mathbb E[B^2]=1/3$. Its <variance> is therefore $1/3-(1/3)^2=2/9$. Existence and <continuity of a characteristic function> alone do not authorize these <moment> calculations: the standard <Cauchy distribution>, whose <expected value> is undefined, still has a <characteristic function>. The stated finite <absolute moment> hypothesis is a sufficient condition for the <derivative> formula, not an assumption built into the definition of a <characteristic function>.
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