derivative of characteristic function I came across an interesting problem but unable to see how to approach it. How do I use the dominated convergence theorem (LDCT), to show that first derivative of the characteristic function of the probability distribution at $t = 0$, $\phi^′(0)=iE[X]$? Any ideas?
References:
http://en.wikipedia.org/wiki/Characteristic_function_(probability_theory)
http://en.wikipedia.org/wiki/Dominated_convergence_theorem
 A: First, one has $$e^{ir}-1=O(|r|) $$ for all $r\in \mathbb{R}.$ This means that there exist constant (independent of $r$) $A>0$ such that $$|e^{ir}-1|\le A|r| \mbox{ for all } r\in \mathbb{R} $$ (In fact, you can take $A=1$, as $|e^{ir}-1|$ is the length of the chord from 1 to $e^{ir}$ and $r$ is the corresponding arc length on the unit circle.)
Let us prove your question in the case $X $ is contiuous. (The discrete case can be proved similarly.) Suppose $f(x)$ is the probability density function of $X$.  Then
$$ \phi(t)=E(e^{itX})=\int_{-\infty}^\infty f(x)e^{itx} dx$$ and $\phi(0)=\int_{-\infty}^\infty f(x) dx =1$. So by linearity of integral
$$\phi'(0)=\lim_{t\to 0}\frac{\phi(t)-\phi(0)}{t}=\lim_{t\to 0}\int_{-\infty}^\infty f(x)\frac{e^{itx}-1}{t} dx.$$ Since
$$\left|f(x)\frac{e^{itx}-1}{t}\right|\le f(x) A \frac{|tx|}{|t|}=A f(x)|x|,$$ and $$\lim_{t\to 0}\frac{e^{itx}-1}{t}=\frac{d}{dt} e^{itx}|_{t=0}=ix e^0=ix,$$
it follows from DCT that if $E(|X|)<\infty$, then $$\phi'(0)= \int_{-\infty}^\infty f(x)ix dx=i E(X).$$
A: You need to show that if $(a_n)$ is a sequence of nonzero numbers
which tends to zero, that $(\phi(a_n)-\phi(0))/a_n\to iE[X]$. Now
$$\frac{\phi(a)-\phi(0)}{a}=E\left(\frac{e^{iaX}-1}{a}\right).$$
You need some hypothesis on $X$ for the result to work, for
instance that $|X|$ has finite expectation. Certainly $(e^{iaX}-1)/a\to iX$
so to apply dominated convergence we need a function $f(X)$ with finite
expectation and with $|(e^{iaX}-1)/a|\le f(X)$ at least for $a$ in a
deleted neighbourhood of $0$. Does $f(X)=|X|$ work?
