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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 characteristic function of probability distribution at t = 0, $\phi^′(0)=iE[X]$? Any ideas?


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up vote 2 down vote accepted

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 $$\phi'(0)= \int_{-\infty}^\infty f(x)ix dx=i E(X).$$

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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?

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