Definition

The characteristic function of a probability measure \(\mu\) on \((\mathbb{R},\mathcal{R})\) is the function \(\phi:\mathbb{R}\rightarrow \mathbb{C}\) defined by \[ \phi(t)=\int e^{itx}\mbox{ }d\mu(x)=\int \cos(tx)\mbox{ }d\mu(x)+ i\int \sin(tx)\mbox{ }d\mu(x). \] If \(F\) is the distribution function corresponding to \(\mu\), or if \(X\) is a random variable with distribution \(\mu\), then we say that \(F\) or \(X\) has characteristic function \(\phi\). Of course we then may write \[ \phi(t)=\int e^{itx}dF(x)= \int \cos(tx)\mbox{ }dF(x)+ i\int \sin(tx)\mbox{ }dF(x), \] or \[ \phi(t)=E\left[e^{itX}\right]=E\left[\cos(tX)\right]+iE\left[\sin(tX)\right]. \]



Proposition (Elementary Properties of the Characterisric Function)

Let \(\phi\) be the characteristic function of a random variable \(X\). Then,

  1. \(\phi(0)=1\);

  2. \(|\phi(t)|\le 1\) (Modulus Inequality);

  3. \(\phi\) is uniformly continuous;

  4. \(\text{Re}\{\phi(t)\}=E[\cos(tX)]\) is an even function of \(t\), and \(\text{Im}\{\phi(t)\}=E[\sin(tX)]\) is odd;

  5. \(\phi(-t)=\overline{\phi(t)}\);

    • \(e^{-itX}=\cos(-tX)+i\sin(-tX)=\cos(tX)-i\sin(tX)=\overline{e^{itX}}.\)

  6. For \(a,b\in\mathbb{R}\), \(\phi_{aX+b}(t)=e^{ibt}\phi_X(at)\), in particular, \(\phi_{-X}(t)=\phi_X(-t)=\overline{\phi_X(t)}\).

    • \(e^{it(aX+b)}=e^{it(aX)}e^{ibt}=e^{i(at)X}e^{ibt}=e^{ibt}\phi_X(at)\).

  7. \(\phi_X\) is a real valued function \(\iff\) \(X\) is symmetrically distributed about the origin i.e., \(-X\sim X\).

    • 6번에 의해 \(\phi_X(t)\stackrel{\text{symm}}=\phi_{-X}(t)=\phi_X(-t)=\overline{\phi_X(t)}\implies \phi_X(t)=\overline{\phi_X(t)}.\)

  8. If \(X\) and \(Y\) are independent, then \(\phi_{X+Y}(t)=\phi_X(t)\phi_Y(t).\)



Lemma

For any \(x\in \mathbb{R}\), \[\begin{eqnarray*} \left|e^{ix}-\sum_{k=0}^n\frac{(ix)^k}{k!}\right|&\le& \min\left\{\frac{|x|^{n+1}}{(n+1)!}, \frac{2|x|^{n}}{n!} \right\}. \end{eqnarray*}\] Thus, we have at \(n=0\), and \(n=1\), \[\begin{eqnarray*} \left|e^{ix}-1\right| &\le& \min\left\{|x|, 2 \right\},\\ \left|e^{ix}-(1+ix) \right| &\le& \min\left\{\frac{|x|^2}{2}, 2|x| \right\}. \end{eqnarray*}\]



Lemma

If \(E[|X|^n]<\infty\) for some integer \(n\ge 1\), then \(\phi(t)=E[e^{itX}]\) satisfies

\[ \left| \phi(t)-\sum_{k=0}^n\frac{(it)^k}{k!}E(X^k) \right| \le E\left[\min\left\{\frac{|tX|^{n+1}}{(n+1)!}, \frac{2|tX|^{n}}{n!} \right\}\right]. \]


\[\begin{eqnarray*} \left| \phi(t)-\sum_{k=0}^n\frac{(it)^k}{k!}E(X^k) \right| &=&\left| E(e^{itX})-\sum_{k=0}^n\frac{(it)^k}{k!}E(X^k) \right| \\ &=& \left| E\left[e^{itX}-\sum_{k=0}^n\frac{(itX)^k}{k!}\right] \right| \\ &\le& E\left[\left| e^{itX}-\sum_{k=0}^n\frac{(itX)^k}{k!}\right|\right] \\ &\le& E\left[\min\left\{\frac{|tX|^{n+1}}{(n+1)!}, \frac{2|tX|^n}{n!} \right\}\right]. \end{eqnarray*}\]



Theorem

If \(E[|X|^n]<\infty\) for some integer \(n\ge 1\), then \[ \phi(t)=\sum_{k=0}^n\frac{(it)^k}{k!}E(X^k)+R_n(t), \] where \(R_n(t)=o(|t|^n)\) as \(t\rightarrow 0\).


\[ |R_n(t)|\le E\left[\min\left\{\frac{|tX|^{n+1}}{(n+1)!}, \frac{2|tX|^{n}}{n!} \right\}\right]= |t|^nE\left[\min\left\{ \frac{|t||X|^{n+1}}{(n+1)!}, \frac{2|X|^{n}}{n!} \right\}\right]. \] Now, \[ 0\le \min\left\{ \frac{|t||X|^{n+1}}{(n+1)!}, \frac{2|X|^{n}}{n!}\right\}\le \begin{cases}\frac{2|X|^{n}}{n!} &\mbox{integrable} \\ \frac{|t||X|^{n+1}}{(n+1)!}\rightarrow0 & \mbox{ as } t\rightarrow 0.\end{cases} \] Thus, by D.C.T \[ \frac{R_n(t)}{|t|^n}=E\left[\min\left\{ \frac{|t||X|^{n+1}}{(n+1)!}, \frac{2|X|^{n}}{n!} \right\}\right]\rightarrow 0 \mbox{ }\mbox{ }\mbox{ as }t\rightarrow 0. \]



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