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Characteristic function

$\newcommand{\rightarrowD}{\stackrel{d}{\longrightarrow}}$

Characteristic function

1. Characteristic function (CF)

We have seen various functions connected with a distribution, PMF, PDF, CDF and MGF. Here is the definition of MGF that you have learned in Probability I:
Definition: Moment generating function (MGF) The MGF of a random variable $X$ is defined as the function $M_X(t) = E(e^{tX})$ for whatever $t\in{\mathbb R}$ the expectation is finite. (Since $e^{tX}$ is a positive random variable, it's expectation is always defined.)
Out of these only CDF is guaranteed to exist for any random variable. And also uniquely determines a distribution (i.e., if the CDFs of two random variables match, then their distributions must also match). Unfortunately, CDF does not "play well" with convolution, i.e., if $X,Y$ are independent then there is no nice formula expressing the CDF of $X+Y$ in terms of those of $X$ and $Y.$ There is, however, one such function that combines all the good properties: it exists finitely for all random variables, it uniquely determines a distribution and "plays well" with convolution. Its definition is given below.
Definition: Characteristic function (CF) The characteristic function (CF) of a random variable $X$ is defined as the function $\xi_X:{\mathbb R}\rightarrow{\mathbb C}$ as $\xi_X(t) = E(e^{itX})$ for $t\in{\mathbb R}.$
You may be scared by the unexpected appearance of complex numbers inside the expectation! Let's learn about complex random variables.

1.1. Complex random variables

A complex random variable $Z$ means $Z = X+i Y,$ where $X,Y$ are (real) random variables. We define $E(Z)=E(X)+iE(Y)$ (and say $E(Z)$ does not exist if at least one of $E(X), E(Y)$ does not).

Since we have $e^{itX} = \cos (tX)+i\sin(tX)$, the characteristic function is just $\xi_X(t) = E(\cos(tX))+i E(\sin(tX)).$ Since $\cos$ and $\sin$ are both bounded, finite existence of the expectation is not a problem.

1.2. Complex calculus

For $f:{\mathbb R}\rightarrow{\mathbb C}$ write $f(x) = g(x) + i h(x) $ for $g,h:{\mathbb R}\rightarrow{\mathbb R}.$ Then differentiation and integration are defined in the obvious way: $$\begin{eqnarray*} f'(x) & = & g'(x) + i h'(x),\\ \int f(x)\, dx & = & \int g(x)\, dx + i\int h(x)\, dx. \end{eqnarray*}$$ From this it immediate follows (check!) that $\frac{d}{dx}e^{ix} = i e^{ix}$ and $\int e^{ix}\, dx = \frac 1ie^{ix}+$ arbit constant.

1.3. Examples

EXAMPLE 1: Find the CF of the degenerate distribution at $c.$

SOLUTION: Here $X = c $ with probability 1. So $\xi_X(t) = E(e^{it X}) = e^{itc}$ for $t\in{\mathbb R}.$ ■

EXAMPLE 2: Find $\xi_X(t)$ if $X\sim Poi(\lambda).$

SOLUTION: Take any $t\in{\mathbb R}.$ Then $$\begin{eqnarray*} \xi_X(t) & = & E(e^{it X})\\ & = & e^{-\lambda} \sum_{k=0}^ \infty e^{itk} \frac{\lambda^k}{k!}\\ & = & e^{-\lambda} \sum_{k=0}^ \infty \frac{(e^{it}\lambda)^k}{k!}\\ = e^{-\lambda} e^{\lambda e^{it}}\\ = e^{\lambda(e^{it}-1)}. \end{eqnarray*}$$ ■

EXAMPLE 3:  Find the CF of $X$ having density $f(x) = \left\{\begin{array}{ll} 3 e^{-3x}&\text{if }x>0\\ 0&\text{otherwise.}\end{array}\right. $

SOLUTION: $$E(e^{itX}) = 3\int_0^ \infty e^{itx}e^{-3x}\, dx = 3\int_0^\infty e^{(it-3)x}\, dx = \frac{3}{3-it}$$ for $t\in{\mathbb R}.$ ■

There is a "trick" called contour integration which may be used to find integrals of complicated complex functions. While the trick itself is beyond the scope of this course, we shall use the following two results which are easily proved using that trick.

TheoremCF of $N(0,1)$ distribution is $e^{-t^2/2}$ for $t\in{\mathbb R}.$ CF for $N(\mu,\sigma^2)$ distribution is $e^{it \mu-\sigma^2 t^2/2}$ for $t\in{\mathbb R}.$

Proof:Skipped.[QED]

TheoremCF for the Cauchy distribution is $e^{-|t|}$ for $t\in{\mathbb R}.$

Proof:Skipped.

[Outline of proof avoiding contour integration...]
This outline is not part of the syllabus. But let's have it anyway! $$\begin{eqnarray*} \xi(t) & = & \frac 1\pi\int_{-\infty}^\infty \frac{e^{itx}}{1+x^2}\, dx\\ & = & \frac 1\pi \left[ \int_{-\infty}^\infty \frac{\cos(tx)}{1+x^2}\, dx+i\int_{-\infty}^\infty \frac{\sin(tx)}{1+x^2}\, dx\right]. \end{eqnarray*}$$ We want to say that the second integral is 0 by virtue of being the integral of an odd function over an interval symmetric around 0. However, since the interval is unbounded, we need to ensure that each half of the integral is finite (else we run into $\infty-\infty$ or undefined$-$undefined). This is indeed true, since $|\sin (tx)|\leq 1,$ and so the right hand half is bounded by $\int_0^\infty \frac{dx}{1+x^2} = \frac \pi2.$

Now we attack the first integral. Since the integrand is an even function, so it is enough to compute $\int_0^\infty \frac{\cos(tx)}{1+x^2}\, dx.$ We shall use a trick here. We start by noticing that $$\frac{1}{1+x^2} = \int_0^\infty e^{-s(1+x^2)}\, ds.$$ So $$\begin{eqnarray*} \int_0^\infty \frac{\cos(tx)}{1+x^2}\, dx\\ & = & \int_0^\infty \cos(tx) \left[\int_0^\infty e^{-s(1+x^2)}\, ds\right]\, dx\\ & = & \int_0^\infty \int_0^\infty \cos(tx) e^{-s(1+x^2)}\, ds dx\\ & = & \int_0^\infty \int_0^\infty \cos(tx) e^{-s(1+x^2)}\, dx ds~~\left[\mbox{interchanging integrals!}\right]\\ & = & \int_0^\infty e^{-s} \int_0^\infty \cos(tx) e^{-sx^2}\, dx ds. \end{eqnarray*}$$ Now we may compute the inner integral by passing back to complex numbers as $$\begin{eqnarray*} \int_0^\infty \cos(tx) e^{-sx^2}\, dx & = & Re\left(\int_0^\infty e^{itx} e^{-sx^2}\, dx\right)\\ & = & Re\left(\int_0^\infty e^{-sx^2+itx}\, dx\right), \end{eqnarray*}$$ which is like integrating a normal density.

This outline will turn into a rigourous proof once we justify two points:
  1. interchanging the integrals
  2. integrating a complex function "like a normal density"
[QED]

Incidentally, the Cauchy distribution does not have an MGF (as $E(e^{tX})$ does not exist finitely for any value of $t$ except $t=0$).!

1.4. Problem set

EXERCISE 1: Find $\xi_X(t)$ if $X\sim Bern(p).$

EXERCISE 2: Find $\xi_X(t)$ if $X\sim Binom(n,p).$

EXERCISE 3: Find CF for the uniform distribution over $(-1,1).$

EXERCISE 4: Find CF for the Double Exponential distribution with rate $\lambda.$ This distribution has a density made by combining two exponential densities back to back (and halving to keep the total integral 1): $$f(x) = \frac \lambda2e^{-\lambda |x|},~~x\in{\mathbb R}.$$

Double exponential densities

2. Properties of CF

We start with two simple properties.
Theorem Let $\xi(t)$ be the CF of some random variable $X.$ Then $\xi(0) = 1$ and $\forall t\in{\mathbb R}~~|\xi(t)|\leq 1.$

Proof:The first one is trivial.

For the second, note that
Apply Pythagoras to the blue triangle
$$\begin{eqnarray*} |\xi(t)| & = & |E(e^{itX})|\\ & = & |E(\cos tX) + iE(\sin tX)|\\ & = & \sqrt{E^2(\cos tX)+E^2(\sin tX)}\\ & \leq & \sqrt{E(\cos^2 tX) + E(\sin^2 tX)}\\ & & [\because E^2 (U) \leq E(U^2)\mbox{ for any }U]\\ & = & \sqrt{E(\cos^2 tX + \sin^2 tX)} = 1. \end{eqnarray*}$$ [QED]

The following two theorems are what make CF useful.

Theorem If $X,Y$ are two random variables such that $\xi_X(t) \equiv \xi_Y(t)$, then $X$ and $Y$ must have the same distribution.

Proof:Follows from Levy inversion formula given in the next section.[QED]

Theorem If $X,Y$ are independent random variables, then $\xi_{X+Y}(t) = \xi_X(t)\xi_Y(t)$ for $t\in{\mathbb R}.$

Proof: Since $X,Y$ are independent, hence so are their functions $e^{itX}$ and $e^{itY}.$

Since their expectations are finite, so $E(e^{itX}\times e^{itY}) = E(e^{itX})\times E(e^{itY}).$ Hence the result.

[QED]

If we know a list of CFs for some standard distributions, then these two results often help us to identify if the convolution[?]
Given two distributions $D_1, D_2$ their convolution is the distribution of $X+Y$ where $X\sim D_1$ and $Y\sim D_2$ are independent. The convolution is often denoted by $D_1 * D_2.$
of two distributions in our list again belong to the list. Here is an example.

EXAMPLE 4:  Suppose that you are told that, for $a>0$, the distribution with density $f_a(x) = \left\{\begin{array}{ll}c x^{a-1}e^{-x}&\text{if }x>0\\ 0&\text{otherwise.}\end{array}\right.$ has CF $\xi_a(t) = (1-it)^{-a}$ for $t\in{\mathbb R}.$

Show that for $a,b>0$ we have $f_a* f_b = f_{a+b}.$

SOLUTION: You can of course show this directly using the definition of convolution. But that would require you to compute an integral. But it is trivial using CF: $\xi_a(t)\xi_b(t) = (1-it)^{-a} (1-it)^{-b} = (1-it)^{-(a+b)}$ for $t \in{\mathbb R}.$

Since CF uniquely determines the distribution, we get the result. ■

Theorem Let a random variable $X$ have characteristic function $\xi(t)$ for $t\in{\mathbb R}.$ Then $\xi(t)$ is a continuous function.

Proof:To show: $$\forall t\in{\mathbb R}~~\forall (t_n)\subseteq {\mathbb R} ~~(t_n\rightarrow t\Rightarrow \xi(t_n)\rightarrow \xi(t)).$$

Take any $t\in{\mathbb R}$ any $(t_n)\subseteq{\mathbb R}$ with $t_n\rightarrow t.$

To show $\xi(t_n)\rightarrow \xi(t),$ i.e., $E(e^{it_n X}) \rightarrow E(e^{it X}).$

Let $Y_n = e^{it_n X}$ and $Y = e^{it X}.$

Then $Y_n\rightarrow Y$ and $|Y_n|,|Y|\leq 1.$

Hence by DCT $E(Y_n)\rightarrow E(Y),$ as required. [QED]

2.1. Problem set

EXERCISE 5: Assuming the validity of term-by-term differentiation, suggest how you may find $E(X^n)$ from the characteristic function $\xi(t)$ of $X.$

EXERCISE 6: Let $X$ have CF $\xi_X(t).$ Let $Y = ax+b.$ Find $\xi_Y(t),$ the CF of $Y.$

EXERCISE 7: Let $X,Y$ be iid Cauchy. Let $\alpha\in[0,1]$ be any fixed number. Show that $\alpha X+(1-\alpha)Y\sim$ Cauchy. Hence conclude that if $X_1,...,X_n$ are iid Cauchy, then $\bar X_n\sim$ Cauchy, as well.

Hint:

Express $\bar X_n$ using $\bar X_{n-1}$, $X_n$ and $n.$

EXERCISE 8: Let $X$ be a random variable with characteristic function $\phi(t).$ Let $(a_n),(b_n)$ be two real sequences with $a_n-b_n\rightarrow 0.$ Then show that $\phi(a_n)-\phi(b_n)\rightarrow 0.$

Hint:

Imitate the proof of the continuity theorem above.

3. Inversion, density, continuity theorems

There are ways to recover the distribution function and density (if exists) of a distribution from its characteristic function. The following theorem gives a way to recover the density.

Fourier inversion Let $\phi$ be the CF corresponding to the the distribution function $F.$ If $\int_{-\infty}^\infty |\phi(t)|\,dt< \infty$, then $F$ has a density $f$ given by $$f(x) = \frac{1}{2\pi}\int_{-\infty}^\infty e^{-itx} \phi(t)\, dt.$$
Unfortunately, it is not always easy to compute the complex integral required in this theorem (except by using tricks like contour integration). This and this exercises in the problem set below mention two manageable cases.

Not all distributions have a density. So the above theorem is not general. However, the next theorem is.
Levy inversion Suppose that $X$ is a random variable with characteristic function $\phi.$ Then for all $a < b$ in ${\mathbb R},$ we have $$\lim_{T\rightarrow \infty}\frac{1}{2\pi}\int_{-T}^T\frac{e^{-ita}-e^{-itb}}{it}\phi(t)\, dt = P(a < X < b) + \frac 12\big(P(X=a)+P(X=b) \big).$$

Proof:SKipped.[QED]

A simple corollary to this theorem shows that characteristic functions indeed characterise distributions, i.e.,
distribution $\mapsto$ characteristic function
is one-one.

Characteristic functions behave nicely in terms of convergence in distribution, as the following theorem shows.
Continuity theorem Let $(X_n), X$ be random variables with characteristic functions $(\phi_n), \phi,$ respectively. Then $X_n\rightarrowD X$ if and only if $\phi_n\rightarrow \phi$ pointwise.

Proof: Skipped.[QED]

This theorem is a crucial ingredient for proving the central limit theorem (CLT).

The "if part" of this result may be strenghthened to some extent: If $(\phi_n)$ converges pointswise to some function $\phi$ that is continuous in a neighbourhood of 0, then $\phi$ must be the characteristic function of some random variable $X$ and $X_n\rightarrowD X.$

3.1. Problem set

EXERCISE 9: Check the Fourier inversion formula for $N(0,1).$

Hint:

We know from $N(\mu,1)$ density that $$\int_{-\infty}^\infty e^{-\frac 12(t-\mu)^2}\, dt = \sqrt{2\pi}.$$ This holds for all $\mu\in{\mathbb R}.$ Well, it also holds for all $\mu\in{\mathbb C}.$ You need to use this fact here.

EXERCISE 10: Check the Fourier inversion formula for the Cauchy distribution.

EXERCISE 11: Let $X_n$ be a random variable with characteristic function $\phi_n(t) = \left(1-\frac{it}{n}\right)^{-n}$ for $t\in{\mathbb R}.$ Show that $(X_n)$ converges in distribution. Identify the limiting distribution.

EXERCISE 12: Check the Levy inversion formula for $X\sim Bern\left(\frac 13\right)$ with $a = -1$ and $b=0.$