Showing characteristic function is infinitely differentiable

Question

Suppose that X is a real-valued random variable. Suppose that there exists a constant M > 0 such that the support of X lies entirely in the interval $[-M,M]$. Let $\phi$ denote the characteristic function of X. Show that $\phi$ is infinitely differentiable.

If infinitely differentiable is equivalent to absolutely continuous, then

$$\int_{-M}^{M}| \phi(t)|\;dt < \infty$$

Answer 1

I'm not sure what you mean exactly by support here, but if X lies in $[-M,M]$ a.s. then the following works:

First, have a look at the statement of Theorem 6.6 (pg 107) of: http://stat.uconn.edu/~boba/stat6894/probabilityI.pdf

As X is bounded a.s., $E|X|^k$ is finite for every $k>0$. Hence, by the theorem, $\phi^{(k)}(t)$ exists for every $k>0$.

Answer 2

Here is an elementary demonstration that uses only the definition of derivative and the simplest form of Taylor's theorem. Let's begin, therefore, by quoting these:

(1) The derivative of a function $f$ at the value $t,$ written $f^\prime(t),$ is any number for which, when $|h|$ is sufficiently small, $$\left|f(t+h)-f(t) - hf^\prime(t)\right| \le c |h|^2$$ for some number $c$ (not dependent on $h$).

Taylor's Theorem applied to the exponential function states

(2) Given $M\gt 0$, there exists a number $C_M$ for which $|e^z - 1 - z| \le C_M|z|^2$ for all complex numbers $z$ for which $|z|\le M.$

Let's generalize the question by considering functions $g$ that are integrable and bounded in size on $[-M,M]$ by, say, the number $G.$ Define

$$f_{g,X}(t) = E\left[g(X) e^{itX}\right].$$

Since the support of $X$ is on $[-M,M],$ we may compute all such expectations by integrating from $-M$ to $M.$ Therefore

$$|f_{g,X}(t)| = \left|E\left[g(X) e^{itX}\right]\right| \le E\left[\left|g(X) e^{itX}\right|\right] = E[G] = G.$$

This shows $f_{g,X}(t)$ exists and is finite for all real $t.$

Write $F$ for the distribution function of $X.$ In the following I will abuse notation by writing (for instance) "$ixg$" for the function $x\to ixg(x).$

Using only linearity of expectation, the inequality $|E[f(X)]| \le E[|f(x)|],$ and algebraic properties of the exponential, we may compute

$$\eqalign{\left|f_{g,X}(t+h) - f_{g,X}(t) - hf_{ixg,X}(t)\right| & = \left|E\left[g(x)e^{i(t+h)x} - g(x)e^{itx} - h ixg(x) e^{itx}\right]\right| \\ &= \left|E\left[ixg(x)e^{itx}\left(e^{ihx}-1-ihx\right)\right]\right| \\ &\le E\left[\left|x\right|\left|g(x)\right|\left|e^{ihx}-1-ihx\right|\right] \\ &\le E\left[|x|GC_M\left|ihx\right|^2\right] \\ &= |h|^2 GC_M E\left[|x|^3\right] \\ & \le |h|^2 GC_M M^3. }$$

Thus, in the definition of the derivative $(1)$ we may take $c=GC_M M^3,$ demonstrating that

$$f_{g,X}(t)^\prime = f_{ixg, X}(t).$$

By induction on the number of derivatives it is immediate that

$$f_{g,X}(t)^{(n)} = f_{(ix)^n g, X}(t).$$

Applying this to the case $g=1$ shows that $\phi(t) = f_{1,X}$ has derivatives of all orders and that $\phi^{(n)}(t) = f_{(ix)^n, X}(t),$ QED.

Answer 3

The characteristic function is

$$\varphi(t) = E[e^{itX}] = \int_{-\infty}^{\infty} e^{itx}f(x) dx $$

Let's see what happens when we differentiate

$$ \begin{array}{} \frac{d^n}{dt^n} \varphi(t) &=& \frac{d^n}{dt^n} \int_{-\infty}^{\infty} e^{itx}f(x) dx \\ &=& \int_{-\infty}^{\infty} \frac{d^n}{dt^n} e^{itx}f(x) dx \\ &=& i^n \int_{-\infty}^{\infty} e^{itx}x^nf(x) dx \\ \end{array} $$

For your case (finite domain) this exists for any $n$ and $t$ since it relates to the Fourier transform of the function $x^nf(x)$ which is integrable (because $f(x)$ is nonzero on a finite domain) and thus exists.

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A very common expression is for $t=0$

$$ \begin{array}{} \frac{d^n}{dt^n} \varphi(0) &=& i^n \int_{-\infty}^{\infty} x^nf(x) dx = i^n E[X^n] \\ \end{array} $$

which expresses the equivalence of the n-th derivative of the characteristic function in the point zero with the n-th moment of the variable $X$.

You could reverse this equivalence: If the higher order moments exist (and they exists when the variable is bounded, e.g. from $-M$ to $M$) then the higher order derivatives of the characteristic function must exist.

But, when you relate it to this equivalence expression then this is only true for the derivatives of the characteristic function in the point $t=0$.

(Although you could argue more generally for the existence of the derivative on the entire real line when higher order moments exist. For instance by arguing that $x^nf(x)$ is an integrable function when the n-th moment exists. But it would be using a somewhat more general property of the characteristic function than the simple equivalence $\frac{d^n}{dt^n} \varphi(0) = i^n E[X^n]$ )