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Can anyone explain, how can I prove either $\phi(x) = |\cos t|$ is characteristic function or not? And which random variable has this characteristic function? Thanks in advance.

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Isn't there some result about characteristic functions $\phi$ with $|\phi(t)|=1$ for some $t \ne 0$? –  GEdgar Mar 20 '12 at 12:14
    
Obviously it is a periodic function: ф(t) = 1 with period of pi, but I don't understand what you exactly mean –  alex Mar 20 '12 at 12:35
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@GEdgar: There is. If $\phi$ is the characteristic function of a random variable $X$ and $|\phi(t)|=1$ for some $t\neq 0$ then $X$ is a discrete random variable. Furthermore, since $\phi(t)$ is real for every $t\in\mathbb{R}$, then $X$ must also be symmetric. –  Stefan Hansen Mar 20 '12 at 12:40
    
@StefanHansen: It is trivial to understand that random variable X is discrete with real values, but I have learned, that it isn't a characteristic function, and I need help to prove it –  alex Mar 20 '12 at 12:51

3 Answers 3

up vote 4 down vote accepted

Factoid 1: If a characteristic function is infinitely differentiable at zero, all the moments of the corresponding random variable are finite.

Factoid 2: If all the moments of a random variable are finite, the corresponding characteristic function is infinitely differentiable everywhere on the real line.

Factoid 3: The function $t\mapsto|\cos(t)|$ is infinitely differentiable at $t=0$ but not everywhere on the real line, for example not at $t=\pi/2$.

Ergo.

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Can you explain fact #2 please? –  alex Mar 20 '12 at 14:17
    
I mean why characteristic function must be differentiable everywhere on the real line, if all moments are finite? –  alex Mar 20 '12 at 14:33
    
@alex Because $\partial_t^m \phi_X(t) = \mathbb{E}( (i X)^m \mathrm{e}^{i t X})$, and $| \partial_t^m \phi_X(t) | \leqslant \mathbb{E}(|X|^m)$ which is finite by assumption. –  Sasha Mar 20 '12 at 16:48

W. Feller, An Introduction to Probability Theory and Applications, Volume I, XIX.4, Theorem 1.
A continuous function $\phi$ with period $2\pi$ is a characteristic function iff its Fourier coefficients (4.2) satisfy $\phi_k \ge 0$ and $\phi(0) = 1$.

$$ \phi_k = \frac{1}{2\pi}\int_{-\pi}^{\pi} e^{-i k \zeta} \phi(\zeta)\,d\zeta \tag{4.2} $$

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Period of $|cost|$ is $\pi$, not $2\pi$ –  alex Mar 20 '12 at 14:36
    
@alex If $\phi_X(t)$ is a characteristic function periodic with period $2 \pi$, $\phi_{2 X}(t) = \phi_X(2 t)$ is periodic with period $\pi$. –  Sasha Mar 20 '12 at 16:32

Since $\phi(t) = | \cos(t) |$ is periodic with period $\pi$ and even and if it is valid, it should correspond to a symmetric discrete random variable.

It is not hard to establish that: $$ | \cos(t) | = \frac{2}{\pi} + \frac{4}{\pi} \sum_{m=1}^\infty \frac{(-1)^{m-1}}{4 m^2-1} \cos(2 m t) $$ enter image description here

Comparing this to $\phi_X(t) = \sum_{m=-\infty}^\infty c_m \mathrm{e}^{i t m}$ we see that $c_4 = - \frac{2}{\pi} \cdot \frac{1}{15}$ is negative, thus can not be a probability of any random variable.

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