# Characteristic functions

• Dec 28th 2010, 12:18 PM
slavert
Characteristic functions
Which of following functions are characteristic functions? Explain why.
a)$\displaystyle \varphi_{1}(t)=e^{i|t|}$
b)$\displaystyle \varphi_{2}(t)=(cos(t))^2$
c)$\displaystyle \varphi_{3}(t)=cos(t^2)$
d)$\displaystyle \varphi_{4}(t)=(1-|t|)I_{[-1,1]}$

Thanks for any help.
• Dec 29th 2010, 08:03 AM
Moo
Hello,

Consider the properties of a characteristic function !
- in the case of the characteristic function of a unidimensional rv -

- $\displaystyle \varphi(0)=1$
- $\displaystyle |\varphi(t)|\leq 1~,~ \forall t\in \mathbb R$
- $\displaystyle \varphi$ is a continuous function
• Dec 29th 2010, 08:37 AM
slavert
All of them satisfy those properties.
Now I know that $\displaystyle \varphi(-t)=\bar{\varphi(t)}$, so a) is not cf.
Characteristic function is uniformly countinuous, so c) either.
From Polya theorem d) either.
But what with b)?
• Dec 30th 2010, 12:11 AM
Moo
Polya's theorem states conditions for a characteristic function to be one of a continuous distribution (wrt Lebesgue's measure).
And by the way, this theorem provides sufficient conditions, not necessary conditions, so if a function doesn't satisfy one of the conditions, it doesn't mean that it's not a characteristic function.

So restate the conditions :
- $\displaystyle \varphi(0)=1$
- Hermitian
- bounded by 1
- uniform continuity

It seems like b) is a characteristic function : points 1,2,3 are satisfied and for the uniform continuity, it's differentiable and its derivative bounded by 1 (2cos(t)sin(t)=sin(2t))
• Dec 30th 2010, 01:06 AM
slavert
But Polya theorem states as follows:
" If φ is a real-valued continuous function which satisfies the conditions

1. φ(0) = 1,
2. φ is even,
3. φ is convex for t>0,
4. φ(∞) = 0,

then φ(t) is the characteristic function of an absolutely continuous symmetric distribution." (from wikipedia)
And then 3,4 are not satisfied for b).

I didn't know that other version of "Polya theorem". From which book or website is this one?
• Dec 30th 2010, 01:54 AM
Moo
Well, I think it's just a matter of me not expressing well enough...

I took the same version as wikipedia.

What I mean is that if the conditions of Polya's theorem are satisfied then fine, it's the cf of an absolutely continuous symmetric distribution.
But there are many other types of distributions.

If it doesn't satisfy these conditions, it doesn't mean it's not the characteristic function (it could be the caracteristic function of some discrete random variable, who knows)