Is this a characteristic function?

[malami]

1. If \phi is a characteristic function, than is e^{\phi-1} also a characteristic function?

I know some general rules like that a product or weighted sum of characteristic functions are also characteristic functions, also a pointwise limit of characteristic functions is one if it's continuous at 0.

So I think the answet is yes, because e^{\phi-1} is continuous at 0 and it's a limit of the product \phi_n(t)^n
where
\phi_n(t)=1+\frac{\phi(t)-1}{n},
and \phi_n is obviously a characteristic function.

Is this correct?

2. Is \phi(t)=\frac{e^{-t^2}}{1+\sin^2(t)} a characteristic function?
Here I can only prove, that it's not a characteristic function from a discrete distribution. I tried integrating it to get the inverse Fourier transform, but it's too difficult.

 

[HallsofIvy]

I know of several different definitions of "characteristic function". What is your definition?

 

[mathman]

I know of several different definitions of "characteristic function". What is your definition?

In probability theory, the characteristic function is the Fourier transform of the density function. More generally it is ∫e^itxdF(x), where F(x) is the distribution function.

 

[mathman]

The answer for 1. is yes. You can use the weighted sum, where the weights are > 0 and the sum = 1.
e^φ-1 = 1/e{1 + φ + φ²/2 + ...φⁿ/n! ...}.
Each φⁿ is a characteristic function (note 1 is the ch. f. of unit dist. at 0) and 1/n! sums to e.