Is this a characteristic function?

[malami]

1. If [tex]\phi[/tex] is a characteristic function, than is [tex]e^{\phi-1}[/tex] 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 [tex]e^{\phi-1}[/tex] is continuous at 0 and it's a limit of the product [tex]\phi_n(t)^n[/tex]
where
[tex]\phi_n(t)=1+\frac{\phi(t)-1}{n}[/tex],
and [tex]\phi_n[/tex] is obviously a characteristic function.

Is this correct?

2. Is [tex]\phi(t)=\frac{e^{-t^2}}{1+\sin^2(t)}[/tex] 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.

 

[blue_raver22]

1. Yes, your answer is correct. The product of two characteristic functions is also a characteristic function, and since e^{\phi-1} is a continuous function at 0 and a limit of characteristic functions, it is also a characteristic function.

2. No, \phi(t)=\frac{e^{-t^2}}{1+\sin^2(t)} is not a characteristic function. A characteristic function must satisfy certain properties, such as being continuous and having a value of 1 at 0. Your attempt to find the inverse Fourier transform is a good approach, but as you mentioned, it may be too difficult. Another approach could be to check if \phi(t) can be written as the Fourier transform of a probability distribution, which would also indicate that it is a characteristic function. However, in this case, it does not seem to have a corresponding distribution. Therefore, \phi(t) does not satisfy the necessary conditions to be a characteristic function.