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I know the result that says "There is a bijection between distribution functions and characteristic functions".

So I was wondering if there are necessary or sufficient conditions on the characteristic functions so that it is a char. function of a valid random variable. I need conditions that are easy to verify.

For instance, is $\varphi_{X_1,X_2}(x_1,x_2) = (1 + |x_1 x_2| e^{-\frac{(x_1^2 + x_2^2)}2 \sigma^2} )$ a valid characteristic function?

Thank you.

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    You might find [this](http://math.stackexchange.com/questions/54806/how-can-i-prove-this-expression-not-to-be-a-characteristic-function) interesting2011-09-29

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Bochner's theorem characterizes if a characteristic function corresponds to the random variable, but it's not an easy criterion.

It says that for all $x_1, \ldots, x_m \in \mathbb{R}^2$ and $c_1,\ldots,c_m \in \mathbb{C}$ the characteristic function $\phi$ should verify $ \sum_{i=1}^m \sum_{j=1}^m \bar{c}_i c_j \phi(x_i - x_j) \ge 0 $

For the characteristic function in question, choose $m=2$, and $c_1 = -1$, $c_2=1$, and $x_1 = \{1, \frac{1}{2}\}$ and $x_2 = \{ \frac{1}{2}, 1 \}$. Then the resulting sum is $-\frac{1}{2} \exp(-\frac{\sigma^2}{4} )$.

Here is Mathematica code:

In[210]:= \[Phi][{w1_, w2_}] :=   1 + Abs[w1*w2]*Exp[(-(w1^2 + w2^2)/2)*sigma^2]  In[211]:= With[{cvec = {-1, 1}, xvec = {{1, 1/2}, {1/2, 1}}},     Expand[   Conjugate[cvec] . Outer[\[Phi][#1 - #2] & , xvec, xvec, 1] . cvec]]  Out[211]= -(1/2)/E^(sigma^2/4) 
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    @DidierPiau: Thanks! I was looking for that :)2011-09-29