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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 interesting –  Sasha Sep 29 '11 at 5:16

1 Answer 1

up vote 4 down vote accepted

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}}}, 
  Conjugate[cvec] . Outer[\[Phi][#1 - #2] & , xvec, xvec, 1] . cvec]]

Out[211]= -(1/2)/E^(sigma^2/4)
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Thank you. I wanted to make a pair of r.vs that are marginally Normal but not jointly Normal. I know there are standard examples, but I wanted to fabricate one through char functions. Anyway the positive definite condition in Bochners theorem is one hard nut to satisfy :( –  Isomorphism Sep 29 '11 at 5:36
@Isomorphism, try $\varphi_{X_1,X_2}(x_1,x_2)=\mathrm e^{-(x_1^2+x_2^2)/2}\cosh(x_1x_2)$. –  Did Sep 29 '11 at 5:59
@DidierPiau: Thanks! I was looking for that :) –  Isomorphism Sep 29 '11 at 19:28

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