Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Suppose $X$ is a continuous random variable with pdf $f_X(x)$. We can compute its characteristic function as $\varphi_X(t)=\mathbb{E}[e^{itX}].$

Question: Given a function, say $\psi(t)$, how does one show that it is a characteristic function?

(Typed this on my phone - my apologies if there's poor formatting)

share|improve this question
3  
See e.g. this under Bochner's theorem. –  Stefan Hansen Jan 9 '13 at 6:42
1  
Look up Bochner’s theorem. A function $\phi$ is a characteristic function of some random variable iff $\phi$ is positive definite with $\phi(0) = 1$ and continuity at $0$. –  user17762 Jan 9 '13 at 6:43
    
Thanks! I'll read up on that. –  pedrosuavo Jan 9 '13 at 6:44

1 Answer 1

up vote 3 down vote accepted

Let me just state the theorem I linked to in my comment, so that this question does not go unanswered.

Bochner's theorem

If $\varphi:\mathbb{R}^d\to \mathbb C$ is a complex-valued function with $\varphi(0)=1$, continuous at $0$ and nonnegative-definite in the sense that for $n\geq 1$ we have that $$ \sum_{j=1}^n\sum_{k=1}^n\varphi(z_j-z_k)\,\xi_j\bar{\xi}_k\geq 0,\quad \text{for }\;z_1,\ldots,z_n\in\mathbb{R}^d,\;\xi_1,\ldots,\xi_n\in\mathbb C, $$ then $\varphi$ is the characteristic function of a distribution (random variable) on $\mathbb{R}^d$.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.