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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)

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See e.g. this under Bochner's theorem. –  Stefan Hansen Jan 9 '13 at 6:42
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 $\mathbb{R}^d$.

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