Discussion
Suppose \(\phi(t) = \exp(-|t|)\) is a characteristic function of a random variable X. Then the distribution of X is :
Answer: C
X has Student's t distribution with 1 degree of freedom, that is, the standard Cauchy distribution.
The standard Cauchy density \(f(x) = \frac{1}{\pi(1 + x^{2})}\) has characteristic function \(e^{-|t|}\), and the t distribution with one degree of freedom is exactly this Cauchy law. The non-differentiability of \(e^{-|t|}\) at 0 matches the fact that the Cauchy distribution has no mean.
The other options are the usual mix-up of roles. The normal has characteristic function \(e^{-t^{2}/2}\). The Laplace density \(\frac{1}{2}e^{-|x|}\) has characteristic function \(1/(1 + t^{2})\): density and characteristic function are swapped relative to the Cauchy. A gamma law has \((1 - it/\beta)^{-\alpha}\), which is complex-valued, while \(e^{-|t|}\) is real, as it must be for a symmetric law.
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