Discussion
Which of the following is a characteristic function of a random variable ?
Answer: B
The characteristic function is \(\Phi(t) = \cos(t/2)\).
If X takes the values \(\pm\frac{1}{2}\) with probability \(\frac{1}{2}\) each, then \(E[e^{itX}] = \frac{1}{2}(e^{it/2} + e^{-it/2}) = \cos(t/2)\).
The others fail. A characteristic function satisfies \(\Phi(0) = 1\) and \(|\Phi(t)| \le 1\): \(t/(1 + t^{2})\) gives \(\Phi(0) = 0\), and \(2/(1 + 2t)\) gives 2 at \(t = 0\). \(\cos(t^{2})\) passes those tests but has \(\Phi''(0) = 0\), which would mean \(E[X^{2}] = 0\), so X = 0 and \(\Phi \equiv 1\), a contradiction.
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