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Suppose \(\phi(\cdot)\) is a characteristic function of a random variable with distribution function F. Then :

  • A.\[ \begin{aligned} F(b) - F(a) &= \lim_{u \to \infty} \frac{1}{2\pi} \\ &\int_{-u}^{u} \frac{[\exp(-iua) - \exp(-iub)]\, \phi(u)}{iu}\, du \end{aligned} \]
  • B.\[ \begin{aligned} F(b) - F(a) &= \lim_{u \to \infty} \\ &\int_{-u}^{u} \frac{[\exp(-iua) - \exp(-iub)]\, \phi(u)}{iu}\, du \end{aligned} \]
  • C.\[ \begin{aligned} F(b) - F(a) &= \lim_{u \to \infty} \frac{1}{2\pi} \\ &\int_{-u}^{u} \frac{[\exp(iua) - \exp(iub)]\, \phi(u)}{iu}\, du \end{aligned} \]
  • D.\[ \begin{aligned} F(b) - F(a) &= \lim_{u \to \infty} \frac{1}{2\pi} \\ &\int_{-u}^{u} \frac{[\exp(-iub) - \exp(-iua)]\, \phi(u)}{iu}\, du \end{aligned} \]

Answer: A

The correct inversion formula is option A:

\[ \begin{aligned} F(b) - F(a) &= \lim_{u \to \infty} \frac{1}{2\pi} \\ &\int_{-u}^{u} \frac{e^{-iua} - e^{-iub}}{iu}\phi(u)\, du \end{aligned} \]

This is Levy's inversion theorem, valid when \(a\) and \(b\) are continuity points of F. It comes from \(\frac{1}{2\pi}\int_{a}^{b} e^{-iux}\, dx = \frac{e^{-iua} - e^{-iub}}{2\pi iu}\) and Fubini's theorem.

Option B lacks the factor \(1/2\pi\), option C has the wrong sign in the exponents, and option D reverses the order of the two terms, which gives \(F(a) - F(b)\).

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