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Let \(\phi_{X}(t)\) be the characteristic function of \(X \sim N_{3}(\mu, \Sigma)\). Then \(EX_{1}X_{2}^{2}X_{3}\) is given by :

  • A.\((-1)^{4}\partial^{4}\phi_{X}(t)\) / \((\partial t_{1}\partial t_{2}^{2}\partial t_{3})\) at \(t = 0\)
  • B.\((-1)^{3}\partial^{3}\phi_{X}(t)\) / \((\partial t_{1}\partial t_{2}\partial t_{3})\) at \(t_{1} = 1, t_{2} = 2, t_{3} = 1\)
  • C.\((-1)^{3}\partial^{4}\phi_{X}(t)\) / \((\partial t_{1}\partial t_{2}^{2}\partial t_{3})\) at \(t_{1} = 1, t_{2} = 2, t_{3} = 1\)
  • D.\(\partial^{4}\phi_{X}(t)\) / \((\partial t_{1}\partial t_{2}^{2}\partial t_{3})\) at \(t_{1} = 1, t_{2} = 2, t_{3} = 1\)

Answer: A

The moment is \((-1)^{4}\partial^{4}\phi_{X}(t)/\partial t_{1}\partial t_{2}^{2}\partial t_{3}\) at \(t = 0\).

For \(\phi_{X}(t) = E[e^{i t' X}]\), each differentiation with respect to \(t_{j}\) brings down a factor \(iX_{j}\). Differentiating once in \(t_{1}\), twice in \(t_{2}\) and once in \(t_{3}\) and setting \(t = 0\) gives

\[ \begin{aligned} &\frac{\partial^{4}\phi_{X}}{\partial t_{1}\partial t_{2}^{2}\partial t_{3}}\Big|_{0} \\ &= i^{4}E[X_{1}X_{2}^{2}X_{3}] \end{aligned} \]

Since \(i^{4} = 1 = (-1)^{4}\), the moment equals the fourth derivative at the origin, as option A states. Moments come from derivatives at \(t = 0\), not at \(t = (1, 2, 1)\), and the order of the derivative must match the total power 4, which rules out the third derivative in option B.

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