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

  • A.\(\dfrac{(-1)^4\partial^4\phi_X(t)}{\partial t_1\partial t_2^2\partial t_3}\) at \(\underline{t} = \underline{0}\)
  • B.\(\dfrac{(-1)^3\partial^3\phi_X(t)}{\partial t_1\partial t_2\partial t_3}\) at \(t_{1} = 1, t_{2} = 2, t_{3} = 1\)
  • C.\(\dfrac{(-1)^3\partial^4\phi_X(t)}{\partial t_1\partial t_2^2\partial t_3}\) at \(t_{1} = 1, t_{2} = 2, t_{3} = 1\)
  • D.\(\dfrac{\partial^4\phi_X(t)}{\partial t_1\partial t_2^2\partial t_3}\) at \(t_{1} = 1, t_{2} = 2, t_{3} = 1\)

Answer: A

\(E X_{1}X_{2}^{2}X_{3}\) is \((-1)^{4}\,\partial^{4}\phi_{X}/\partial t_{1}\partial t_{2}^{2}\partial t_{3}\) at \(\underline{t} = \underline{0}\).

Differentiating \(\phi_{X}(t) = E e^{i t'X}\) brings down a factor \(iX_{j}\) for each derivative in \(t_{j}\). A fourth-order mixed derivative at \(t = 0\) is \(i^{4}E[X_{1}X_{2}^{2}X_{3}]\), so the moment is \(i^{-4} = (-i)^{4}\) times the derivative at 0.

Since \((-i)^{4} = 1 = (-1)^{4}\), the expression in option A equals the moment. Moments come from derivatives at the origin only; the options evaluated at (1, 2, 1) give something else, and a third-order derivative cannot produce a fourth-order moment.

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