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Let X and Y be two random variables defined on the same probability space such that for each \(y \gt 0\), the conditional density function of X given \(Y = y\) is :

\[ \begin{aligned} f(x/y) &= \sqrt{\frac{y}{2\pi}}\, e^{-yx^{2}/2}, \\ &y \gt 0,\ x \in R. \end{aligned} \]

Let

\[ g(y) = \frac{1}{\sqrt{2\pi y}}\, e^{-y/2},\ y \gt 0. \]

Which of the following statements is more appropriate ?

  • A.E(X|Y) = 0
  • B.E(E(X|Y)) = 0
  • C.E(X) = 0
  • D.\(E(X) \ne E(E(X|Y))\)

Answer: A

The most appropriate statement is E(X|Y) = 0.

Given \(Y = y\), X is normal with mean 0 and variance \(1/y\), so \(E(X|Y) = 0\) exactly. The marginal of Y is \(g(y)\), the chi-square density with 1 degree of freedom. Integrating out \(y\):

\[ \begin{aligned} f_{X}(x) &= \frac{1}{2\pi}\int_{0}^{\infty} e^{-y(1 + x^{2})/2}dy \\ &= \frac{1}{\pi(1 + x^{2})} \end{aligned} \]

So X has the standard Cauchy distribution, and \(E(X)\) does not exist. That rules out \(E(X) = 0\) and \(E(X) \ne E(E(X|Y))\), and makes \(E(E(X|Y)) = 0\) misleading: the iterated expectation is 0 although the expectation it should equal does not exist, since the tower property needs \(E|X| \lt \infty\).

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