求证明John Hull书中提到的ln[E(x)] > E[ln(x)]ln[E(x)] equal E[ln(x)]+

John Hull在讲解B-S模型一章，在the expected return一小节中用到一个结论：“ in fact,ln[E(x)] > E[ln(x)]”，如何证明啊

This is an application of Jenson's Inequality, which says that the expectation of a strictly concave-down function of a random variable is smaller than the function of the expectation of the random variable.
A function f is said to be strictly concave-down if for any x and y within its domain and any a between 0 and 1, f(a*x+(1-a)*y)>a*f(x)+(1-a)*f(y). Of course, the domain must be a convex set so that a*x+(1-a)*y falls in the domain of f.
Jenson's Inequality follows from the definition above, though not really directly. However, please note that a convex combination (the form a*u+(1-a)*v) is the expectation of a two-point distribution taking value either u or v. Expand two-point case to finitely many point case, then to countably infinitely many point case, and finally to Lebesgue integration case. Just apply definitions and note that expectation is a special integral, and Jenson's Inequality will be reached.
For this particular case, note that the function ln is strictly concave-down.

简单讲，这是由凸函数的性质决定的。简单说，函数值的平均大于平均的函数值（你可以想象一个凸函数），而Ln是凸函数。
具体证明，请参阅http://en.wikipedia.org/wiki/Jensen's_inequality

谢谢两位这么及时的解答，但最佳答案只能给一人，就给KevinOu朋友，同样感谢jerryren

之前用不等式证明陷入困境，看来要补的数学还好多！