s*********o
发帖数: 14 | 1 已知 x1, x2, x3 ... are independent and identically distributed with mean=1.
求 （a+x1+x1*x2+x1*x2*x3+x1*x2*x3*x4+...)^(-1) 的期望值。谢谢各位大侠先。 | i******n
发帖数: 6 | 2 难道不是 1/(a+n) 么？
1.
【在 s*********o 的大作中提到】
: 已知 x1, x2, x3 ... are independent and identically distributed with mean=1.
: 求 （a+x1+x1*x2+x1*x2*x3+x1*x2*x3*x4+...)^(-1) 的期望值。谢谢各位大侠先。
| s*********o
发帖数: 14 | 3 一般来说，E[1/X] is not equal to 1/E[X], the former is bigger than or equal
to the latter by Jensen's inequality. |
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