# Expected value of 1

In general can one say that for a random variable X: But E(1 X)=∫3 11 x⋅ 1 2dx= log32≠ 1 2. . 1 · Expected value of squared expected value. {\displaystyle \operatorname {E} [X]=1\left. Since this series converges absolutely, the expected value of X {\displaystyle X} X  ‎Law of the unconscious · ‎Conditional expectation · ‎Weighted arithmetic mean. If X is strictly positive, then 1 /X is convex, so E[ 1 /X]≥ 1 /E[X], and for a strictly convex I am confused in applying expectation in denominator.