If I selected a real number at random from the interval (0.0,1.0), assuming a uniform distribution, the "expected value" would be 0.5. (I am not certain I am using the phrase correctly; I mean, if I took a large number of sample, I would expect the average of the samples to converge on 0.5 as I took more and more.)
But, if I looked at the reciprocal of that number, what would it converge on? My first guess, 2.0, is patently wrong. Perhaps there is no limit? Perhaps it's infinity (my bet)?
On
In general, for a positive random variable $X$, an application of Jensen's inequality to the convex function $1/x$ yields
$$E(1/X) \geq 1/E(X),$$
if the expectation exists.
In the case of the $U(0,1)$ uniform random variable
$$E(X) = \int_{0}^{1}xdx= \frac{1}{2}$$
$$E(X^{-1}) = \int_{0}^{1}x^{-1}dx= \infty$$
Where the second integral diverges due to the log singularity.
This just means that for the ideal model of a uniformly distributed random variable, the first moment of the reciprocal RV does not exist.
The pdf of a RV with inverse uniform distribution is given by
$$p(x)=\frac{1}{(b-a)x^2},\quad x\in[b^{-1},a^{-1}]$$
where $[a,b]$ is the range of the original uniformly distributed variable. In your case you have
$$p(x)=\frac{1}{x^2},\quad x\in[1,\infty)$$
with mean
$$m_X=\int_1^{\infty}\frac{dx}{x}\rightarrow\infty$$
So your bet was correct.
However, it's median is 2:
$$\int_1^2\frac{dx}{x^2}=\int_2^{\infty}\frac{dx}{x^2}$$