Let's say we have a random variable $X$ of which the distribution is unknown. Now there are these general rules like $E[X + Y] = E[X] + E[Y]$ etc. But what if we would define
$ \quad Y = \dfrac{1}{X} $
and we would be interested in the expected value $E[Y]$ and the variance $Var(Y) = E[(Y-\bar{Y})^2]$?
Now, I do realize that $X$ might be zero and therefore it is undefined. But what if we would assume $X \neq 0$? My quick solution was to assume $X$ being a log-normal random variable. But I would prefer something more general. Maybe a power series expansion is possible?
3 Answers
If you write $X=\bar X+Z$, with $E[Z]=0$, then $$\begin{align} 1/X &= \sum_{k\geq0}(-1)^k\frac{Z^k}{\bar X^{k+1}} \\&= (1/\bar X)(1 - Z/\bar X + O(Z/\bar X)^2),\qquad (Z/\bar X \to0) \end{align}$$ and this series will converge by the ratio test if $|Z|<|\bar X|$ (in which case $X\neq0$ also). So then $$\begin{align}E[1/X]&=\sum_{k\geq0}(-1)^k\frac{E[Z^k]}{\bar X^{k+1}} \\&= \frac{1}{E[X]}+O\left(V[X]/\bar X^3\right),\qquad (Z/\bar X)\to0.\end{align}$$ The variance, in the limit $Z/\bar X\to0$, satisfies $$V[1/X]\to\frac{V[X]}{\bar X^4}.$$
Let's say that $X$ takes values in the interval $[a,b]$. (I realize that this is not the most general case, but this is an illustrative example.) We can then write
$$E[Y] = \int_a^b \frac{p(x)}{x} dx$$
where $p(x)$ is the PDF of $X$. The variance of $Y$ is then
$$ \mathrm{Var{Y}} = E[Y^2] - E[Y]^2 = \int_a^b \frac{p(x)}{x^2}dx - \left [ \int_a^b\frac{p(x)}{x} dx \right ]^2 $$
If you restrict $X$ to be a positive random variable, you can apply Jensens to conclude that $E[\frac {1}{X}] \geq \frac {1}{E[X]}$. Equality holds if $X$ is a constant. There is no upper bound of $E[X]$.
There is no general formula. Given just $E[X], Var(X)$, you can construct probabilities such that $E[Y], Var(Y)$ take on any value.
