I am well aware (as for instance discussed here https://math.stackexchange.com/questions/910846/isittrueingeneralthate1x1ex) that for an arbitrary random variable $X$ it does not hold that $\mathbb{E}[1/X]=1/\mathbb{E}[X]$. However, when one cannot calculate $\mathbb{E}[1/X]$ directly, is there a good (rigorous) way to approximate $\mathbb{E}[1/X]$ (ideally, regardless of the distribution of $X$)?

3$\begingroup$ What information do you have available regarding $X$? If you have central moments up to some order, and they fall off quickly compared to $\mu^n$, then you can use a series expansion of $\frac{1}{X}$ around $X = \mu$, for example. $\endgroup$ – student Jan 9 at 11:27

$\begingroup$ Unfortunately I do not have much information about $X$  I only really know its expectation, which is why I am primarily interested in approximations that are independent of its distribution. However any inputs or ideas would be helpful. $\endgroup$ – Silvia Jan 9 at 12:59

2$\begingroup$ Pick e.g. $X \sim \Gamma(\frac{\mu \nu}{\mu\nu  1}, \frac{\nu}{\mu\nu  1})$, then $\mathrm{E}[X] = \mu$ but $\mathrm{E}[1/X] = \nu$. $\endgroup$ – student Jan 9 at 13:54

$\begingroup$ @student, are you trying to imply that no approximations in term of E[X] can exist? $\endgroup$ – Silvia Jan 11 at 10:25

$\begingroup$ Yes, and even in the bounded case you can find such examples using basically any twoparameter distribution, e.g. the beta distribution with $\alpha = \frac{\mu (\nu1)}{\mu\nu1}, \beta = \frac{(1  \mu)(\nu1)}{\mu\nu1}$. The bounds from Iosif's answer are likely as good as it gets. $\endgroup$ – student Jan 11 at 11:37
Assume that $X\ge0$. Then, by Jensen's inequality for the convex function $x\mapsto1/x$, \begin{equation} E\frac1X\ge\frac1\mu, \end{equation} where $\mu:=EX$. Thus, we have the lower bound $\frac1\mu$ on $E\frac1X$; this bound is exact, as it is attained when $X$ is a constant.
On the other hand, it is easy to see that there is no finite upper bound on $E\frac1X$ in terms of $\mu$ only. Indeed, for any real $\mu>0$, let $P(X=0)=1/2=P(X=2\mu)$. Then $E\frac1X=\infty$ while $EX=\mu$. Even if we assume that $X>0$, still there is no finite upper bound on $E\frac1X$ in terms of $\mu$ only. Indeed, for any real $\mu>0$ and any $a\in(0,\mu)$, let $P(X=a)=1/2=P(X=2\mua)$. Then $EX=\mu$ while $E\frac1X=\frac1{2a}+\frac1{2(2\mua)}\to\infty$ as $a\downarrow0$.
If $X$ is known to be bounded away from $0$, so that $X\ge a$ for some real $a>0$, then trivially $1/X\le1/a$ and hence \begin{equation} E\frac1X\le \frac1a. \end{equation} The latter upper bound on $E\frac1X$ is exact here for each feasible value of $\mu=EX$, even though this bound does not even involve $\mu$. Indeed, take any real $\mu\ge a$. For any real $u>\mu$, let $P(X=u)=p=1P(X=a)$, where $p:=\frac{\mua}{ua}\in[0,1)$, so that $EX=\mu$. On the other hand, letting $u\to\infty$, we have $p\to0$ and hence $E\frac1X=\frac1a\,(1p)+\frac1u\,p\to\frac1a$, so that the upper bound $\frac1a$ on $E\frac1X$ is attained in the limit.
Finally, if $X$ is known to be bounded away both from $0$ and $\infty$, so that $a\le X\le A$ for some positive real $a,A$ such that $a<A$, then, by the convexity of the function $x\mapsto1/x$, we have $\frac1X\le\frac{A+aX}{Aa}$ and hence \begin{equation} E\frac1X\le\frac{A+a\mu}{Aa}. \end{equation} The latter bound on $E\frac1X$ is again exact, as it is attained when $P(X=A)=p=1P(X=a)$, where $p:=\frac{\mua}{Aa}\in[0,1]$, so that $EX=\mu$.

$\begingroup$ Thank you for your answer. Unfortunately my X is not bound away from zero  it runs between $0$ and $\sqrt{2}$ (included), although for practical purposes I can consider it strictly between $0$ and $\sqrt{2}$, however I still do not have a nonzero lower bound. $\endgroup$ – Silvia Jan 11 at 10:27

$\begingroup$ @Silvia : Then, as shown in my answer, there is no finite upper bound on $E\frac1X$ in terms of $\mu$ only. That is, to be able to have a finite upper bound, you need to know more than just $\mu$. $\endgroup$ – Iosif Pinelis Jan 11 at 14:06