# Approximating $\mathbb{E}[1/X]$

I am well aware (as for instance discussed here https://math.stackexchange.com/questions/910846/is-it-true-in-general-that-e1-x-1-ex) 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 $$X$$)?

• 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. – student Jan 9 at 11:27
• 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. – Silvia Jan 9 at 12:59
• Pick e.g. $X \sim \Gamma(\frac{\mu \nu}{\mu\nu - 1}, \frac{\nu}{\mu\nu - 1})$, then $\mathrm{E}[X] = \mu$ but $\mathrm{E}[1/X] = \nu$. – student Jan 9 at 13:54
• @student, are you trying to imply that no approximations in term of E[X] can exist? – Silvia Jan 11 at 10:25
• Yes, and even in the bounded case you can find such examples using basically any two-parameter distribution, e.g. the beta distribution with $\alpha = \frac{\mu (\nu-1)}{\mu\nu-1}, \beta = \frac{(1 - \mu)(\nu-1)}{\mu\nu-1}$. The bounds from Iosif's answer are likely as good as it gets. – student Jan 11 at 11:37

Assume that $$X\ge0$$. Then, by Jensen's inequality for the convex function $$x\mapsto1/x$$, $$$$E\frac1X\ge\frac1\mu,$$$$ where $$\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\mu-a)$$. Then $$EX=\mu$$ while $$E\frac1X=\frac1{2a}+\frac1{2(2\mu-a)}\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$$ and hence $$$$E\frac1X\le \frac1a.$$$$ The latter upper bound on $$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=1-P(X=a)$$, where $$p:=\frac{\mu-a}{u-a}\in[0,1)$$, so that $$EX=\mu$$. On the other hand, letting $$u\to\infty$$, we have $$p\to0$$ and hence $$E\frac1X=\frac1a\,(1-p)+\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, then, by the convexity of the function $$x\mapsto1/x$$, we have $$\frac1X\le\frac{A+a-X}{Aa}$$ and hence $$$$E\frac1X\le\frac{A+a-\mu}{Aa}.$$$$ The latter bound on $$E\frac1X$$ is again exact, as it is attained when $$P(X=A)=p=1-P(X=a)$$, where $$p:=\frac{\mu-a}{A-a}\in[0,1]$$, so that $$EX=\mu$$.
• 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 non-zero lower bound. – Silvia Jan 11 at 10:27
• @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 $\mu$. – Iosif Pinelis Jan 11 at 14:06