# $\mathbb{E} \left[ \frac{X}{c + X} \right]$ for known constant $c > 0$?

In general, describing expectations of ratios of random variables can be tricky. I have a ratio of random variables, but thankfully, it's nicely behaved due to known structure. Specifically, I have a univariate random variable $$X$$ whose support is non-negative reals $$\mathbb{R}_{\geq 0}$$, and I want to compute

$$\mathbb{E} \Bigg[ \frac{X}{c + X} \Bigg]$$

where $$c > 0$$ is a known constant. Is there an exact expression for this expectation, perhaps in terms of the moments (or centered moments) of $$X$$?

Edit: The first-order Taylor series approximation is $$\frac{\mathbb{E}_X[X]}{c + \mathbb{E}_X[X]}$$ but I was hoping for something better / exact if possible.

• Not sure if this is what you would consider "closed form", but perhaps you might consider taking the Taylor expansion of the function $\frac{X}{c+X}$ around $X = 0$? Under some regularity conditions, you should be able to swap the infinite sum and the expectation. Feb 25, 2022 at 1:53
• The first order Taylor series expansion yields E[X] / (c + E[X]). That's good, but I was hoping for an exact answer? Feb 25, 2022 at 1:56
• Find $1 - E[1/(1 + X/c)].$ How you do that depends on the details of the distribution: in some cases it can be worked out. Since usually there is some chance that $|X/c|\ge 1,$ it is hopeless to find an approximation based on (positive integral) moments of $X.$
– whuber
Feb 25, 2022 at 13:55
• As the current answer points out, the Taylor series does not converge for such values. As the degree of the moment grows, those values play larger roles in the moments, just making a bad situation worse and worse. I know--I have run into this in a great many situations. It's always a temptation to use the moments; it can work when $X/c$ has a truly tiny chance of exceeding $1;$ but I can't recall any actual problem where this approach worked decently.
– whuber
Feb 25, 2022 at 16:59
• There's very little relationship between the behavior of this expectation and convergence (or lack thereof) of the series.
– whuber
Feb 26, 2022 at 14:37

Breaking down $$\frac{X}{c+X} = 1 - \frac{1}{1+X/c},$$ and expanding in Taylor series, its expectation is $$E \left[ \frac{X}{c+X} \right] = 1 - E \left[ 1 - \frac{X}{c} + \frac{X^2}{c^2} - \frac{X^3}{c^3} + \cdots \right] \\ = E \left[ \frac{X}{c} - \frac{X^2}{c^2} + \frac{X^3}{c^3} - \cdots \right] \\ = \sum_{k=1}^\infty \frac{(-1)^{k-1}}{c^k} E[X^k]$$ Now, that Taylor series only converges for $$-c < X < c$$, and so if the support of $$X$$ lies outside of this range, this expansion is invalid.
$$E{\frac {X}{X+c}}=1-cE{\frac {1}{X+c}}=1-c\int {\frac 1 {x+c}}p_X(dx)=1-cH_{p_X}(-c)$$
where $$H_{p}$$ denotes the Hilbert transform of the density $$p$$. The wikipedia page gives a few examples of distributions for which the Hilbert transform has a closed form (in particular, the case of uniform distributions), but there is not any kind of explicit expression in general.
A general approximation without knowing more about the distribution of $$X\ge 0$$ seems difficult, you might be better of trying to approximate the distribution with some known distribution family, and then evaluating the expectation for that family. I will give a few examples below, but without giving details pf the integration.
Half-Cauchy distribution $$\frac{2 c \ln \! \left(c \right)+\pi}{\pi \left(c^{2}+1\right)}$$ Uniform distribution on $$[0,a]$$ $$-\frac{c \ln \! \left(a +c \right)-c \ln \! \left(c \right)-a}{a}$$ Exponential distribution with mean 1 $$1+\frac{\left(-2 \,\mathrm{Ei}_{1}\! \left(c \right)-\ln \! \left(c \right)-\ln \! \left(\frac{1}{c}\right)\right) {\mathrm e}^{c} c}{2}$$ where $$\mathrm{Ei}$$ is the Exponential integral.
Gamma distribution with scale 1 and shape $$k>0$$ $$c^{k} {\mathrm e}^{c} k \Gamma \! \left(-k , c\right)$$ and such results can be extended further, for instance with mixture modeling.