# Expected value of the shifted inverse of a binomial random variable, and application

Here is an exercise given by a colleague to a student :

Let $$X\hookrightarrow B(n,p)$$ and $$Y=\frac{1}{X+1}$$. Find $${\rm E}(Y)$$.

It is not very difficult to prove that the answer is $${\rm E}(Y) = \frac{1-q^{n+1}}{p(n+1)}$$ where $$q=1-p$$. But the answer can also be written $${\rm E}(Y) = \frac{1+q+q^2+\dots+q^n}{n+1}$$

First question: Is there any meaning to this form, which looks very much like a mean value of some sort? Or maybe another proof of this result which explains it in a more direct way?

Second question : Is there some context which could make this exercise more "concrete"?

• In other words, is there a "natural" (whatever that means) explanation of the identity $$E\left(\frac1{X+1}\right)=E(q^U)$$ for $X$ binomial $(n,p)$, $U$ uniform on $\{0,1,\ldots,n\}$, and $q=1-p$? Interesting question... – Did Jun 21 '18 at 5:55

Here is an application of your question to a setting that highly interests me. In queueing there is the notion of utilization which is the long-run fraction of time a server is busy serving demand. Consider a Markovian service setting where demand arrives according to Poisson with $$\lambda=1$$ and there are $$N+1$$ servers with Exponential service time and mean rate $$\mu=1$$, where $$N\sim\text{Bin}\left(n,p\right)$$.
An application of this is the utilization of an Uber driver; the number of Uber drivers on a given instance is uncertain as it cannot be mandated or enforced by the firm. Given $$k$$ drivers choose to drive on the streets, their utilization would be $$\frac{\lambda}{\mu\cdot k}=\frac{1}{k}$$. So, in this setting if we assume the above model by the law of total probability the utilization of an Uber driver would be $$E\left[\frac{1}{1+N}\right]$$. Given this, I would be interested to know of an interpretation of the RHS? Any ideas?