Let $X_1, X_2, \ldots, X_n$ be i.i.d. Bernoulli random variables with $P(X_i = 1) = p$ for every $i \in \{1, \ldots, n\}$. Let $Y = \sum_{i=1}^n X_i$ and let $c$ be a positive number.

I am interested in computing $\mathbb{E}\left[\frac{1}{c + Y}\right]$. I know that the expected value can be lower-bounded using Jensen's inequality: $\mathbb{E}\left[\frac{1}{c + Y}\right] \geq \frac{1}{\mathbb{E}\left[c + Y\right]} = \frac{1}{c + pn}$.

But is it possible to compute $\mathbb{E}\left[\frac{1}{c + Y}\right]$ exactly?

This answer seems to be very relevant, but I am not sure I understand how to correctly extend it to the above case. I would be grateful for any hints.


Let's attack this using the nice linked answer.

Here we have

$$\mathbb{E}\left(\frac{1}{c+X_1+\cdots+X_n}\right)=\int_0^\infty e^{-tc} \, \mathbb{E}\left(\exp\left(-t X \right) \right)^n \mathrm{d}t \tag{1}$$

and $\mathbb{E}\left(\exp\left(-t X \right) \right)=p e^{-t} + q$ with $q=1-p$. Hence the integrand in $(1)$ equals

$$ e^{-tc} (p e^{-t} + q)^n=e^{-tc} \sum_{k=0}^n p^k e^{-kt}q^{n-k} \binom{n}{k} \tag{2}$$

Then the expectation is

$$ \sum_{k=0}^n p^k q^{n-k} \binom{n}{k} \int_0^\infty e^{-t(c+k)} dt = \sum_{k=0}^n p^k q^{n-k} \binom{n}{k} \frac{1}{c+k} \tag{3}$$

I'm afraid you cannot simplify this further.

Notice that this could be obtained by a much simpler approach: just write down the expectation of $\frac{1}{c+Y}$,where $Y=X_1+\cdots+X_n$ is a Binomial $(n,p)$

Update (inspired by this)

An alternative expression, can be obtained by noticing that our expectation as in $(3)$ (let's call it $G$) can be expressed as

$$G=\int_0^1 (p x+q)^n x^{c-1} dx \tag{4}$$

or , doing $u=px+q$:

$$G= \frac{1}{p^c}\int_q^1 u^n (u-q)^{c-1} du \tag{6}$$

Further, if $c$ is a positive integer :

$$G= \left(\frac{q}{p}\right)^c \sum_{j=1}^c \frac{(-1)^{c-j}}{n+j} \binom{c-1}{j-1} \left( q^{-j} - q^{n}\right) \tag{7}$$

This might be more convenient than $(3)$ for $c \ll n$

  • $\begingroup$ leonbloy, many thanks for the answer! I derived the sum in (3) directly writing down the expectation, exactly as you noticed, and my hope was that there could be another way of computing the expectation that could lead to a closed formula for the sum. Nevertheless, the answer is very useful to me, as now I understand better how to use the Laplace generating function approach. $\endgroup$ – Victor Oct 12 '18 at 14:59
  • 1
    $\begingroup$ You're welcome. Notice that if you want an approximate solution (or tighter bounds) you can improve the Jensen bound by computing (at least one or two more terms) of the Taylor expansion of the function, as here math.stackexchange.com/questions/1536459/… $\endgroup$ – leonbloy Oct 12 '18 at 15:13

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