Given a binomial random varible $N \sim \text{Bin}(n,p)$, I want to find an upper bound for the value:

$$\phi \equiv \sup_{n \cdot p \geq 1} \mathbb{E} \bigg( \frac{np}{\max(N,1)} \bigg).$$

(Note that $n,p$ are arbitrary.) I have proved that $1 \leq \phi \leq 2$. I think I can show that $\phi=1$, but I am not sure.

Edit: Using Wolfram Alpha, it is shown that the supremum is larger than 1.3.


1 Answer 1



Trying out a few values we see that the values get higher for larger $n$.

(we have drawn a red line based on the bound computed below)


f = function(n,p) {
  x = 0:n
  px = dbinom(x,n,p)
  x[1] = 1
f = Vectorize(f)
p = seq(0,1,0.001)

plot(-1,-1, xlim = c(0,1), ylim = c(0,2),
     xlab = "p", ylab = "E[np/x]")
title("exploring the bound")
for (ni in c(1:10)) {

lines(c(0,1),opt*c(1,1), col = 2)

Computing/approximating attempt

Based on the above (tendency to high n and small p) we can try and approximate the sum by replacing the binomial distribution with a Poisson distribution with $\lambda = np$

$$f(k) \approx \frac{\lambda^k e^{-\lambda}}{k!}$$

and let Wolfram alpha compute

$$\sum_{k=1}^\infty \frac{\lambda}{k} \frac{\lambda^k e^{-\lambda}}{ k!} = -e^{-\lambda}\lambda(log(\lambda)-Ei(\lambda)+ \gamma)$$

We arrive at

$$E[np/k] \approx e^{-np}np(1-log(np)+Ei(np)- \gamma) $$

(where we added an extra term $np e^{-np}$ to capture the contribution of $k=0$)

Maybe this can be differentiated, but I had it computed and arrive at a bound of approximately

$$\phi \approx 1.386317$$

x <- seq(0,6,0.1)
y <- -exp(-x)*x*(log(x)-expint::expint_Ei(x)+0.57721)+x*exp(-x)
opt <- max(y, na.rm = TRUE)
### opt = 1.386317

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.