# Bounds over expected value of reciprocal binomial random variable

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.

### Exploring

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
return(sum(px*n*p/x))
}
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(p,f(ni*10,p))
}

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 $$\lambda = np$$

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

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

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)
plot(x,y)
opt <- max(y, na.rm = TRUE)
opt
### opt = 1.386317