Poisson approximation to binomial distribution: $f(x)\geq g(x)$ or $f(x) \leq g(x)$

Question

We have a random variable $X$ which has a Binomial Distribution Bin(n,p) and a random variable $Y$ which has a Poisson Distribution Poisson(np).

We are interested in $$f(x):=Pr[X \geq x].$$ For this we consider $$g(x):=Pr[Y \geq x]=\sum_{i=x}^{\infty} \frac{(np)^i}{i!}e^{-np}.$$

I know that the distance between These two is at most $$|f(t)-g(t)|\leq d_{TV}(Bin(n,p), Poisson(np))<p.$$

But can I say something about whether $f(x) \geq g(x)$ for all $x$ or $f(x) \leq g(x)$ for all $x$? Or is there a boundary for which $f(x)>g(x)$ or $f(x)<g(x)$?

Answer 1

See this question. I don't know if we can say anything pointwise about $f(x)$ vs. $g(x)$. What we can say is: $$ \large \lim_{n \to \infty, p \to 0, np = \lambda}\textrm{Binom}(n,p,k) = \textrm{Pois}(\lambda,k) $$

As a matter of fact, here are some examples which show that the Poisson and Binomial probability mass functions can be larger or smaller than the other. $$ \begin{align} \textrm{Binom}(10, .1, 3) \approx 0.057396 &< 0.61313 \approx \textrm{Pois}(1, 3)\\ \textrm{Binom}(2, .5, 1) = 0.5 &> 0.367879 \approx \textrm{Pois}(1, 1)\\ \end{align} $$

Answer 2

Since binomial random variables are bounded and Poisson random variables are not, asking that $f(x)\geqslant g(x)$ for every $x$ is hopeless. On the other hand, the inequality $f(x)\leqslant g(x)$ is realized for some parameters $(n,p,x)$, but not all, for example, for every $n\geqslant1$ and $p\lt1$, one sees that $f(1)=1-(1-p)^n\gt1-\mathrm e^{-np}=g(1)$.

A universal inequality in this context is that each binomial distribution with parameter $(n,p)$ is stochastically dominated by a Poisson distribution of parameter $n\lambda$, provided $\lambda\geqslant-\log(1-p)$. To sum up, $$ f_{n,p}\leqslant g_{n\lambda(p)},\qquad\lambda(p)=-\log(1-p). $$