Trying to solve this problem of mine (you don't actually need to read the linked problem, the problem is rephrased below with a different assumption):

Estimating a sample size such that its sum has some probability of not crossing some upper bound

I started with the assumption that the mean $\lambda$ is known, to see how far I can get, but I got a result whose semantics are confusing me. Such confusing semantics is the reason of doing a separate question.

Rephrasing the problem under such assumption of known $\lambda$:

Given a natural number $L$ and a real number $p\in[0, 1]$, find the biggest natural number $n$ such as, given $n$ random variables $X_1, \ldots, X_n$, where each $X_i\sim \mathcal{P}(\lambda)$ and with $\lambda\gt 0$, $$\mathbb{P}\biggl[\sum_{i=1}^nX_i\leq L\biggr] = p$$

Knowing that the sum of poisson variables is a poisson variable whose mean is the sum of means, and a poisson variable can be approximated by a normal distribution when its mean is big (which in my case is):

$$\sum_{i=1}^nX_i\sim \mathcal{P}(n\lambda)\approx \mathcal{N}(n\lambda, n\lambda)$$


$$ \mathbb{P}\biggl[\frac{\sum_{i=1}^nX_i-n\lambda}{\sqrt{n\lambda}}\leq \frac{L-n\lambda}{\sqrt{n\lambda}}\biggr] = \mathbb{P}\biggl[Z\leq \frac{L-n\lambda}{\sqrt{n\lambda}}\biggr]=p $$

Now I find the constant $c$ such that $\mathbb{P}[Z\leq c]=p$, name $\alpha^2=n\lambda$, and solve:

$$\frac{L-n\lambda}{\sqrt{n\lambda}} = c\quad\Rightarrow\quad\frac{L-\alpha^2}{\alpha} = c$$ $$\Rightarrow\quad\alpha^2+c\alpha -L=0\quad\Rightarrow\quad\alpha=\frac{-c+\sqrt{c^2+4L}}{2}$$ $$\Rightarrow\quad \boxed{n = \frac{\alpha^2}{\lambda} = \frac{(-c+\sqrt{c^2+4L})^2}{4\lambda}}$$

But according to the formula, if I want a higher probability, e.g., a larger $c$, it calculates a larger $n$. In other words, asking for more probability implies asking for more data to sum, and that confuses me.

I mean, informally speaking, If I sum $n=0$ numbers the result will be below any $L$ for sure. Adding more numbers should increase the probability of being above $L$, not the other way around, but the equation above says otherwise. Did I do something wrong?

  • $\begingroup$ I made the question a bit shorter now. $\endgroup$ Feb 15 at 9:38

2 Answers 2


You don't need a normal approximation for this. You just look at the cumulative distribution function $F$ and the quantile function $F^{-1}$ of a $\text{Poisson}(n \times \lambda)$ Poisson distribution.

You basically want to solve $F(L; n \lambda) \leq p$ or equivalently $F^{-1}(p; n \lambda) \leq L$ in terms of $n \in \text{Integers}$ (or solve the equality in terms of $n \in \text{Reals}$ and then take the floor).

  • $\begingroup$ Ok I will try to look at those functions to know how I should use them. But even if I don't need a normal approximation, is it wrong to use it? In case it is not wrong, what did I do wrong then? Because my final result makes no sense to me. $\endgroup$ Feb 15 at 10:26

Ok, all I did looks fine. My misunderstanding was based on my thought that the function

$$ n=f(c)=\frac{(-c+\sqrt{c^2+4L})^2}{4\lambda} $$

was increasing on $c$. But it's not, it's actually a decreasing function, which means that the highest the requested probability, the lowest the number of terms I have to add; exactly what I thought.

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