Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Given a random variable $X \sim \text{Poisson}(\lambda)$ such that $\lambda > D$, with $\lambda, D \in \mathbb{N}$, what is the probability that a sample obtained from $X$ is greater than $\lambda$?

In other words, what is the value of $\mathbb{P}(X > \lambda)$?

I think that calculating the probability of $X$ being greater than the said value $D$ analysing this for every possible mean greater than $\lambda$ is a possible path to follow, but I'm not sure of how difficult to calculate this would be.

Poisson's CDF states that, for $X \sim \text{Poisson}(\lambda)$ and $k \in \mathbb{N}$, $$ \mathbb{P}(X \leqslant k) = \mathsf{e}^{-\lambda} \sum\limits_{i = 0}^{k} \frac{\lambda^i}{i!} $$

So, the value I'm asking should be the summation of the above probability's complement from $D+1$ to infinity, or even $$ 1 - \sum\limits_{p = 0}^{D} \left( 1 - \mathsf{e}^{-p} \sum\limits_{i = 0}^{D} \frac{p^i}{i!} \right) $$

Is this the correct value? I already know that $\lambda > D$, so maybe there should be some conditional probability involved, but I'm not sure.

If this is correct, what I'm asking is if there isn't a more concise way to calculate this, with less summations or none at all. This is because I'll need to calculate this value extensively in a computer program and computational time is very, very precious.

share|cite|improve this question
It approaches $1/2$ as $\lambda\to\infty$, but I think its exact value for small $\lambda$ might be a bit delicate. – Michael Hardy Jun 10 '13 at 18:55

I am unsure where you use $D$. For a Poisson-distributed random variable: $$ P(X > x) = 1 - P (X \leq x) = 1 - \sum_{i=0}^x \frac{e^{-\lambda}\lambda^i}{i!} $$ It does not matter what $x$ is as long as $x$ is a nonnegative integer. If $x = \lambda$ the answer is still: $$ 1 - \sum_{i=0}^\lambda \frac{e^{-\lambda}\lambda^i}{i!} $$ Unless I have misunderstood you.

share|cite|improve this answer
Capital letters such as $X$ are used for random variables while small letters such as $x$ are used for real numbers. Why invert these? – Did Sep 29 '15 at 8:20
Unintentionally; now switched. – Avraham Sep 30 '15 at 1:41

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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