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.
Best Answer
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.
Update - 2018-10-08
It climbs to $0.5$ very slowly. Here is some R code that will calculate entries. The first function calculates a single entry, the second will calculate a sequence for plotting.
For the sequence $10^0, 10^1, \cdots, 10^8$ the values of the Poisson being greater than the mean are:
A plot of the first 10,000 entries is below.