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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.

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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
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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. If $X = \lambda$ the answer is still: $$ 1 - \sum_{i=0}^\lambda \frac{e^{-\lambda}\lambda^i}{i!} $$ Unless I have misunderstood you.

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