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Given two Poisson distributions with different λ values, if each were to produce a single random variable, is there closed-form expression for calculating the probability of one random variable being greater than the other?

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    $\begingroup$ There isn't a nice closed-form expression for this, as you will find if you attempt to compute e.g. $$\sum _{n=1}^{\infty } \sum _{k=0}^{n-1} \frac{e^{-\lambda _1-\lambda _2} \lambda _2^k \lambda _1^n}{k! n!} $$ $\endgroup$ – Math1000 Jun 28 '16 at 22:34
  • $\begingroup$ Yeah, should have mentioned in the question I was looking for a closed-form expression. Thanks! $\endgroup$ – samblake Jun 29 '16 at 8:42
  • $\begingroup$ Please edit your Question if the wording can be improved. For example the current phrasing could be interpreted in a trivial sense, that one of the two random variables will exceed the other with probability one (because the alternative is for the random variables to be exactly equal). $\endgroup$ – hardmath Jun 29 '16 at 11:05
  • $\begingroup$ Perhaps of interest: the difference of two Poisson random variables follows a Skellam distribution. You'll notice that Wikipedia doesn't give an expression for the CDF. Also, this question is related: math.stackexchange.com/questions/1297299 $\endgroup$ – lum Jun 29 '16 at 13:40
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Take two poisson random variables $A$ and $B$ with means $\lambda_A$ and $\lambda_B$ respectively. We see that

$$P(A > B) = \sum_{k = 0}^{\infty} P(A > B | B = k)P(B = k)$$ $$=\sum_{k=0}^{\infty} P(A \geq k + 1)P(B = k) = \sum_{k= 0}^{\infty} \left(\sum_{l=k+1}^{\infty} \frac{\lambda_A^{l} e^{-\lambda_A}}{l!} \right)\frac{\lambda_B^k e^{-\lambda_B}}{k!}.$$

Generally, this is difficult to calculate for a general result, but the key idea behind this is to condition on the value of one of the variables.

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    $\begingroup$ Conditioning is unnecessary (and certainly not "key" here). $\endgroup$ – Did Jun 29 '16 at 8:58
  • $\begingroup$ why? please provide some argumentation $\endgroup$ – Daniel Hitzel Sep 27 '18 at 11:19

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