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I have two independent Poisson random variables $A \sim \text{Poisson}(\lambda_A)$ and $B \sim \text{Poisson}(\lambda_B)$. For a fixed given integer $k$, I'd like to determine

$$P(A < k \leq A + B).$$

Is there an analytical expression for this probability as a function of $k, \lambda_A, \lambda_B$?

Edit 1: Is the following approach valid?

$$P(A < k \leq A + B) = 1 - P(A < k \leq A + B)^C $$

where $\cdot^C$ denotes the complement. The complement here is given by

$$P(A < k \leq A + B)^C = P(A \geq k) + P(k > A + B) $$

Plugging the complement in gives us:

\begin{align*} P(A < k \leq A + B) &= 1 - P(A < k \leq A + B)^C\\ &= 1 - P(A \geq k) - P(k > A + B)\\ &= 1 - (1 - P(A < k)) - P(A + B < k)\\ &= P(A \leq k - 1) - P(A + B \leq k-1) \end{align*}

where the final expression is the CDF of $A$ up to $k-1$ minus the CDF of $A+B$ up to $k-1$.

Is this correct? And if so, can I simplify further?

Ben
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Rylan Schaeffer
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1 Answers1

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For simplicity, you can write:

$$\begin{align} \mathbb{P}(A < k \leqslant A+B) &= \mathbb{P}(A+B \geqslant k) - \mathbb{P}(A \geqslant k) \\[6pt] &= [1-\mathbb{P}(A+B \leqslant k-1)] - [1-\mathbb{P}(A \leqslant k-1)] \\[6pt] &= \mathbb{P}(A \leqslant k-1) - \mathbb{P}(A+B \leqslant k-1) \\[6pt] &= F_\text{Pois}(k-1 | \lambda_A) - F_\text{Pois}(k-1 | \lambda_A+\lambda_B). \\[6pt] \end{align}$$

The CDF of the Poisson distribution does not simplify any more than writing it as a sum of the mass values, so this is really can't simplify any more than this. (As whuber notes in the comments, you can "simplify" it to an integral if you prefer, but this will still require computer evaluation.) You can compute this probability in R using standard probability functions for the Poisson distribution. Here is an example taking $k=6$, $\lambda_A=3$ and $\lambda=6$.

#Set parameters
K <- 6
LAMBDA.A <- 3
LAMBDA.B <- 5

#Compute probability
ppois(K-1, lambda = LAMBDA.A + LAMBDA.B, lower.tail = FALSE) - 
  ppois(K-1, lambda = LAMBDA.A, lower.tail = FALSE)

[1] 0.724846
Ben
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  • When I simplify my final expression for the difference of two CDFs, I get a much nicer expression: $F(A \leq k) - F(A+B\leq k) = e^{-\lambda_A}\Big[\sum_{x=0}^k \frac{\lambda_A^x - e^{-\lambda_B}(\lambda_A + \lambda_B)^x}{x!} \Big]$. I'm having trouble showing this to be equivalent to your final expression. Could you check my math? – Rylan Schaeffer Mar 08 '21 at 04:19
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    I have made a major edit to simplify the solution. – Ben Mar 08 '21 at 04:42
  • I'm going to implement some Monte Carlo simulations to check the correctness, and if the formula hold, I'll accept your answer. Thanks for your help! – Rylan Schaeffer Mar 08 '21 at 04:54
  • Ben, is there no way to make use of the Skellam distribution or one of its properties to further simplify? https://en.wikipedia.org/wiki/Skellam_distribution – Rylan Schaeffer Mar 08 '21 at 05:06
  • It won't make things any simpler, since you will still end up with an equivalent expression. – Ben Mar 08 '21 at 05:10
  • You can simplify the answer to $(\Gamma(k)^{-1})\int_{\lambda_A}^{\lambda_A+\lambda_B} t^{k-1} e^{-t}\,\mathrm{d}t.$ – whuber Mar 08 '21 at 17:20
  • @whuber thank you! Can you explain how you got there? – Rylan Schaeffer Mar 08 '21 at 17:46
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    Rylan, it's based on the connection between the counts of events in a Poisson process and the waiting times between them. See https://stats.stackexchange.com/a/287025/919. – whuber Mar 08 '21 at 18:18
  • Ben, that's a great simplification but it could use just a little more explaining, because the first step is not generally true. It relies on $B$ being non-negative. – whuber Mar 10 '21 at 15:35