Splitting a Poisson total tells you nothing new

Question

Two equal time windows each record an independent Poisson $(\lambda)$ count, $X_1$ and $X_2$ . You are told only that the combined count is $X_1 + X_2 = 4$ .

Given the total of four, what is the probability that the first window held exactly two, and what does the answer reveal about sufficiency of the total for $\lambda$ ?

Accepted Answer

Given the sum, the split is binomial. A known fact about independent Poissons: given that their total is s, the first count is Binomial(s, 1/2) when the two windows are equally sized, because each of the s events independently fell in window one with probability 1/2. Crucially, this conditional law does not involve \lambda. Compute the probability. With s = 4 and success probability 1/2: P(X1 = 2 \mid X1 + X2 = 4) = \binom{4}{2}\Big(\frac12\Big)^4 = 6 \cdot \frac{1}{16} = \frac{6}{16} = \frac{3}{8}. The punchline: \lambda is absent. The answer 3/8 holds for every rate \lambda. Once the combin…

Splitting a Poisson total tells you nothing new

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