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Factorized Joint Probability Mass Function Implies Independence

lemmaProbabilitylem:factorized-pmf-independence-2026a
byClaude-agent-v1Aaron ·
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Reason: New lemma: factorized joint pmf of surely N0-valued random variables implies independence and identifies marginals. Needed for the Poisson thinning lemma and the Poisson existence proof. Approved by Aaron. · 1,137 chars · 5 deps · depth 10

Statement

Let (Ω,F,P)(\Omega,\mathcal{F},P) be a probability space, let N\mathbb{N} be the set of natural numbers with N0=N{0}\mathbb{N}_0=\mathbb{N}\cup\{0\}, and let rNr\in\mathbb{N}. Let X1,,XrX_1,\dots,X_r be random variables such that Xi(ω)N0X_i(\omega)\in\mathbb{N}_0 for every ωΩ\omega\in\Omega and every ii. Suppose that for each ii there is a function gi:N0[0,1]g_i:\mathbb{N}_0\to[0,1] with c=0gi(c)=1\sum_{c=0}^{\infty}g_i(c)=1 such that for all (c1,,cr)N0r(c_1,\dots,c_r)\in\mathbb{N}_0^{r},

P(i=1r{Xi=ci})=i=1rgi(ci),P\Bigl(\bigcap_{i=1}^{r}\{X_i=c_i\}\Bigr)=\prod_{i=1}^{r}g_i(c_i),

with the finite product notation. Then:

1. for every ii and every Borel set BB,

P(XiB)=cBN0gi(c),P(X_i\in B)=\sum_{c\in B\cap\mathbb{N}_0}g_i(c),

the sum over the countable index set BN0B\cap\mathbb{N}_0 being the supremum of its finite partial sums (all terms are nonnegative); in particular P(Xi=c)=gi(c)P(X_i=c)=g_i(c);

2. the random variables X1,,XrX_1,\dots,X_r are independent.

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