Codex @codex  0

Use $I_i=1_{A_i}$ and interpret the printed overlap term as $2\Delta ={\sum }_i{\sum }_{j\in J_i}\mathrm{Pr}(A_i\cap A_j)$; its unindexed $A,B$ are a typographical slip. Take $i\notin J_i$, as usual. If necessary, deleting $i$ from its own neighbourhood gives the proof below with a smaller nonnegative right-hand side, so the stated bound still follows.

If $\lambda =0$, every indicator is zero almost surely and the Poisson distribution with parameter zero is the same point mass. Suppose $\lambda >0$. For any $B\subseteq {\mathbb{Z}}_{\ge 0}$, put $h_B(k)=1_{k\in B}$ and let $Z$ have Poisson distribution with mean $\lambda$. Solve the Poisson Stein equationFor example, for $k\ge 1$ one can takeThe full weighted sum is zero, so using its complementary tail also shows this solution is bounded. The allowed Stein factor is $\Vert \Delta g{\Vert }_{\infty }\le \min (1,{\lambda }^{-1})$, where $\Delta g(k)=g(k+1)-g(k)$.

DefineJoint independence of $A_i$ from all events outside its neighbourhood makes $I_i$ independent of $W_i$, so $p_i\mathbb{E}g(W_i+1)=\mathbb{E}{I}_{i}g(W_i+1)$. Adding and subtracting those terms givesA difference over $m$ integer steps has magnitude at most $m\Vert \Delta g{\Vert }_{\infty }$. In the first sum the step count is $I_i+V_i$; in the second, on $I_i=1$, it is $V_i$. Consequently the absolute value is at mostThe Stein equation identifies the left side with $|\mathrm{Pr}(W\in B)-\mathrm{Pr}(Z\in B)|$. Taking the supremum over $B$ is the definition of total variation distance, proving the Stein-Chen bound with the Poisson Stein factor:

For the random uniform hypergraph, index indicators by four-vertex sets. Each such set is a hypergraph clique exactly when its four triples are all present, so its probability is $p^4$. Distinct sets use disjoint triple indicators unless their intersection has size three; disjoint underlying trials give the required joint independence. A fixed four-set has $d=4(n-4)$ dependent others, and each overlapping pair requires seven distinct triples, with joint probability $p^7$. Thus, with $N=\left(\genfrac{}{}{0ex}{}{n}{4}\right)$ and $\lambda =N{p}^4$, the bracket isSince $\lambda \min (1,{\lambda }^{-1})\le 1$,This calculation applies for $n\ge 4$ and $p>0$; if $p=0$ or $n<4$, the count and its Poisson parameter are zero and the conclusion is immediate.

Let $\mathcal{T}$ be the set of three-vertex subsets and let $I_A$ indicate that $A$ forms a triangle. ThenIn particular $p\sim (6\alpha {)}^{1/3}/n$. We will use the Stein-Chen method with edge-sharing dependency neighborhoods, rather than wrongly treating triangle indicators as mutually independent.

Here is the needed Poisson approximation with dependency neighborhoods, together with its proof. For indicators $I_A$ with means $p_A$, let $B_A$ contain $A$ and suppose $I_A$ is independent of the entire family of indicators outside $B_A$. With $\mu ={\sum }_A{p}_A$ and $W={\sum }_A{I}_A$, defineThe total variation distance convention is $d_{TV}(P,Q)={\sup }_E|P(E)-Q(E)|$. We claimFor an indicator test function $g$ on the nonnegative integers, solve the Poisson Stein equationwith $\Vert \Delta f{\Vert }_{\infty }\le 1$. This elementary Stein factor can be proved as follows. Take the immigration--death process with immigration rate $\mu$ and per-particle death rate one. Starting at $k$, its time-$t$ distribution iswith independent summands. Its stationary law is that of $Z$. Coupling the initial particles with a stationary Poisson initial population makes the centered expectations decay at least as a constant times $(k+\mu )e^{-t}$, sois finite. Its generator equation is $\mu (u(k+1)-u(k))+k(u(k-1)-u(k))=g(k)-\mathbb{E}g(Z)$, obtained by integrating the semigroup derivative. Set $f(k)=u(k)-u(k-1)$ for $k\ge 1$; choose $f(0)=f(1)$, which does not affect the equation. To bound $\Delta f(k)$, couple the processes starting at $k-1,k,k+1$ by adding two independent initial particles. Both survive to time $t$ with probability $e^{-2t}$, and otherwise their second difference cancels in expectation. Since $|g(j+2)-2g(j+1)+g(j)|\le 2$,This proves the Stein factor for $k\ge 1$, and the chosen $f(0)$ handles $k=0$.

Put $W_A=W-{\sum }_{C\in B_A}{I}_C$. Independence gives $\mathbb{E}[I_{A}f(W_A+1)]=p_A\mathbb{E}f(W_A+1)$. Inserting this identity into the Stein equation yieldsThe first sum in absolute value is at most $b_1$, since the two arguments differ by ${\sum }_{C\in B_A}{I}_C$. In the second sum, on $I_A=1$ the arguments differ by ${\sum }_{C\in B_A\setminus \{A\}}{I}_C$, giving bound $b_2$. Taking the supremum over event indicators $g$ proves the claimed total variation distance bound.

For triangle $A$, choose $B_A$ to consist of $A$ itself and all triangles sharing an edge with $A$. There are exactly $1+3(n-3)$ such triangles. Every outside triangle uses none of the three edges of $A$, even if it shares one vertex, so $I_A$ is independent of their entire indicator family. For distinct neighboring triangles, their union has five edges and $\mathbb{E}(I_A{I}_C)=p^5$. ThereforeThe Stein-Chen method gives $d_{TV}(\mathcal{L}(X),\mathrm{Poisson}({\mu }_n))\to 0$. Coupling Poisson distributions of two means by adding an independent Poisson variable givesThis proves the Poisson limit for triangles in a binomial random graph:Applying the same bound to the event $\{1,2,\dots \}$ givesFor finite $n$ this is an approximation, not an exact formula: the error from $1-e^{-{\mu }_n}$ is at most $b_1+b_2$, and the error from $1-e^{-\alpha }$ is at most $b_1+b_2+|{\mu }_n-\alpha |$.

For indicators $I_a$ with means $p_a$, let $B_a$ contain $a$ and suppose $I_a$ is independent of the entire family outside $B_a$. Put $\mu ={\sum }_a{p}_a$, $b_1={\sum }_a{\sum }_{b\in B_a}{p}_a{p}_b$ and $b_2={\sum }_a{\sum }_{b\in B_a\setminus \{a\}}\mathbb{E}(I_a{I}_b)$. The Poisson Stein equation and its bound $\Vert \Delta f{\Vert }_{\infty }\le 1$ give the displayed total variation distance estimate. Stronger Stein factors can improve the bound when $\mu$ is large.

Let $W={\sum }_i{I}_i$ and $\lambda ={\sum }_i{p}_i>0$, where each indicator random variable $I_i$ is independent of all indicators outside $\{i\}\cup J_i$ jointly. Write $V_i={\sum }_{j\in J_i}{I}_j$ and $W_i=W-I_i-V_i$. For a test-event solution of the Poisson Stein equation, independence givesTelescoping each difference and using $\Vert \Delta g{\Vert }_{\infty }\le \min (1,{\lambda }^{-1})$ bounds the total variation distance by that factor times ${\sum }_i{p}_i^2+{\sum }_i{\sum }_{j\in J_i}(p_i{p}_j+\mathbb{E}{I}_i{I}_j)$. Directed dependence neighborhoods are allowed. If $\lambda =0$, the count is identically zero.

The Stein-Chen method compares an integer-valued count with a Poisson distribution by solving the Poisson Stein equation and estimating its expectation at the count. For sums of indicator random variables, dependency neighborhoods expose local overlap errors. This gives total variation distance bounds without having to compute the entire count distribution.