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An immigration--death process has constant birth rate $b$ and linear death rate $dy$ in state $y$. Its stationary distribution is Poisson with mean $b/d$.

This is an immigration--death process, equivalently an M-M-infinity queue: admission occurs at rate $\alpha$ regardless of occupancy, and each of $n$ animals is discharged independently at rate $\delta$. Its state-transition diagram through four animals isEvery upper arrow points right and every lower arrow points left. Thus $q_{n,n+1}=\alpha$ and $q_{n,n-1}=n\delta$ for $n\ge 1$, with no discharge from zero. Four is only the limit of the requested sketch, not an absorbing state or a capacity boundary.

Use the parameters from (e), and interpret overflow as the number needing simultaneous care exceeding capacity at a randomly chosen time in stationary distribution. The occupancy of the immigration--death process is not fixed at its mean. Detailed balance givesNormalizing yields ${\pi }_0=e^{-25}$, so stationary occupancy has a Poisson distribution with mean 25. Thus integer capacity $K$ must satisfyThe adjacent tails areSince the tail decreases with $K$, the minimum capacity is $\boxed{32\text{animals}}$. The small excess above $0.1$ at 31 matters; a normal approximation rounded too early can choose the wrong integer.

This is a stationary occupancy criterion. If the event is instead an offered arrival finding at least $K$ animals already present in this unrestricted process, Poisson arrivals see time averages gives $P(X\ge K)$, whose minimum qualifying capacity is 33. That uses a different event definition. From an initially empty centre, occupancy at a finite time is Poisson distributed with mean $N(t)\le 25$, so the stationary criterion also bounds the fixed-time overflow probability during initial filling. A probability of ever exceeding capacity over an unlimited future is a different question: this irreducible recurrent process will eventually exceed every finite capacity with probability one. If instead arrivals are refused when all places are occupied, that changes the model to an loss network, whose blocking probability is not the unrestricted occupancy tail calculated here.

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 $b_1+b_2+|{\mu }_n-\alpha |$.

This is an immigration--death process. New individuals enter independently of the current population at the constant rate $\lambda$, producing transitions $n\to n+1$. Each existing individual dies at rate $\beta$, so the total transition rate $n\to n-1$ is $\beta n$.

At leading order, ${\mathcal{L}}_0^{\ast }[p_0(y\mid x)p_0(x,t)]=0$. For fixed $x$, the conditional stationary law therefore satisfiesThis is an immigration--death process, whose stationary distribution is Poisson with meanThe mean is finite only for $x>0$. The assumption $X(0)\ne 0$ and the pair-coalescence propensity $x(x-1)$ ensure that the slow process cannot remove its last $X$ molecule, so the reduced model remains in that domain.

For indicator test functions $g$, the equation has a solution with $\Vert \Delta f{\Vert }_{\infty }\le 1$. One proof uses an immigration--death process with immigration rate $\mu$ and per-particle death rate one. Its potential $u(k)=-{\int }_0^{\infty }(\mathbb{E}g(Z_t^k)-\mathbb{E}g(Z))dt$ satisfies its generator equation, and $f(k)=u(k)-u(k-1)$. coupling two additional initial particles bounds the second difference of $u$ by ${\int }_0^{\infty }2e^{-2t}dt=1$.