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Let $A,B$ be indicator random variables with conditional independence given a random variable $T$, and suppose $\mathbb{E}[A\mid T]=\mathbb{E}[B\mid T]=g(T)$. Then the law of total expectation gives $\mathbb{E}[AB]=\mathbb{E}[g(T{)}^2]$ and $\mathbb{E}[A]=\mathbb{E}[B]=\mathbb{E}[g(T)]$. Thus their covariance is the variance of $g(T)$ and is nonnegative. The result explains correlated component evolutionary states in a coeval binary population without correlation between initial masses.

If a fixed graph $H$ has $v$ vertices, $e$ edges, and $N_H(n)$ unlabelled copies in $K_n$, then the number $X_H$ of its copies in $G(n,p)$ satisfiesThis follows by writing $X_H$ as a sum of indicator random variables.

It suffices to prove the assertion after replacing $\omega$ by $\min \{\omega ,\sqrt{\log n}\}$: that replacement gives a tighter interval. Thus assume $\omega \to \infty$ and $\omega =o(\log n)$, and putAt $p_{-}$, let $J$ count isolated vertices in $L$. The same indicator random variable calculation as for all isolated vertices givesBy the Chebyshev inequality, $J>0$ with high probability. Since $\ell \ge 2$ eventually, the marked vertices are then not all connected.

At $p_{+}$, the logarithmic-regime giant component result applies with $k=2$ and any fixed ${\gamma }_0\in (1/3,1/2)$. Its unique giant component misses $(1+o(1))n^{1/2}{e}^{-\omega /4}$ vertices. On the event that it misses at most twice that number, exchangeability makes the missed vertex set uniform conditional on its size. A union bound therefore givesThus all of $L$ is connected at $p_{+}$ with high probability.

For the uniform random graph process, assign uniform labels to the edges of the complete graph as independent random variables and reveal them in label order. The binomial random graph $G(n,p)$ is the prefix containing the $M(p)$ labels at most $p$, where $M(p)$ has a binomial distribution with parameters $N=\left(\genfrac{}{}{0ex}{}{n}{2}\right)$ and $p$. Its variance at $p_{\pm }$ is $O(n\log n)$, whereas the gap between $\mathbb{E}M(p_{\pm })$ and the relevant endpoint $m_{\pm }$ is $(1+o(1))\omega n/8$. The Chebyshev inequality gives $M(p_{-})\ge m_{-}$ and $M(p_{+})\le m_{+}$ with high probability, with harmless integer rounding. The monotone graph property that all marked vertices are connected now impliesThe marked-set connectivity threshold is lower than the threshold for connecting every vertex, because only about $\sqrt{n}$ specified vertices need to avoid the small graph components.

Write $p=p(n)$ and let $T$ be the triangle count in a binomial random graph. For every three-element set $S$ of vertices, let $I_S$ be the indicator random variable that its three edges are present. ThenFor two distinct triples $S,R$, their indicator random variables are independent random variables unless they share an edge. Sharing just one vertex still leaves the sets of relevant edges disjoint. If they share an edge, their union has five edges, and thereforeThere are $\left(\genfrac{}{}{0ex}{}{n}{2}\right)\left(\genfrac{}{}{0ex}{}{n-2}{2}\right)$ unordered pairs of distinct triangles in a graph sharing an edge: choose that edge, and then choose their two additional vertices. The variance of a sum consequently gives the exact formulaFor sufficiently large $n$, the hypothesis $np\to \infty$ ensures $p>0$, so $\mu >0$. Dropping negative terms and simplifying the binomial coefficients,Here $n^{2}p=n(np)\to \infty$ as well. The event $T\ge 2\mu$ implies $|T-\mu |\ge \mu$. Applying the Chebyshev inequality from part (i) gives the requested conclusion:Thus the triangle count is concentrated around its expectation throughout this range of the binomial random graph.

The common-age conditional probabilities giveHere $GW$ includes both possible component orders. Hence the binary star fractions have the concise ratioAs a check on the role of shared age, $p_{GG}=4p_G^2/3$ rather than $p_G^2$. This is covariance induced by a shared latent variable: the red giant indicator random variables have covariance equal to the variance of $g(Y)$, namely $1/97200$.

For a starburst galaxy with age $t$, put $y_b=t/(10\mathrm{Gyr})$. There is no age averaging. The component masses remain independent, so the fraction of systems with two red giants isThe last comparison is ambiguous in the printed question. Literally, the fraction of individual red giants in the first galaxy is $1/180$. Since $p_{GG}^{\mathrm{burst}}<1/8100<1/180$, no allowed starburst age satisfies that literal comparison. If the intended comparison is instead between systems containing two red giants in the two galaxies, compare with $p_{GG}=1/24300$; the answer isFor completeness, the individual red giant fraction in the burst is $g(y_b)=t/(900\mathrm{Gyr})$, which exceeds $1/180$ precisely for $5\mathrm{Gyr}<t<10\mathrm{Gyr}$. This is a third, different comparison; it does not change the computed fraction of two-giant systems.

Pointwise, an integer-valued nonnegative random variable has the representationwhen $T=k$, exactly the terms $n=0,\dots ,k-1$ equal one. Applying the Tonelli theorem, or the monotone convergence theorem to the finite sums of indicator random variables, gives the tail-sum formula for expectationBoth sides may be infinite; no integrability assumption is necessary.

Let $N(u)$ count all jumps of the continuous-time Markov chain. Its Markov-chain entry count into state $s$ isThusThis is a Lebesgue-Stieltjes integral against a counting process, so each jump contributes one. The initial occupation of state $s$ is not counted as an entry. An ordinary time integral of an indicator random variable asserting a jump at exactly time $u$ would be zero, since individual jump times have zero Lebesgue measure. The requested integral must therefore use this counting-measure interpretation. Assume non-explosion and finite expected jump counts on bounded intervals, as holds for a finite-state continuous-time Markov chain.

Write $Z(u)$ for the continuous-time Markov chain and ${\mathbb{E}}_r$ for expected value conditional on $Z(0)=r$. The occupation time of a continuous-time Markov chain in state $s$ is the integral of its indicator random variable. By Tonelli theorem,The interchange is valid because the integrand is nonnegative; indeed the random integral is bounded by $t$.

Let $X_i=1_{A_i}$ be the indicator random variable of $A_i$. Then $X={\sum }_i{X}_i$, so linearity of expectation givesUsing variance of a sum andwe obtain

For the circular array, let $X_i$ indicate that bulbs $i$ and $i+1$ are both on, with indices read modulo $n$. Then $\mathbb{E}{X}_i=p^2$. Non-neighbouring indicators are independent, whileThere are $n$ individual terms and $2n$ ordered neighbouring pairs in the double covariance sum, so

The event $B$ is exactly $\{X\ge 1\}$. If $p=n^{-0.6}$, Markov inequality givesIf $p=n^{-0.4}$, then $B^c=\{X=0\}$ implies $|X-\mathbb{E}X|\ge \mathbb{E}X$. By Chebyshev inequality,Therefore $\boxed{\mathbb{P}(B)\to 1}$ in the second regime.

Let $R$ be the number of returns to $v$ after time $0$. Since $v$ is a recurrent state, the probability of another return after each visit is one. By the stated geometric distribution description, ${\mathbb{P}}_v(R\ge k)=1$ for every positive integer $k$. Thus $R=\infty$ almost surely and ${\mathbb{E}}_{v}R=\infty$.

On the other hand, writing each visit as an indicator random variable and applying the monotone convergence theorem to the partial sums givesThe left side is infinite, proving the recurrence criterion by return probabilities in the required direction:

A fixed edge is monochromatic exactly when none of the $r$ hyperplanes separates its endpoints. Each hyperplane fails to cut it with probability at most $1/3$, and the hyperplanes are independent. ThereforeLet $B={\sum }_{ij\in E}{1}_{\{c(i)=c(j)\}}$ count monochromatic edges. Applying linearity of expectation to these indicator random variables givesIndependence between different edges is unnecessary; only the independently sampled hyperplanes are used. The $r$ signs provide at most $2^r$ colors, whether or not every sign region is nonempty.

For each edge $e$, let $I_e$ be the indicator random variable of the event that precisely one endpoint lies in $U$. Independent inclusion of the two endpoints with probability $1/2$ givesBecause $X={\sum }_{e\in E(G)}{I}_e$, linearity of expectation givesSome realization therefore has $X\ge m/2$. Retaining its crossing edges gives a subgraph that is bipartite, with vertex classes $U$ and $V(G)\setminus U$, proving the random cut lower bound.

Now suppose $n$ is an even number and instead choose $U$ uniformly among all subsets of size $n/2$. For a fixed edge $uv$,and the reverse orientation has the same probability. ThusAnother application of linearity of expectation givesAt least one such balanced choice attains this expectation, and its crossing edges give the required subgraph that is bipartite. This is the balanced random cut lower bound.