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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 autocovariance vanishes beyond lag $q$. For an ordinary consecutive sum, variance of a sum givesThe last sum is finite. For $h\ge 0$, $\gamma (h)={\sigma }^2{\sum }_{j=0}^{q-h}{\theta }_j{\theta }_{j+h}$. Counting all coefficient pairs givesFor the odd-indexed sample the lag-$h$ autocovariance is $\gamma (2h)$, so the same finite-sum argument gives ${\sum }_h\gamma (2h)$. Only coefficient pairs of the same parity contribute. If $E={\sum }_{j\mathrm{even}}{\theta }_j$ and $O={\sum }_{j\mathrm{odd}}{\theta }_j$, the odd-subsample limit isThis is the odd-subsample long-run variance of a moving average. Equivalently it is $[f_X(0)+f_X(1/2)]/2$ in the cycles convention. Under the printed invertibility assumption and positive noise variance, the polynomial is nonzero at both $1$ and $-1$, so both limits are positive. The same formulas also hold without invertibility, when cancellations can make a limit zero.

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.