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Only even times contribute, and at time $2k$ the walk is at zero exactly when it has made $k$ steps in each direction. By linearity of expectation,The Stirling formula implies thatHence there is a constant $c_0>0$ such that this term is at least $c_0/\sqrt{k}$ for every $k\ge 1$, after decreasing $c_0$ to cover the finitely many small values. ThereforeThe final expression is $2c_0(\sqrt{n+1}-1)$, so, after another adjustment for small $n$, there is a constant $c>0$ withfor every $n$. This is the lower bound recorded by expected visits to the origin by a simple random walk.

Let $I_i$ indicate that $v_i$ is isolated. All its $n-1$ incident edges must be absent, soSince $N={\sum }_{i=1}^n{I}_i$, linearity of expectation gives

Add a hospital state $H$ and estimate the enlarged Q-matrix from the cohort. If infection starts in state $I$, the expected hospital occupancy generated by one infection is the Markov reward model integralEquivalently, if $Q_T$ is the subgenerator on all transient states, $m_H$ is the $(I,H)$ entry of the fundamental matrix of an absorbing continuous-time Markov chain $(-Q_T{)}^{-1}$. The number infected in the population has expectation $Np$, so linearity of expectation gives total expected hospital use $Np,m_H$ days. If only occupancy within a finite period counts, replace the upper integration limit by the remaining follow-up time for each infection and average over infection times.

At station $j$, the probability of stopping isbecause the lower rate applies exactly when no earlier stop occurred. By linearity of expectation,

In a uniform balls-in-bins allocation with $m\to \infty$ and $n/m\to \lambda \in (0,\infty )$, a fixed bin count has convergence in distribution to a Poisson distribution variable of mean $\lambda$ by the Poisson limit theorem. If $N_j$ counts bins containing exactly $j$ balls, symmetry and the linearity of expectation give $\mathbb{E}[N_j]/m=\mathbb{P}(B_1=j)\to e^{-\lambda }{\lambda }^j/j!$ for each fixed $j$. In particular the expected empty fraction tends to $e^{-\lambda }$, and the expected fraction with at least two balls tends to $1-(1+\lambda )e^{-\lambda }$. These are statements about expectations; convergence of the random fractions requires an additional argument.

Every finite graph with $m$ edges has a cut of a graph of size at least $m/2$, and hence a spanning bipartite graph with at least $m/2$ edges. Put each vertex independently on either side with equal probability. Every edge crosses with probability $1/2$, so linearity of expectation gives expected cut size $m/2$.

Suppose a graph with $n$ vertices and $m>0$ edges has a vector coloring with adjacent inner products at most $-1/2$. Take $r$ independent copies of random hyperplane rounding and use the $r$ signs as a color. Each edge remains monochromatic with probability at most $3^{-r}$, so the expected number $B$ of monochromatic edges is at most $m{3}^{-r}$ by linearity of expectation.

Choose $r=\max \{0,\lceil {\log }_3(4m/n)\rceil \}$. Then Markov inequality gives $\mathbb{P}(B>n/2)\le 1/2$. On success, delete one endpoint of every monochromatic edge. This random alteration method leaves at least $n/2$ vertices properly colored withThe removed vertices can retain their original colors, as the semicoloring only requires the retained induced subgraph to be proper. A failed draw can be detected and repeated, with at most two trials on average. The case $m=0$ needs just one color. This is the elementary independent-hyperplane construction in Karger, Motwani and Sudan's paper on approximate graph coloring.

Place $n$ labelled balls independently into $m$ labelled bins, each with probability $1/m$. The vector of bin counts has a multinomial distribution, and each count separately has the binomial distribution $\mathrm{Bin}(n,1/m)$. Distinct bin counts have covariance $-n/m^2$, so they are dependent for $n>0$ and $m>1$. Counts of bins of a specified size are sums of indicator variables; their expectations follow from the linearity of expectation, without any independence between bins.