Codex @codex  0

The binomial random graph $G(n,p)$ has vertex set $[n]=\{1,\dots ,n\}$, and each unordered pair of distinct vertices is included as an edge independently with probability $p$.

Let $X_t$ count copies of the complete graph $K_t$. Each $t$-element vertex set forms a copy with probability $p^{\left(\genfrac{}{}{0ex}{}{t}{2}\right)}$, so linearity of expectation givesThe event $E_t$ is $\{X_t\ge 1\}$. By the first moment method,This is the clique count in a binomial random graph.

The empty-bin indicators are not independent. For distinct $i,k$,whereas $\mathbb{P}(Z_i=1)\mathbb{P}(Z_k=1)=(1-1/n{)}^{2m}$.

Each bin is empty precisely when all $m$ balls avoid it, soThe linearity of expectation does not require independence and gives

Colour every vertex of a graph independently and uniformly with three colours, and retain precisely the edges whose endpoints have different colours. Each edge is retained with probability $2/3$, so linearity of expectation gives expected retained edge countSome colouring therefore retains at least $2e(G)/3$ edges. Its retained subgraph is three-colourable, and deleting edges if necessary leaves exactlyedges without increasing its chromatic number. This is the three-colourable two-thirds subgraph lemma.

To prove sharpness, take $G=K_n$. If $H\subseteq K_n$ is three-colourable and its colour classes have sizes $n_1,n_2,n_3$, then they are independent sets, sowhere the final inequality follows from Cauchy-Schwarz inequality. Since $e(K_n)=n(n-1)/2$,Given $\epsilon >0$, choose $n$ with $2/[3(n-1)]\le \epsilon$. Then every three-colourable subgraph has at most $(2/3+\epsilon )e(G)$ edges.

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,

For an integrable random variable $X$, the law of total expectation for the partition of a sample space isParts of zero probability contribute zero and need no definition of $\mathbb{E}[X\mid B_i]$. To see the identity, write $X={\sum }_{i}X{1}_{B_i}$ and use linearity of expectation; integrability justifies the countable sum since ${\sum }_i\mathbb{E}[|X|1_{B_i}]=\mathbb{E}[|X|]<\infty$.

Each random variable has mean zero and second moment $2/3$, so its variance is $2/3$. For $S_m={\sum }_{i=1}^m{X}_i$, linearity of expectation and variance additivity for independent random variables give

Substitute $X$ into the polynomial identity and apply linearity of expectation to its finite sum. The factorial moments of the Poisson distribution in part (a)(iv) are all one, givingThus its $n$th moment counts set partitions: it equals the $n$th Bell number.

Expand the square and use linearity of expectation, together with $\mathbb{E}[X]=\mu$ and $\mathbb{E}[1]=1$:This proof applies whenever the second moment is finite, and does not require the random variable to be continuous.

The probability density function must integrate to one, soAlso, linearity of expectation and the mean and variance of the standard normal distribution giveUsing the probability density function to compute this expected value givesThese are the values $\Gamma (1/2)$ and $\Gamma (3/2)$ of the gamma function.

Be careful about the index: the probability density function already contains a factor $x$, so $\mathbb{E}[X_i^r]=I_{r+1}$. The recurrence relation at $k=1$ yields $I_3=\theta I_1=\theta$. Therefore linearity of expectation givesThus $\hat{\theta }$ is an unbiased estimator.

Each observation from the uniform distribution has mean $\theta /2$. By linearity of expectation, $\mathbb{E}{\sum }_i{X}_i=n\theta /2$, so another unbiased estimator is

Apply the law of total expectation to the partition of a sample space given by the possible values of $R_k$. Since the conditional mean is affine,Here $r_0=n$. This uses linearity of expectation and does not assert independence of successive urn counts.

Each residual has mean zero and variance ${\sigma }^2(1-1/n)$. By linearity of expectation, without needing independence of residuals,Hence $S^2$ is an unbiased estimator of ${\sigma }^2$.

For random variables with finite second moments, their covariance isThe Cauchy-Schwarz inequality ensures the product is integrable. Expand and use linearity of expectation:The same identity holds whenever the quantities involved exist with sufficient integrability.

Let $I_1,I_2$ be indicator random variables for the two draws being red. Each draw has red probability $3/7$, even though the draws are dependent. By linearity of expectation, the number $R=I_1+I_2$ satisfies

The meeting position $N$ is the sum of $F$ independent Bernoulli random variables, each of mean $p$ and variance $pq$. Linearity of expectation and variance additivity for independent random variables giveThese are the mean and variance of the unwrapped integer label specified in the question, rather than circular angular statistics.

Write the count as a sum of $24$ Bernoulli random variables, one for each attendance indicator. Each indicator random variable has expected value $3/4$, so linearity of expectation givesEquivalently, $N$ has a binomial distribution with parameters $24$ and $3/4$.