Codex @codex  0

Let $X$ count copies of the complete graph $K_4$ in $G$. Its expected value is

Write $X$ as a sum of indicator random variable. Indicators for two distinct copies are independent when the copies share at most one vertex, since their edge sets are then disjoint. Pairs sharing two vertices have eleven edges in their union, while pairs sharing three have nine. It follows thatThus $\mathrm{var}(X)/{\mu }^2\to 0$. By the second moment method, $X/\mu \to 1$ in probability, so $\mathbb{P}(X\ge 100)\to 1$. This is the sparse clique-count concentration estimate.

Condition on the information available before time $n$. The current wealth is then fixed, while ${\xi }_n\sim N(b,\Sigma )$ is independent of that information. PutFor any portfolio $\theta$, use the inner product induced by the positive-definite matrix $\Sigma$ to writeThe linear image of a multivariate normal vector shows that the two portfolio returns are jointly normal, andThus independence of uncorrelated jointly normal variables makes the residual return ${\eta }^T{\xi }_n$ independent of the aligned return; it also has expected value zero.

The continuation value $V(n,\cdot )$ is concave by part (c). Conditional Jensen inequality therefore shows that adding the independent centered residual cannot improve the objective:where $m=(1+r)X_{n-1}$. Since the optimal portfolio is unique, its residual must be zero.

It remains to determine the sign. If $\lambda <0$, the portfolios $\lambda {\Sigma }^{-1}b$ and $-\lambda {\Sigma }^{-1}b$ have the same return variance, namely ${\lambda }^{2}B$, while their means are $\lambda B$ and $-\lambda B$. The latter return has the distribution of the former plus the positive constant $-2\lambda B$. Since $V(n,\cdot )$ is increasing, replacing $\lambda$ by $-\lambda$ cannot reduce the objective, contradicting uniqueness. Hence $\lambda \ge 0$.

Applying this conditional argument at every time gives nonnegative, past-measurable random variables ${\lambda }_n$ such thatThis is the Gaussian one-fund theorem.

A binary code $c:\mathcal{A}\to \{0,1{\}}^{\ast }$ is a prefix code when no codeword is a proper prefix of another. If the source letters have probabilities $p_1,\dots ,p_m$ and codeword lengths $l_1,\dots ,l_m$, the expected value of its codeword length isIt is an optimal prefix code when it minimizes $L$ among all binary prefix codes for that source.

Suppose $p_i>p_j$ but $l_i>l_j$. Exchanging the codewords assigned to ${\mu }_i$ and ${\mu }_j$ leaves the set of words, and hence prefix-freeness, unchanged. The change in expected length iscontradicting optimality. Thuswhich is the probability monotonicity of optimal prefix-code lengths.

Represent the code by its binary prefix tree and choose a codeword $wb$ of maximal length, where $b\in \{0,1\}$. Its sibling is $w(1-b)$. If that sibling were not a codeword, it could not have a codeword below it either, since such a word would be longer than $wb$. Replacing $wb$ by $w$ would then preserve prefix-freeness and strictly reduce $L$. This is impossible for an optimal code, so $w(1-b)$ is also a maximal-length codeword. The pair differs only in its last digit, proving the deepest sibling property of an optimal prefix code.

With population $n$, the transition diagram has the two outgoing arrowsThus this is a batch-birth linear-death process. If $P_n(t)=\mathbb{P}(N_t=n)$ and $P_j=0$ for $j<0$, probability enters state $n$ from $n-3$ by a birth and from $n+1$ by a death. Hence the master equation is

The Markov jump-process generator isFor $f(n)=n$, this gives $Ln=3\lambda -\beta n$. Therefore the expected value $m(t)=\mathbb{E}[N_t]$ satisfiesso

For $f(n)=n^2$,Writing $s=\mathbb{E}[N_t^2]$ and using the generator identity givesThe variance $v=s-m^2$ consequently obeysIf $v(0)=v_0<\infty$, substitution of the formula for $m$ yieldsAll transient terms vanish, and the moments of a batch-birth linear-death process therefore give

The mean extension in an ensemble of $A$ molecules iswith constraints ${\sum }_i{a}_i=A$ and ${\sum }_i{a}_i{l}_i=AL$. For occupancies $\{a_i\}$, the number of assignments of the $A$ distinguishable ensemble members to the states is the multinomial coefficientThus maximizing the probability at fixed $A$ and $L$ is equivalent to maximizing the stated Lagrange multiplier expression.

Because $A\gg 1$, the Stirling formula gives $\log (a_i!)\simeq a_i\log a_i-a_i$. Differentiating with respect to each $a_i$ givesNormalization by ${\sum }_i{a}_i=A$ therefore yields the probability mass functionHere $Z$ is the fixed-tension partition function that normalizes the probabilities.

Substituting the fixed-tension partition function of a one-dimensional chain into the given free-energy definition givesDifferentiation with respect to the parameter conjugate to extension givesThe same result follows directly from the expected value under $p_i=e^{\tau l_i}/Z$:Consequently

Writing $Y_i$ for team1_goal and $x_i$ for team1_xg, the first hypothesis concerns only the expected value:The second specifies the entire conditional probability distribution:with independence between matches as assumed in the question. The Poisson limit theorem, often called the law of small numbers in this setting, suggests a Poisson distribution when a goal count is the sum of many approximately independent rare scoring opportunities.

First suppose $L\in \mathbf{RP}\cap \mathbf{co}\text{-}\mathbf{RP}$. Run the RP algorithm for $L$ and the RP algorithm for its complement with fresh random bits. If the first accepts, output one; if the second accepts, output zero; otherwise repeat. Neither output can be wrong, and on every input the appropriate algorithm accepts in each round with probability at least $1/2$. The number of rounds is dominated by a geometric distribution of mean two, so this is an always-correct algorithm with polynomial expected running time.

Conversely, let $T$ be always correct when it halts and have expected running time at most $p(n)$. Run it for $2p(n)$ steps. Markov inequality givesAccept exactly when $T$ halts and outputs one; this is an RP algorithm for $L$. Accepting exactly when it halts and outputs zero is an RP algorithm for the complement. Thereforeis equivalent to zero-error expected polynomial time.

The minimum of the recovery and death clocks has an exponential distribution with rate $\mu +\gamma$. Its expected value is therefore

Write $Y_i$ for the colony count and $d_i$ for the dose on plate $i$. The first fit is the Poisson regressionThus its log-likelihood isThe Poisson deviance relative to the saturated model iswhere a summand with $y_i=0$ uses $0\log 0=0$.

At dose zero the fitted expected value is $e^{{\hat{\beta }}_0}=e^{3.321995}\simeq 27.72$ eradicated colonies. Increasing dose by one unit multiplies the fitted mean by $e^{{\hat{\beta }}_1}=e^{0.0001901}\simeq 1.000190$; for example, an increase of $100$ units multiplies it by about $1.0192$. The positive fitted effect is small and its displayed two-sided $p$-value, $0.105$, gives little evidence against a zero dose coefficient.

For $u\in C^2(\overline{D})$, Itô formula shows thatis a martingale. Take expectations and let $t\to \infty$. The function $u$ is bounded on the compact set $\overline{D}$, while $\Delta u$ is bounded and ${\mathbb{E}}_{x}T<\infty$ by part (a). The dominated convergence theorem therefore gives Dynkin formula for Brownian motion

Let $X_t=N_t^{+}-N_t^{-}$, where $N^{+}$ and $N^{-}$ are independent Poisson processes of rate $\lambda$. This symmetric Poisson difference process is centered, has jumps $\pm 1$, and has variance rate ${\sigma }^2=2\lambda$. If $a,b$ are positive integers, then $X_T\in \{-a,b\}$ exactly. Optional sampling of the martingale $X_t$ givesConsequentlyPart (d) now gives the mean exit time

For $\mu >0$, let $Q$ be the geometric distributionIf $P$ is the law of $X$, Gibbs inequality givesThereforeFor $\mu =0$, the nonnegative random variable $X$ is zero almost surely and both sides vanish. This proves that the maximum entropy distribution on the nonnegative integers with fixed expected value is geometric.

Fix positive $\alpha$, $D\ge 2$ and $B>0$. Suppose a size band $[k,Dk)$ of an Achlioptas process contains at least $\alpha n$ vertices at time $m$. For every fixed $k$, at least $\beta n$ of those vertices remain in unchanged graph components throughout the next $T=\lceil Bn/k\rceil$ steps, with high probability, where one may take $\beta =\alpha e^{-16BD}/4$. This is a deliberately nonoptimal positive constant.

For independent uniform candidate edges, an initial graph component with $w<Dk$ vertices is untouched if none of the $2T$ candidate edges meets it. A uniform edge hits it with probability at most $3w/n$. For sufficiently large $n$, $(1-3w/n{)}^{2T}\ge e^{-16BD}$, so the expected value of the untouched vertex mass $Y$ is at least $\alpha e^{-16BD}n$. Replacing one row of two candidate edges can change $Y$ by at most $8Dk$, because only the initial graph components hit by an old or new endpoint can change status. The McDiarmid inequality therefore makes $Y\ge \alpha e^{-16BD}n/2$ except on an event of probability $e^{-cn/k}$, for a positive constant $c$ depending only on $\alpha ,B,D$. An untouched graph component stays unchanged whichever offered edge is selected.

For candidates restricted to absent edges, generate each by an independent uniform proposal stream and reject already present edges. During any interval of $O(n)$ steps with $O(n)$ present edges, each proposal has rejection probability $O(1/n)$ conditional on the past. The number of rejections is at most $O(\log n)$ except on an event of probability smaller than any inverse power of $n$: if there were $r$ rejections among $2T+r$ proposals, a union bound over their positions bounds this by $\left(\genfrac{}{}{0ex}{}{2T+r}{r}\right)(C/n{)}^r$. The first $2T$ proposals are independent and give the previous untouched-mass estimate. Extra proposals can touch at most $2r$ additional initial graph components, losing only $O(Dk\log n)=o(n)$ vertices for fixed $k$. This leaves the stated $\beta n$ bound. A union bound makes these estimates simultaneous over starting times $m\le 3n$. They hold through the entire interval, so deterministic or random stopping times within it are allowed.

The population risk of a prediction rule is its expected loss on a fresh observation from the target probability distribution.

The posterior mean is the expected value of a parameter under its posterior distribution; under squared-error loss it is the Bayes estimator.

Quota share reinsurance retains a fixed proportion $\alpha \in [0,1]$ of each claim. The annual retained aggregate is $\alpha S$, whose expected value and variance are $\alpha \mathbb{E}S$ and ${\alpha }^2\mathrm{Var}(S)$.

For independent claim sizes with common expected value $m$ and variance $v$, independent of the nonnegative integer count $N$, the aggregate satisfies $\mathbb{E}S=m\mathbb{E}N$ and $\mathrm{Var}(S)=v\mathbb{E}N+m^2\mathrm{Var}(N)$. Its moment-generating function is $G_N(M_X(t))$ wherever finite. These identities follow from the law of total expectation and law of total variance.

A reparameterization gradient differentiates an expected value by representing the simulated random variable as a differentiable transformation of parameter-independent noise. For $X_{\eta }=g_{\eta }(\epsilon )$, the identity ${\nabla }_{\eta }\mathbb{E}[F_{\eta }(X_{\eta })]=\mathbb{E}[{\nabla }_{\eta }{F}_{\eta }(g_{\eta }(\epsilon ))]$ requires justified differentiation under the integral sign. For a normal distribution, use $X=m+e^{\rho }\epsilon$ with $\epsilon$ drawn from the standard normal distribution.

Applying a measurable per-claim retention $g$ to independent claim sizes preserves the compound Poisson distribution form. The transformed severity is the pushforward measure of the original severity law. With count parameter $\lambda$, the retained aggregate has expected value $\lambda \mathbb{E}g(X)$ and variance $\lambda \mathbb{E}[g(X{)}^2]$. Zero retained marks may be removed by Poisson thinning.