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The expected value of an indicator random variable is the probability of its event. For a simple symmetric random walk absorbed at $0,k$, the gambler's ruin success probability from $i$ is $i/k$. HenceThe boundary values are $0$ at the left exit and $1$ at chamber $k$.

Interpret existence of the expectations as finite expected values, equivalently $\mathbb{E}|X|<\infty$ and $\mathbb{E}|Y|<\infty$. First justify the summations rather than assuming that a signed double series can be rearranged. Independence and Tonelli theorem giveConsequently $XY$ is an integrable random variable, and Fubini theorem permits summing the signed terms in either order:If extended infinite expectations are allowed instead, the phrase “exist” needs additional care: expressions such as $0\cdot \infty$ do not define an expectation. The finite-expectation convention avoids that ambiguity.

For a probability density function $f$, the finite expected value isThe absolute integrability condition makes the signed integral unambiguous. One can also allow an extended expectation when exactly one of the positive and negative parts has infinite integral, obtaining $+\infty$ or $-\infty$. If both parts have infinite integral, the expectation is undefined; subtracting two infinities is not a definition.

The expected value is finite because $t>2$:The condition on $t$ is precisely the first moment threshold for this parameterization of the Pareto distribution.

Part (a) gives $\mathbb{P}(X\in \mathbb{Z})=0$. Therefore the ceiling function and floor function satisfy $\lceil X\rceil =\lfloor X\rfloor +1$ almost surely. Both are integrable, since $0\le \lfloor X\rfloor \le X$ and $\lceil X\rceil \le X+1$. Thus $\mathbb{E}[\lceil X\rceil ]=\zeta (t-1)+1$.

Now take expectations in $\lfloor X\rfloor \le X\le \lceil X\rceil$:Writing $s=t-1$, which ranges over every $s>1$, and rearranging yieldsThese are the zeta bounds from rounded Pareto variables: rounding below and above converts one elementary expected value into bounds on an infinite series.

Test homogeneity of the four category probabilities among the three samples. The null hypothesis is $p_{iA}=p_{iB}=p_{iC}$ for every category $i$; the alternative allows at least one difference. This is a Pearson chi-squared test of homogeneity, assuming independently sampled grains with consistent classification.

The row totals are $(195,52,64,16)$, the sample totals $(131,135,61)$ and the grand total $N=327$. Under the null hypothesis, $E_{ij}=O_{i+}{O}_{+j}/N$, giving the expected value tableFor example, $E_{\mathrm{undulatory},C}=52(61)/327=9.7003$, so that cell contributes $(4-9.7003{)}^2/9.7003=3.34974$. The complete cell contributions areso $X^2=6.87754$. There are $(4-1)(3-1)=6$ statistical degrees of freedom. The supplied $5\%$ critical value is $12.592$, and $6.87754<12.592$, so do not reject homogeneity at $\alpha =0.05$. The approximate upper-tail p-value is $0.3323$.

One of twelve expected cells is below five, with none below one; the asymptotic approximation is broadly defensible by a common expected-count guideline but is not exact. An exact/conditional simulation could check it, especially if grain counts are clustered. Do not pool categories solely to obtain a desired result. The conclusion is that these counts do not demonstrate different category proportions; they do not prove that the rocks share one geological formation. Different formations can share such proportions, and real grains within one rock can fail the independent-sampling assumption.

The Pearson correlation coefficient is the covariance divided by the product of the standard deviations. For real random variables with finite, strictly positive variances,It is the expected value of the product of the two centered variables after dividing each by its standard deviation. The Cauchy-Schwarz inequality gives $-1\le {\rho }_{X,Y}\le 1$. This removes measurement units, while retaining the direction and strength of the linear relationship. Constant random variables have zero standard deviation, so their Pearson correlation coefficient is undefined.

For the covariance model $Y=2X+Z$, with $X,Z$ independently uniform on $\{-1,0,1\}$, the calculation givesThe positive correlation is less than one because the independent $Z$ term introduces extra variation. For $Y=-2X+Z$ the correlation is $-2/\sqrt{5}$. Both examples appear in the covariance diagram.

For nonzero real $a,c$, the correlation of $aX+b$ and $cY+d$ is $\mathrm{sgn}(ac){\rho }_{X,Y}$. Positive rescaling and added offsets therefore leave the correlation unchanged; reversing one scale reverses its sign. Equality $|{\rho }_{X,Y}|=1$ means that one random variable is an affine function of the other almost surely, with a nonzero slope of the corresponding sign. This is the equality condition in the Cauchy-Schwarz inequality, and includes perfect positive correlation.

Zero correlation says that the centered variables are orthogonal in their square-integrable inner product. It permits nonlinear statistical dependence, as the example $Y=X^2$ in uncorrelated random variables demonstrates. More generally, a correlation is a statement about a joint probability distribution; identifying effects of interventions is the separate task of causal inference. This distinction keeps an observed association from being mistaken for a demonstrated mechanism.

An early source for standardized association is Francis Galton's 1888 paper “Co-relations and their Measurement, chiefly from Anthropometric Data”. He compared stature and forearm measurements and changed each measurement to units of its own variability so the two regression directions could be compared. His scale used probable error, so it should not be identified word for word with the modern standard deviation formula above. ^[ref].

For a dividend-free Black-Scholes model with $r>0$, $\sigma >0$, $K>0$ and positive integer $n$, the optimal stopping value of the power put option isIn the continuation region the stationary Black-Scholes equation has powers $S$ and $S^{-\gamma }$. Boundedness at infinity removes $S$. Value matching gives the coefficient $(K-b{)}^n{b}^{\gamma }$, maximized at the displayed $b$; equivalently this is the smooth fit boundary.

For a verification beyond constant-trigger policies, let $g(S)=(K-S{)}_{+}^n$. The maximum property of $b^{\gamma }(K-b{)}^n$ gives $V\ge g$. For the infinitesimal generator $\mathcal{L}=rS{\partial }_S+{\sigma }^2{S}^2{\partial }_{SS}/2$, $(\mathcal{L}-r)V=0$ above $b$. Below $b$, set $y=S/(K-S)\le \gamma /n$; thenThe candidate is continuously differentiable across $b$, so the generalized Itô formula has no derivative-jump local time of a semimartingale term. The discounted candidate is a nonnegative supermartingale, bounding every stopping reward by optional stopping. First hitting of $b$ attains the bound; the remainder at infinite time vanishes since $V\le K^n$ and $r>0$. For $n=1$ this recovers the perpetual put option. At $r=0$ the value is $K^n$ for $S>0$, approached but not attained at a finite exercise time. At $r<0$ the perpetual value is infinite, because exercise at deterministic times tending to infinity has unbounded discounted expected value.

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.

Photon statistics is the probability distribution of the number operator in specified field modes, or the distribution of registered counts in a specified gate. Those distributions agree only for an appropriate ideal measurement. Mode selection, losses, and detector dead time matter. For a one-mode density matrix, $p_n=⟨n|\rho |n⟩$ andThese two sums already distinguish preparations with the same expected value.

For free planar turning with constant $D_r$, no imposed torque and negligible inertia, model an unwrapped angle by the stochastic differential equationHere $W$ is standard one-dimensional Brownian motion, with increment variance $dt$, and brackets denote expected values over repeated realizations. The angular variance is conventionally reported in $\mathrm{ra}{\mathrm{d}}^2$. “Unwrapped” means recording accumulated turns continuously, rather than resetting the angle at each revolution. Directly subtracting wrapped camera angles creates artificial jumps; for an unmarked ellipsoid the wrapping period is $\pi$, and resolving successive turns additionally requires sufficiently frequent images.

For a planar arrow $\mathbf{u}=(\cos \varphi ,\sin \varphi )$, the orientational autocorrelation isFor an unoriented planar axis, use an autocorrelation unchanged by replacing either endpoint angle by that angle plus $\pi$:The same $D_r$ thus gives arrow and axis correlation times $1/D_r$ and $1/(4D_r)$, respectively, where a correlation time here is the lag at which the relevant curve reaches $1/e$. A short explanation is enough: $\Delta \varphi$ has a Gaussian distribution of variance $2D_{r}t$, so its characteristic function gives $⟨\cos (n\Delta \varphi )⟩=e^{-n^2{D}_{r}t}$. The integer $n$ selects what the observable can distinguish; it does not change the underlying angular spreading.

For an original numerical example, choose $D_r=0.25{\mathrm{s}}^{-1}$. After $2\mathrm{s}$, the angular variance is $1\mathrm{ra}{\mathrm{d}}^2$ and its root mean square displacement is $1\mathrm{rad}$, although the mean displacement is zero. The planar arrow autocorrelation is $e^{-0.5}\simeq 0.607$, while the planar axis autocorrelation is $e^{-2}\simeq 0.135$. Calling both “the orientation correlation” without specifying the observable would hide a factor of four in decay rate.

A point process records random locations by a random measure that counts them. Examples include arrival times of a queue, positions of stars, and the finite eigenvalue configuration of a random matrix. The count in an observation window is a random variable; its expected value is supplied by the intensity measure of a point process, while its fluctuations and relationships with other windows require more information.

For a precise spatial convention, take $E={\mathbb{R}}^d$ with its Borel sigma-algebra and a probability space $(\Omega ,\mathcal{F},\mathbb{P})$. A locally finite point process $\Pi$ is a measurable function from $\Omega$ to the space ${\mathcal{N}}_{\mathrm{lf}}(E)$ of locally finite point measures, equipped with the sigma-algebra generated by evaluations $\nu \mapsto \nu (A)$ for Borel sets $A$. Each realization has the formwhere $J(\omega )$ is finite or countable and repeated locations are allowed; ${\delta }_x$ is a Dirac measure. Here a point measure is a counting measure with possible multiplicities, rather than the counting measure of a set of distinct locations.^[ref]

Local finiteness means ${\Pi }_{\omega }(B)<\infty$ for every bounded Borel set $B$, on a single almost sure event. In ${\mathbb{R}}^d$ it suffices to check the countable collection of balls of integer radius centered at the origin, and this is equivalent to finiteness on every compact set. On a general measurable space there is no notion of boundedness without additional structure: state a metric or an explicit family of observation windows. Another convention on locally compact spaces requires finiteness on compact sets; this need not coincide with finiteness on all bounded sets outside ${\mathbb{R}}^d$. Local finiteness does not require finitely many points in all of $E$.^[ref]

The time of the $n$-th event of a Poisson process. At positive rate $\lambda$, the Poisson interarrival times $T_j-T_{j-1}$, with $T_0=0$, are independent and identically distributed random variables with the rate-$\lambda$ exponential distribution. In particular, $T_n$ has expected value $n/\lambda$ and variance $n/{\lambda }^2$. The inverse-count identity is $\{T_n\le t\}=\{N_t\ge n\}$.

The sum of independent Bernoulli random variables with possibly unequal success probabilities $p_i$ has a Poisson binomial distribution. Its probability generating function is $G_K(z)={\prod }_i(1-p_i+p_{i}z)$: expand the product to read off the probability mass function. Linearity of expectation and additivity of variance for independent random variables give $\mathbb{E}K={\sum }_i{p}_i$ and $\mathrm{Var}(K)={\sum }_i{p}_i(1-p_i)$. Equal $p_i=p$ recover the binomial distribution.

For three trials with success probabilities $1/10,3/10,4/5$, expanding $(9/10+z/10)(7/10+3z/10)(1/5+4z/5)$ gives masses $0.126,0.572,0.278,0.024$ at $0,1,2,3$. Their expected value is $1.2$ and variance is $0.46$. Replacing all three success probabilities by their average $0.4$ preserves the expected value but gives binomial distribution variance $0.72$, so it does not preserve the law.

Let $N$ be a nonnegative integer-valued random variable with finite expected value $\mu$, and independently label each of its items with either of two equally likely labels. If the two resulting counts are independent random variables, then $N$ has a Poisson distribution. Indeed, its probability generating function obeys $G(s)=G((1+s)/2{)}^2$. With $H(t)=G(1-t)$, iteration gives $H(t)=H(t/2^k{)}^{2^k}$. Finite expected value gives $H(h)=1-\mu h+o(h)$, hence $H(t)=e^{-\mu t}$ and $G(s)=e^{\mu (s-1)}$. Conversely, Poisson thinning produces independent Poisson distributions for the two counts.

For a rate-one Poisson process with Poisson arrival times $T_n$, suppose $t_n\to \infty$ and $t_n/n\to \lambda \in [0,\infty ]$. Thenwhere $1/\sqrt{\infty }=0$ and $\mathrm{BN}(\rho )$ denotes the bivariate normal distribution with standard normal marginal distributions, including the singular cases. In particular, equal scales give $\rho =-1$ and separated scales give $\rho =0$.

To prove this, fix real $x,y$ and put $s_n=n+y\sqrt{n}>0$ for large $n$. The event $(T_n-n)/\sqrt{n}\le y$ is exactly $N_{s_n}\ge n$. Counts at the two deterministic times $t_n,s_n$ are sums of a shared count on $[0,\min (t_n,s_n)]$ and an independent count on the remaining interval. The central limit theorem for a centered Poisson random variable of diverging parameter follows by writing a count at a real time $u$ as its count at $\lfloor u\rfloor$, a sum of independent unit increments, plus an independent remainder with expected value at most one. Markov's inequality makes that remainder negligible after division by $\sqrt{u}$.

Apply this central limit theorem to the two independent count blocks whenever their lengths diverge. A block with bounded length, or length negligible compared with the scale of its coordinate, contributes a term tending to zero in probability. The standardized pair of counts therefore has a joint normal limit whose correlation coefficient isThis includes $\lambda =1$: the difference block then has negligible variance after normalization. In the arrival event the standardized count at $s_n$ must be at least $(n-s_n)/\sqrt{s_n}\to -y$. Reversing this inequality changes the sign of the second limiting normal random variable, giving the asserted negative correlation coefficient. Continuity of the marginal distributions permits the moving thresholds and the weak inequalities, including at the singular limits.

Every $\rho \in [-1,0)$ is realized by $t_n=\lambda n$ with $\lambda ={\rho }^{-2}\ge 1$, and zero by $t_n=n^4$. Conversely, even when $t_n/n$ has no limit, it has a subsequence converging in the extended interval $[0,\infty ]$. If the original pair has a joint bivariate normal distribution limit, that subsequence has the same limit, forcing its correlation coefficient into $[-1,0]$.

Consider a deliberately simplified model: fatal defects form a spatially uniform Poisson point process with intensity $D$ per unit area. A semiconductor die of area $A$ works exactly when it contains no fatal defect. Its defect count $N$ has a Poisson distribution with expected value $DA$, hence its probability of working, called its yield, isThe product $DA$ is dimensionless. The formula concerns fatal defects, not every visible imperfection. Spatial uniformity, independent defect counts in disjoint areas, and failure from even one fatal defect are model assumptions. Clustering, shared processing failures, repairable regions and defects that affect only some functions can invalidate them. This is a chosen probability model, not historical yield data.

Choose $D=0.005{\mathrm{mm}}^{-2}$. For semiconductor die areas $25$, $100$ and $200{\mathrm{mm}}^2$, the model gives $Y=e^{-0.125}=0.882497$, $e^{-0.5}=0.606531$ and $e^{-1}=0.367879$. Doubling the area squares the yield: $Y(2A)=Y(A{)}^2$. This follows either from the exponential function or by splitting the larger area into two disjoint parts and using independence. More functions on one semiconductor die can save assembly while making the no-defect requirement harder to satisfy.

A Homogeneous Poisson process models arrivals at a constant rate $\lambda >0$. Its counting process $N_t=N(t)$ counts arrivals in $(0,t]$, with $N_0=0$. Its paths are right-continuous nondecreasing functions, take nonnegative integer values, and have unit jumps almost surely. The defining Poisson distribution of each increment isIt has stationary increments: the probability distribution above depends on the duration $t$, not the starting time $s$. It also has independent increments: counts in any finite collection of disjoint time intervals are independent random variables. Together these conditions specify the joint count laws, not merely the Poisson distribution at a single time.^[ref]

The rate has units of arrivals per unit time; $\lambda t$ is dimensionless. The Poisson distribution givesFor a toy rate $\lambda =2$ per minute and a duration of $1.5$ minutes, the expected value and variance of the count are both $3$. The probability of no arrival is $e^{-3}\approx 0.049787$, and the probability of exactly three is $3^3{e}^{-3}/3!=\frac{9}{2}{e}^{-3}\approx 0.224042$. In two consecutive one-minute intervals, the probability of exactly two arrivals in the first and exactly one in the second is $(2e^{-2})(2e^{-2})=4e^{-4}\approx 0.073263$, using independent increments. These are exact model calculations with rounded decimals, not measured frequencies.

The diagram's nine hand-chosen arrivals illustrate a path; the line $\lambda t$ is an expected value, not a prescribed trajectory. Constant rate does not imply evenly spaced arrivals. Poisson interarrival times connect the count description to waiting, while Poisson thinning explains how independently routing arrivals produces separate streams. This makes the Poisson process useful as a simple arrival model for a queue, provided constant rate and independent increments are plausible. For a deterministic changing rate, use an Inhomogeneous Poisson process; bunching or history-dependent arrivals need additional assumptions. Conditional on $N_t=n$, the arrival times are the order statistics of $n$ independent uniform distributions on $[0,t]$, as developed in Poisson process conditional arrival times.

Poisson thinning splits a Poisson process by independently marking each arrival as retained or rejected. Let the original Homogeneous Poisson process have rate $\lambda >0$, and let $p(s)\in [0,1]$ be a deterministic measurable function of time. Attach independent uniform distributions $U_1,U_2,\dots$ on $(0,1)$, independent of the entire original Poisson process. At arrival time $T_i$, retain the arrival if $U_i\le p(T_i)$. Given all arrival times, these decisions are independent and have the specified retention probabilities. Define the two counting processes using an indicator function:The Poisson thinning theorem gives independent retained and rejected Inhomogeneous Poisson processes. Their rates are $\lambda p(s)$ and $\lambda (1-p(s))$, respectively, in the sense of the intensity function of a point process. In particular, for $0\le a<b$,The independence is of the two whole processes: every finite collection of retained counts is independent of every finite collection of rejected counts. It is not merely zero covariance for one interval. For constant $p$, the streams are Homogeneous Poisson processes of rates $\lambda p$ and $\lambda (1-p)$; the endpoints $p=0,1$ include the identically zero stream.^[ref]

For an exact toy calculation, take $\lambda =2$ per minute and retain with $p(s)=1/4$ on $(0,2]$ and $p(s)=3/4$ on $(2,4]$. In the first two minutes, the retained and rejected expected values are $2\cdot 2\cdot 1/4=1$ and $2\cdot 2\cdot 3/4=3$. In the next two minutes they are $3$ and $1$. Over all four minutes, $R_4,D_4$ are independent Poisson distributions with mean $4$ each, and their sum $N_4$ has mean $8$. ThusThese probabilities and expected values describe the model, not empirical measurements. In the diagram, the original nine synthetic arrivals split into five retained and four rejected arrivals; those realized counts need not equal their expected values.

The mechanism explains why mutually exclusive labels can yield independent counts: randomness of the original total offsets the dependence from allocating a fixed total. For constant $p$ and an interval with original mean $m$, the conditional retained count has a binomial distribution given $N=n$. The joint probability generating function isThis factorization proves the two independent Poisson distributions; disjoint original intervals give independent pairs. For varying $p(s)$, the Independent marking theorem for Poisson point processes gives the same conclusion with the two integrated intensities. The joint red and green Poisson count formula gives the corresponding direct calculation with probability mass functions.

Independence here is unconditional. After conditioning on $N=n$, the identity $D=n-R$ gives covariance $\mathrm{Cov}(R,D\mid N=n)=-np(1-p)$ for constant $p$. Thus the retained and rejected counts are dependent for $n>0$ and $0<p<1$, even though their unconditional covariance is zero. Poisson thinning can model independent routing of arrivals, or independent detection of arrivals with a known efficiency schedule; using only a retention fraction without checking the marking mechanism can give a wrong count law.

The expected value of a numerical characteristic under the target population's probability distribution, when it exists. A sample mean estimates this fixed statistical parameter from observed data; it is not itself the population value.