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Let $Y_1,Y_2,\dots$ be real independent and identically distributed random variables with finite expected value $\mu$ and variance $0<{\sigma }^2<\infty$. The central limit theorem statesEquivalently, the cumulative distribution functions of these normalized sums converge at every real point to the standard normal distribution function. The theorem itself does not require a finite moment-generating function.

The given expected value and variance imply $M_X(0)=1$, $M_X^{\prime }(0)=0$ and $M_X^{\prime \prime }(0)=1$. The Taylor expansion therefore givesSince $\log (1+u)=u+o(u)$ as $u\to 0$, it follows thatThis is the quadratic expansion of a cumulant-generating function for a centered variable of variance one.

The unit-time increments $D_j=N_j-N_{j-1}$ are independent and identically distributed random variables, by independent increments and stationary increments of the Poisson process. Each has the Poisson distribution of parameter one, hence expected value one and variance one. The central limit theorem therefore gives

Let $X_f$ have the specified probability density function $f_m$. Splitting the integral into the positive and negative half-lines, and using the Gamma integral ${\int }_0^{\infty }{x}^r{e}^{-sx}dx=r!/s^{r+1}$ for a nonnegative integer $r$ and $s>0$, givesAt $t=2$ or $t=-2$, one term becomes the divergent integral of $x^m$; beyond these endpoints it grows exponentially. Consequently, the full extended-valued moment-generating function isIn particular $M_f(0)=1$, checking normalization of a probability distribution. All absolute moments exist. Direct integration yieldsThe symmetric distribution makes every odd moment zero, so

Now let $X_g$ have probability density function $g_m$. Its moment-generating function would beFor every $t\ne 0$, one summand has exponential growth, which dominates the polynomial decay. HenceThere is no finite moment-generating function on any neighborhood of zero, but some moments still exist. Specifically,ThusFor even $k\ge m+1$, the moment is $+\infty$, so it is not a finite moment. For odd $k\ge m+1$, both the positive and negative contributions have infinite integral, so the expected value is undefined. Cancellation in a Cauchy principal value does not define an expected value. In particular, when $m=0$ there are no finite positive-order moments.

First justify that the moments being used exist. For every $r>0$,The bound $|y{|}^k\le k!r^{-k}{e}^{r|y|}$ gives $\mathbb{E}|Y{|}^k<\infty$ for all $k$.

If $Y$ has a symmetric distribution, then $Y$ and $-Y$ have the same probability distribution, so for odd $k$,which forces $\mathbb{E}{Y}^k=0$.

Conversely, suppose every odd moment vanishes. For fixed $t\in \mathbb{R}$, the absolute values of the partial Taylor series for $e^{tY}$ are bounded by $e^{|t||Y|}$, whose expected value is finite by the preceding inequality. The dominated convergence theorem therefore permits termwise expectation:The uniqueness theorem for moment-generating functions implies equality of the probability distributions of $Y$ and $-Y$. Hence

Suppose a stationary distribution $\pi$ exists. Its stationarity equations areSince ${\sum }_j{\pi }_j=1$, its masses tend to zero as $|j|\to \infty$, and a positive maximum $M={\pi }_{j_0}$ is attained. Both terms on the right are at most $M$. Since $p,q>0$ and $p+q=1$, equality forces ${\pi }_{j_0-a}={\pi }_{j_0+b}=M$. Applying the same argument repeatedly at $j_0-a,j_0-2a,\dots$ givesThese are infinitely many distinct states, contradicting ${\sum }_j{\pi }_j=1$. ThusThis maximum principle proof uses neither a finite expected value of the state nor the coprimality assumption.

The strong law of large numbers states that, for independent and identically distributed random variables $X_1,X_2,\dots$ with $\mathbb{E}|X_1|<\infty$,Thus, outside an event of probability zero, the sample mean converges to the common expected value.

Direct integration of the uniform distribution yields $\mathbb{E}{U}_j=b/2$ and $\mathbb{E}{U}_j^2=b^2/3$. The expectation of a product of independent random variables givesSubtracting the square of the expected value gives the variance:Since $W_n\ge 0$, Markov's inequality implies, for every $\epsilon >0$,Hence $\boxed{W_n\to 0\text{ in probability for }0<b<2.}$ The first moment suffices: the variance need not tend to zero throughout this range.

Fix $0<c<1$ and choose $R_c$, independently of $X$, with the normal distributionThe linear combination of independent normal random variables $cX+R_c$ is normal. Its expected value is $c\mu +(1-c)\mu =\mu$ and its variance is $c^2{\sigma }^2+(1-c^2){\sigma }^2={\sigma }^2$. Hence $cX+R_c\stackrel{d}{=}X$ for every $c$, proving that every specified normal distribution is a self-decomposable distribution.

For independent and identically distributed random variables $X_j\sim \mathcal{N}(0,{\sigma }^2)$, the linear combination of independent normal random variables $n^{-1/2}{\sum }_{j=1}^n{X}_j$ has expected value zero and variance $n^{-1}n{\sigma }^2={\sigma }^2$. ThusThis is exactly the requested $2$-stability. The paper's definition is a strictly stable distribution, since it allows no centering term.

Write $A=g(X)$ and $B=g(Y)$. The two conditional expectation identities supply a $\sigma (Y)$-measurable version of $B$ and a $\sigma (X)$-measurable version of $A$. Work with those versions, which do not change any almost sure equality. Both lie in L2 space, and the Cauchy-Schwarz inequality makes their product an integrable random variable.

Testing the first conditional expectation identity against $B$, and the second against $A$, givesFor completeness, these tests with unbounded multipliers are valid by first using the bounded truncations $B_m=(-m)\vee (B\wedge m)$ and $A_m=(-m)\vee (A\wedge m)$, then passing to the limit with the Cauchy-Schwarz inequality and convergence in L2. Consequently,A nonnegative random variable with zero expected value vanishes almost surely. Therefore $g(X)=g(Y)$ almost surely. This is mutual conditional expectation rigidity in L2; it also follows from the equality case for conditional second moments.

For a counterexample to equality of the original random variables, let $X$ give probability $1/4$ to each of $-2,-1,1,2$, set $Y=-X$, and use the nonconstant function $g(t)=t^2$. Then $g(X)=g(Y)=X^2$ takes both values $1$ and $4$. It is measurable with respect to both generated sigma-algebras, so both required conditional expectation identities hold. Symmetry gives $\mathbb{E}X=\mathbb{E}Y=0$, while neither random variable can be zero. Thus

The L2-bounded martingale convergence theorem states that a martingale with ${\sup }_n\mathbb{E}|X_n{|}^2<\infty$ converges both almost surely and with convergence in L2 to a square-integrable random variable. Its hypotheses hold because $0<X_n<1$. Thus there is $X_{\infty }\in [0,1]$ almost surely with both kinds of convergence. Alternatively, the martingale convergence theorem gives the almost sure convergence, and the bound $|X_n-X_{\infty }{|}^2\le 1$ gives convergence in L2 by the dominated convergence theorem.^[ref]

The calculation of the conditional expectation of the square is especially informative. The martingale property makes the cross term vanish in the square expansion, soSubtracting this from $\mathbb{E}[X_{n+1}\mid {\mathcal{F}}_n]=X_n$ givesTaking expected values and iterating provesBy bounded convergence theorem, $\mathbb{E}[X_{\infty }(1-X_{\infty })]=0$. The random variable inside this expected value is nonnegative, so it is zero almost surely; on $[0,1]$ this forces $X_{\infty }\in \{0,1\}$. Again by bounded convergence theorem, $\mathbb{E}{X}_{\infty }={\lim }_n\mathbb{E}{X}_n=x$. HenceThis is the Bernoulli polarization martingale with update size $1/2$. It also has an exact mean squared error: the terminal representation $X_n=\mathbb{E}[X_{\infty }\mid {\mathcal{F}}_n]$ and the equality case for conditional second moments give

Let $m=\mathbb{E}X$. First, $m$ lies in the open interval $I$. For example, if $I$ has a finite lower endpoint $\ell$, then $X-\ell >0$ almost surely, and its expected value is strictly positive; hence $m>\ell$. The corresponding argument at a finite upper endpoint gives the other strict inequality. Integrability ensures $m$ is finite when either endpoint is infinite.

The tangent affine function at $m$ gives the supporting inequality for a differentiable convex function. Under the stipulated condition $f^{\prime \prime }>0$, contact with that tangent occurs only at $m$: defineThen $h(m)=0$, and $h(t)>0$ for every $t\ne m$. One direct proof uses $h^{\prime }(t)=f^{\prime }(t)-f^{\prime }(m)$: $f^{\prime }$ is strictly increasing, so $h$ decreases towards $m$ on the left and increases away from $m$ on the right. Thereforewith equality exactly when $h(X)=0$ almost surely, equivalently when $X=m$ almost surely. Conversely, an almost surely constant $X=m$ plainly gives equality. This proves the strict Jensen equality criterion.

No extra assumption that $f(X)$ is an integrable random variable is needed to interpret this argument. Its negative part is bounded by the absolute value of the integrable random variable $f(m)+f^{\prime }(m)(X-m)$, since $h(X)\ge 0$. Thus its expected value is defined in $(-\infty ,+\infty ]$; if it equals the finite number $f(m)$, then $\mathbb{E}[h(X)]=0$ and the required integrability follows automatically. Equality holds if and only if $X=\mathbb{E}X$ almost surely.

Yes. The preceding calculation established $\mathbb{E}{L}_{\tau }=-2ab/(b-a)<\infty$. For every $n$, the stopped martingale satisfiesHere $H$ is a single nonnegative integrable random variable that bounds the whole family. Thus, for any $K>0$,The last limit follows from the dominated convergence theorem. This is precisely uniform integrability. It also gives $M_{n\wedge \tau }\to M_{\tau }=|S_{\tau }|-L_{\tau }$ with convergence in L1, and therefore $\mathbb{E}{M}_{\tau }=0$, consistently with the preceding answer. A common bound on the individual expected values alone would not have sufficed; the integrable dominating random variable is what supplies control of every tail.

Let $g_i=(X_{i1},\dots ,X_{in})$ be row $i$ of $X$, viewed as a real vector. These are independent standard Gaussian random vectors, and $(X{X}^T{)}_{ij}=g_i\cdot g_j$. The trace expression for the empirical spectral measure givesWe need the norm momentsFor completeness, $S=\Vert g_i{\Vert }^2$ has the chi-squared distribution with $n$ degrees of freedom. By independence and the Gaussian integral, its moment-generating function is $\mathbb{E}{e}^{sS}=(1-2s{)}^{-n/2}$ for $s<1/2$. Differentiating at zero once, twice and three times proves these three identities.

For the first moment,For the second moment, if $i\ne j$, conditional expectation given $g_i$ gives $\mathbb{E}[(g_i\cdot g_j{)}^2\mid g_i]=\Vert g_i{\Vert }^2$, hence an unconditional value $n$. Separating equal and unequal row indices yields

For the third moment, the summand is $(g_i\cdot g_j)(g_j\cdot g_k)(g_k\cdot g_i)$. There are three disjoint cases.

ConsequentlyThe three limiting expected values are therefore

The correlation function of a point process is even and tends to zero as $|\lambda |\to \infty$. Its derivative factors asConsequently its local maxima occur at $0,\pm \sqrt{3}$, and its local minima at $\pm 1$. The heights are

For an interval $I$, ${\int }_I{R}_3^{(1)}(\lambda )d\lambda$ is the expected value of the number of eigenvalues in $I$. Thus equally short intervals have the highest expected counts nearwith the central peak highest. Each exchangeably labelled eigenvalue has the same marginal probability density function $R_3^{(1)}/3$. The three peaks should not be interpreted as three separate marginal densities for ordered eigenvalues.

In this case the most likely joint configuration also lies at these three positions. On the ordered chamber, the logarithm of the joint probability density function, up to a constant, isIts stationary equations are ${\lambda }_i=2{\sum }_{j\ne i}({\lambda }_i-{\lambda }_j{)}^{-1}$, satisfied by $(-\sqrt{3},0,\sqrt{3})$. For every nonzero real vector $v$, the second directional derivative isThus $L$ is strictly concave on this convex chamber, proving that the stationary configuration is its unique global maximum. Permuting the labels gives the other maxima on ${\mathbb{R}}^3$.

With $H>0$, setThen the first integral gives $1/\sqrt{1+y^{\prime 2}}=s$. On the descending and ascending halves of the physical track choose respectively $0<\varphi <\pi$ and $\pi <\varphi <2\pi$, so that $y^{\prime }=\cot (\varphi /2)$. Solving for $y$ givesThe unsquared equation requires $1-bs>0$. At $s=1$ this says $b<1$; it is not permissible to retain other values merely because the displayed formula for $y$ has a squared denominator.

For completeness, the Euler-Lagrange equation also gives $y^{\prime \prime }=(A^{\prime }(y)/A(y))(1+y^{\prime 2})<0$. Thus $y^{\prime }$ decreases through zero exactly once, so the chosen parameter runs through the whole single arch without an extra branch. Differentiating the formula for $y$ and dividing by $y^{\prime }=\cot (\varphi /2)$ gives, continuously also at $\varphi =\pi$,Using $x(0)=0$ yieldsSince $\sqrt{1+y^{\prime 2}}=1/\sin (\varphi /2)$, the arc length is obtained by cancelling one factor of $\sin (\varphi /2)$ in $dx/d\varphi$. The conditions $x(2\pi )=1$ and $S=L$ becomeThese determine a unique parameter pair for each $L>1$. To see uniqueness, regard $L$ as the expected value of $1/s$ under a probability density function proportional to $s^2/(1-bs{)}^3$ on $0<u<2\pi$. Differentiation under the integral sign gives the covariance identityThe first argument decreases with $s$ and the second increases. The strict sign follows by applying $(f(s)-f(t))(g(s)-g(t))<0$ to two independent unequal values and averaging; the integrable endpoint singularity of $1/s$ causes no problem. As $b\uparrow 1$, the weight concentrates at $s=1$, so $L\downarrow 1$. As $b=-k\to -\infty$, endpoint estimates give numerator $\sim 2/k^2$ and denominator $\sim 4\log k/k^3$, so $L\sim k/(2\log k)\to \infty$. Continuity then covers every $L>1$. At $b=0$ one recovers the cycloid and $L=4/\pi$.

Smoothness qualification. Near $\varphi =0$,and the other endpoint has $y^{\prime }\to -\infty$. These are smooth graphs on $0<x<1$, not $C^1$ graphs on the closed interval. Indeed a finite endpoint slope would make the conserved $H$ diverge as $y\to 0$. The travel time remains finite becauseThus the printed phrase “smooth solution” must be understood on the interior, with improper endpoint limits. The variational calculation supplies necessary stationary arches; it does not by itself prove global minimality in every larger class of tracks.

For Coulomb binding, the virial theorem gives a typical kinetic energy of the order of the binding energy. Taking $p_{\mathrm{typ}}\simeq \sqrt{2m_e|E_1|}$ and using the Heisenberg uncertainty relation estimate $r_{\mathrm{typ}}\sim \hslash /p_{\mathrm{typ}}$ givesThe Bohr model gives $r_1=a_0/Z=a_0/2=2.65\times 10^{-11}\mathrm{m}$ for a heavy helium nucleus, agreeing with the characteristic scale. Equivalently, minimizing the localization estimate $E(r)\sim {\hslash }^2/(2m_e{r}^2)-Z{e}^2/(4\pi {ϵ}_{0}r)$ gives that same scale.

This is an order-of-magnitude radius estimate, not an exact deduction of the cloud's expected value from the inequality alone. Numerical coefficients depend on which spread is used; $\Delta x\Delta p\ge \hslash /2$ does not assert $⟨r⟩=\hslash /p$. For the exact hydrogenic $1s$ state, $⟨r⟩=3a_0/(2Z)=3.97\times 10^{-11}\mathrm{m}$ at $Z=2$, one and a half times the Bohr orbital radius. An electron cloud has a radial distribution, whereas the Bohr model assigns a definite radius.

The wave function is a probability amplitude: by the Born rule, $|\psi (x){|}^{2}dx$ is the probability of a position measurement in an interval of width $dx$. For a normalized bound state $\int |\psi {|}^{2}dx=1$, the expected values areThe integrals run over the allowed domain. If the given state is not normalized, divide both by $\int |\psi {|}^{2}dx$. Its variance is $⟨x^2⟩-⟨x{⟩}^2$; neither $\psi$ itself nor its overall phase is a directly measured position probability.

Each energy eigenstate acquires its own phase under the time-independent Hamiltonian operator:Orthonormality removes all off-diagonal terms from the expected value of the energy, givingReturn up to an overall phase requires the relative phase to return to one. For $E_1\ne E_2$, the shortest positive recurrence time isIf the two energies coincide, the whole superposition is stationary up to a common phase for every time; there is then no smallest strictly positive recurrence time. A half-period changes the relative sign and produces the orthogonal antisymmetric state rather than the original symmetric state.