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Put $\eta =\mathrm{Im}z>0$. For a real eigenvalue $\lambda$,The spectral theorem for real symmetric matrices therefore gives $\mathrm{Im}g>0$ and $\mathrm{Im}{q}_i\ge 0$, since $q_i$ is a sum of $(\lambda -z{)}^{-1}$ with nonnegative coefficients given by squared eigenvector coordinates of $x_i$. This is the positive imaginary part of a Stieltjes matrix resolvent. HenceApply the triangle inequality to the identity above, and add and subtract the minor trace:The allowed principal minor resolvent trace bound, with the printed $N$ normalization, bounds the second average by $c/(\eta N)$. Thus the explicit conclusion isIn particular, the final term is $O((\mathrm{Im}z{)}^{-3}{N}^{-1})$ with a constant independent of $N$. The upper-half-plane restriction gives the displayed positive denominators; in the lower half-plane the analogous bound uses $|\mathrm{Im}z|$.

The chosen centering also has a probabilistic meaning. The column vector $x_i$ is independent of the minor, has mean zero and coordinate variance $1/N$, so its conditional expectation satisfiesThis explains why the bound isolates fluctuations of a quadratic form around the minor trace. Mean centering alone is not a concentration estimate; no unrequested limiting assertion is being assumed.

Vary only the probability density function of $X$, using ${\eta }_t=\eta (1+tb)$ with bounded $b$ and $E_{\eta }b(X)=0$. The nuisance score function is $b(X)$, giving the statistical tangent setIts closed linear span is the nuisance tangent space of all centered $L^2(\eta )$ functions of $X$, by density of bounded centered scores.

The conditional expectation of the parametric score function given $X$ is zero:It is therefore orthogonal to this nuisance tangent space. Its orthogonal projection onto that space vanishes, so the efficient score is unchanged. By independence and $E{\epsilon }^2={\sigma }^2$,These equal the parametric score function and Fisher information when $\eta$ is known. There is no loss of information from the unknown covariate density. This is adaptivity to an unknown covariate distribution; it follows from score orthogonality, without having to estimate the nuisance parameter.

Work under the money-market risk-neutral measure, with $r_0\ge 0$. The zero-coupon bond price is $P(0,T)=\mathbb{E}[\exp (-{\int }_0^T{r}_{s}ds)]$. For this CIR model, write $p(\tau ,r)=A(\tau )e^{-B(\tau )r}$. The Feynman-Kac formula givesMatching the constant and linear coefficients of $r$ yieldsFor CIR bond pricing, solving this Riccati equation and integrating the scalar equation for $A$, put $\gamma =\sqrt{b^2+2{\sigma }^2}$ and $D(\tau )=(\gamma +b)(e^{\gamma \tau }-1)+2\gamma$. ThenSubstitution verifies both equations and their initial conditions. The discounted candidate solves the martingale pricing equation; its boundedness on finite horizons, or Feynman-Kac formula, identifies it with the conditional expectation. The answer is $\boxed{P(0,T)=A(T)e^{-B(T)r_0}}$. More generally the same coefficients give $P(t,T)=A(T-t)e^{-B(T-t)r_t}$.

The CIR model mean reverts to $a/b$ and has volatility $\sigma \sqrt{r}$, so it remains nonnegative and its fluctuation size depends on the rate level. For positive initial rate, zero is inaccessible if $2a\ge {\sigma }^2$; below this threshold zero is reachable but the process remains nonnegative. Its stationary law is a gamma distribution, of shape $2a/{\sigma }^2$ and rate $2b/{\sigma }^2$. Explicit affine bond prices and nonnegative rates are useful features. Limitations include exclusion of negative rates, restricted volatility behavior and a single stochastic factor; the time-homogeneous parameter family cannot fit an arbitrary initial term structure exactly.

The principle terminal correlation does not identify adapted stock volatility exposes a false printed assertion for the correlation coefficient defined in the introduction. Take $r=0$, $S_0=S_0^{\prime }=1$, andThen $d{S}_t=S_{t}d{B}_t$ andHence ${\sigma }_t=1$ and ${\sigma }_t^{\prime }=S_t/(1+S_t)$ are bounded. Both terminal prices have positive finite variance and are perfectly correlated, because one is a positive affine function of the other. Nevertheless $\boxed{{\int }_0^1({\sigma }_t-{\sigma }_t^{\prime }{)}^{2}dt={\int }_0^1\frac{dt}{(1+S_t{)}^2}>0}$ on every continuous finite stock path. Equal initial prices fix the means, not the affine slope.

Here is the exact general conclusion and the intended qualified proof. Bounded volatilities make these zero-rate stochastic exponentials true square-integrable martingales. Writing $s_0=S_0=S_0^{\prime }$, perfect terminal correlation gives $S_1-s_0=c(S_1^{\prime }-s_0)$ for some $c>0$. Conditional expectation therefore gives $S_u-s_0=c(S_u^{\prime }-s_0)$ for every $0\le u\le 1$. Comparing stochastic integrals, the Itô isometry yieldsThis relation need not imply equality of the volatilities.

The terminal proportionality identifies bounded stock volatility criterion repairs the statement. If one additionally assumes equal terminal variances, then $c=1$ and the whole stock paths coincide, so positivity gives $\sigma ={\sigma }^{\prime }$ almost everywhere and the requested integral is zero. The same repair follows if terminal proportionality $S_1=c{S}_1^{\prime }$ is assumed: equal martingale means force $c=1$, after which the conditional-expectation and Itô isometry argument applies. This proves the intended conclusion under a sufficient additional hypothesis without treating centered correlation as uncentered proportionality.

For zero drift, the reflection principle gives the known Brownian first-passage time densityTo obtain a general drift, on a fixed finite horizon use the exponential martingale $M_t=e^{c{B}_t-c^{2}t/2}$ as change-of-measure density. The Novikov condition holds for constant $c$, and the Girsanov theorem says that the coordinate process $B$ has drift $c$ under the new measure. Thus the hitting law of $B$ under that measure is the law of $B+ct$ under the original measure. Conditional expectation of $M_T$ at a hitting time bounded by $T$ equals its stopped value, by the martingale optional sampling theorem. At $H_a=t$ this value is $e^{ca-c^{2}t/2}$. The drifted Brownian first-passage density is obtained by multiplying $h_0$, giving $\boxed{h_c(t)=a{e}^{-(a-ct{)}^2/(2t)}/\sqrt{2\pi t^3}}$.

To verify the integrated formula, put $u=(a-cT)/\sqrt{T}$ and $v=(-a-cT)/\sqrt{T}$. The identity $e^{2ac}\varphi (v)=\varphi (u)$ for the standard normal density givesBoth terms vanish as $T\downarrow 0$, so $\boxed{\mathbb{P}(H_a\le T)=\overline{\Phi }((a-cT)/\sqrt{T})+e^{2ac}\Phi ((-a-cT)/\sqrt{T})}$. The distribution has total mass one for $c\ge 0$ and mass $e^{2ac}$ for $c<0$; in the latter case the remaining probability is an atom at infinity.

Use the usual normalization $Z_0=1$, finite deterministic initial wealth $x=X_0\ge 0$, positive state-price density, and nonnegative consumption. The deflated asset prices $ZP$ are local martingales, so predictable stochastic integrability makes $M=\int H\cdot d(ZP)$ a zero-starting local martingale. Integrating part (b) givesA nonnegative local martingale is a supermartingale: localize to martingales, apply Conditional Fatou lemma, and obtain the conditional expectation inequality and integrability. Therefore $N$, and hence $M=N-x$, are supermartingales. In particular,This is supermartingale control of deflated consumption gains. Nonnegative consumption is essential to the lower bound; it is the ordinary meaning of a consumption rate here. Wealth nonnegativity alone would not supply that bound if arbitrary signed cash flows were allowed.

Construct the densities using the Riesz representation theorem for a Hilbert space, rather than assuming the measure-decomposition conclusion. Put $\rho =\mu +\nu$, a finite dominating measure, and work in the real L2 space $L^2(\rho )$. The linear functionalis well-defined on its equivalence classes because $\nu \le \rho$. By the Cauchy-Schwarz inequality,Thus the Riesz representation theorem supplies $h\in L^2(\rho )$ such that $L(g)=\int ghd\rho$. Since $\rho$ is finite, $h$ is also integrable. Testing $g=1_A$ givesThe first identity and nonnegativity of $\nu$ rule out a positive-measure set where $h<0$; for example, test $\{h\le -1/n\}$. The second identity and nonnegativity of $\mu$ similarly rule out $\{h\ge 1+1/n\}$. Modify a measurable representative on a $\rho$-null set so that $0\le h\le 1$ everywhere. This is the Hilbert-space construction of dominated measure densities.

SetThen $B\in \Sigma$ and $\mu (B)=0$. The indicator identities for $\mu$ extend to integrals of nonnegative measurable functions by approximation with simple functions and the monotone convergence theorem. Consequently,and, for every $A$,HenceThis proves Lebesgue decomposition from a sum-measure density. The argument includes the cases of zero measures; there is no division on $B$, where the denominator vanishes.

Absolute continuity of measures, written $\nu \ll \mu$, means that $\mu (A)=0$ implies $\nu (A)=0$ for every $A\in \Sigma$. If $\nu \ll \mu$, then $\nu (B)=0$ in the construction above, so the singular term vanishes. Conversely, if $\nu (A)={\int }_{A}fd\mu$ with a nonnegative integrable $f$, that integral is zero on every $\mu$-null set. ThusThis is the finite-measure Radon-Nikodym theorem as a consequence of the proved construction.

Finally restrict the base measure to the sub-sigma-algebra ${\Sigma }_0$. For a real $f\in L^1(\mu )$, write $f=f^{+}-f^{-}$, with both parts nonnegative and integrable. On ${\Sigma }_0$ define finite positive measuresThey are absolutely continuous with respect to $\mu {|}_{{\Sigma }_0}$. The result just proved gives nonnegative, ${\Sigma }_0$-measurable densities $g_{\pm }$ satisfyingSet $f_0=g_{+}-g_{-}$. Since $\int (g_{+}+g_{-})d\mu =\int |f|d\mu <\infty$, this is in $L^1(\Omega ,{\Sigma }_0,\mu )$, and subtraction givesFor complex $f$, apply the real construction separately to its real and imaginary parts and combine the two results. This is conditional expectation from finite-measure densities. The resulting conditional expectation is unique almost everywhere: for two real versions, their difference has integral zero on its positive and negative level sets in ${\Sigma }_0$, forcing it to vanish; apply this to both components in the complex case. No probability normalization of $\mu$ is required.

Expose the independent coordinates one at a time and define the Doob exposure martingaleFor a fixed exposed prefix, couple the future coordinates identically under two choices of $Z_i$. The coordinate-change hypothesis bounds the difference of the resulting conditional expectations by $c_i$. Thus the increment $D_i=M_i-M_{i-1}$ has conditional mean zero and lies in a conditional interval of length at most $c_i$. This range bound, rather than merely $|D_i|\le c_i$, gives the sharp constant in the McDiarmid inequality.

Here is a proof of the needed Hoeffding lemma. For a mean-zero random variable $X$ in an interval of length $c$, let $h(u)=\log \mathbb{E}{e}^{uX}$. Its second derivative is the variance under the exponentially tilted distribution. A random variable in $[a,b]$ has variance at most $(b-a{)}^2/4$: the inequality $\mathbb{E}[(X-a)(b-X)]\ge 0$ gives $\mathrm{Var}X\le (\mathbb{E}X-a)(b-\mathbb{E}X)\le (b-a{)}^2/4$. Hence $h^{\prime \prime }(u)\le c^2/4$, and $h(0)=h^{\prime }(0)=0$ imply $h(u)\le u^2{c}^2/8$, for either sign of $u$.

Apply this conditionally to each increment and iterate the conditional expectation:For $u>0$, Markov's inequality bounds the upper tail by $\exp (-ut+u^2\sum c_i^2/8)$. Taking $u=4t/\sum c_i^2$ and then applying the same argument to $-f$ givesAt $t=0$ the bound is immediate. If all $c_i$ vanish, $f$ is constant on the product support and the positive tails are zero, so that degenerate case is handled directly.

For a binomial random graph, let coordinate $i$ be the entire vector of edges from vertex $i$ to vertices of smaller label. The coordinates are independent finite probability spaces. Changing coordinate $i$ changes only edges incident to that one vertex. Deleting the vertex gives the same graph under both outcomes, and each outcome's chromatic number is either that graph's chromatic number or one more. Thus the coordinate range is at most one. Taking all $c_i=1$ and $t=\lambda \sqrt{n}$ proves vertex exposure for chromatic number:Using individual edges as coordinates would give a weaker scale; grouping the incident edges is what yields the required bound.

For the expectation asymptotic, take fixed $0<p<1$, put $b=1/(1-p)$, and write $L={\log }_{b}n$. We outline both bounds and the step that turns probability estimates into an expectation estimate. For each fixed $\gamma >0$, the expected number of independent sets of size $k=\lceil (2+\gamma )L\rceil$ isIts logarithm is $k\log n-\frac{1}{2}k(k-1)\log b-O(k\log k)=-{\Omega }_{\gamma }((\log n{)}^2)$. Thus Markov's inequality gives $\alpha (G)\le (2+\gamma )L$ with probability tending to one, and $\chi (G)\ge n/\alpha (G)$ yields the corresponding lower bound on its expectation.

For the upper bound, set $m_0=\lceil n/(\log n{)}^2\rceil$ and $k=\lfloor (2-\gamma )L\rfloor$, with $0<\gamma <1$. The key uniform fact is that every vertex subset of size at least $m_0$ contains an independent $k$-set with probability tending to one. For a fixed subset of size $m$, let $X$ count its independent $k$-sets and let $\mu =\left(\genfrac{}{}{0ex}{}{m}{k}\right)b^{-\left(\genfrac{}{}{0ex}{}{k}{2}\right)}$. The normalized dependency sum in Janson inequality is bounded byThe $j=2$ term is $O(k^4/m^2)$. The remaining overlap terms give the same order or less: use $\left(\genfrac{}{}{0ex}{}{k}{j}\right)\left(\genfrac{}{}{0ex}{}{m-k}{k-j}\right)/\left(\genfrac{}{}{0ex}{}{m}{k}\right)\le (2k^2/m{)}^j$ for large $n$, and split the sum at $k/2$. For the lower half the terms beyond $j=2$ decrease at the initial endpoint and are exponentially small at the other endpoint; for the upper half, $k\le (2-\gamma /2){\log }_{b}m$ makes every endpoint exponent negative of order $(\log n{)}^2$. Also $1/\mu$ is exponentially small on that scale. The exponential form of Janson inequality therefore gives, uniformly for $m\ge m_0$,for a positive constant depending only on $p,\gamma$. A union bound over at most $2^n$ subsets succeeds, because $m_0^2/k^4$ is of order $n^2/(\log n{)}^8\gg n$. This is independent sets in every large subset of a dense random graph.

On that event, repeatedly remove an independent $k$-set and assign it one new colour until fewer than $m_0$ vertices remain; colour the remainder individually. This gives $\chi (G)\le n/k+m_0$. The failure probability is exponentially small compared with $1/\log n$, and always $\chi (G)\le n$, so the failure event contributes negligibly to the expectation. Combining the upper and lower bounds and then letting $\gamma \downarrow 0$ provesThis is the chromatic number of a binomial random graph asymptotic. The slash in the printed expression must be read with $2{\log }_{1/(1-p)}n$ as the denominator; a multiplicative reading would eventually exceed $n$. At the endpoint probabilities $p=0$ and $p=1$, the chromatic numbers are respectively one and $n$ for $n\ge 1$, so this logarithmic formula is intended for $0<p<1$.

If $T\le s$, then $\varphi (W_T)K$ is ${\mathcal{F}}_s$-measurable. Independence and centring of the future Brownian increment make the desired left side zero; the time multiplier on the right is zero as well.

Suppose $T>s$. Conditional on ${\mathcal{F}}_s$, write $W_T=W_s+U$ and $W_t-W_s=V$. The pair $(U,V)$ is jointly normal and independent of ${\mathcal{F}}_s$, withApply the supplied Gaussian integration by parts formula to $z\mapsto \varphi (W_s+z)$, treating the known $W_s$ as its parameter. This givesMultiply by the bounded ${\mathcal{F}}_s$-measurable $K$ and use the defining property of conditional expectation. Thus the required expectation identity holds for every $s<t$, including intervals crossing or lying after $T$.

Conditional on $\lambda$, a Poisson distribution has both conditional expectation and conditional variance equal to $\lambda$. For the uniform distribution on $(1,3)$,Thus $K=v/a=6$ and the one-year credibility factor is $Z_1=1/7$. With the observed count equal to one, the Bühlmann credibility estimate of the next year's conditional claim expected value isThe estimate gives relatively little weight to one year because the expected process variance is six times the variance of hypothetical means.

Backward induction makes the Snell envelope integrable and adapted: $|U_t|\le |Y_t|+\mathbb{E}(|U_{t+1}|\mid {\mathcal{F}}_t)$. Its definition gives $U_t\ge Y_t$ and $U_t\ge \mathbb{E}(U_{t+1}\mid {\mathcal{F}}_t)$, so it is a supermartingale dominating the reward.

For a stopping time $\tau$ taking values in $\{0,\dots ,T\}$, expand its stopped value asThe indicators are ${\mathcal{F}}_t$-measurable. Taking conditional expectations in each summand makes its expectation nonpositive by the supermartingale property. Since ${\mathcal{F}}_0$ is trivial, $U_0$ is deterministic andThis proves the finite-horizon optional sampling theorem directly in the instance needed here, without assuming nonnegative rewards.

Assume $\mathbb{E}|g(X)|<\infty$, or use nonnegative $g$ with extended expectations. Let $p_{ij}=\mathbb{P}(X=x_i,Y=y_j)$ and $p_j=\mathbb{P}(Y=y_j)$. For $p_j>0$, the discrete conditional expectation is $\mathbb{E}[g(X)\mid Y=y_j]={\sum }_{i}g(x_i)p_{ij}/p_j$. Therefore Fubini's theorem, or Tonelli theorem in the nonnegative case, gives the tower property of conditional expectation:Values of a conditional expectation on zero-probability conditioning events do not affect the sum. An unrestricted signed $g$ whose expectation is undefined is outside this assertion.

Write $M(x)=\mathbb{E}[N(x)]$. The waiting time for two increments larger than $1/2$ dominates $N(x)$ for $0\le x\le 1$ and has mean $4$, so $M$ is finite. Conditioning on the first increment, one step is always consumed; if that increment is below $x$, the remaining independent sequence has the original law. Thus first-step analysis and the tower property of conditional expectation giveThis first makes $M$ continuous and then gives $M^{\prime }=M$, with $M(0)=1$. Hence the uniform-sum crossing time below one hasAn independent explanation uses the tail-sum formula: $\mathbb{P}(N(x)>n)=\mathbb{P}(U_1+\cdots +U_n\le x)=x^n/n!$ for $x\le 1$, since this region is an $n$-dimensional simplex inside the unit cube. Summing from $n=0$ gives $e^x$. In particular the strict-crossing convention gives $N(0)=1$ almost surely and $\mathbb{E}N(1)=e$.

The conditional expectation $\mathbb{E}[X\mid \mathcal{G}]$ is a $\mathcal{G}$-measurable integrable random variable $Z$ such thatIt is defined up to almost sure equality. Measurability with respect to the sub-sigma-algebra and equality of these integrals are both essential: the first expresses that only the information in $\mathcal{G}$ is retained, and the second preserves all averages visible through that information. Existence follows from the Radon-Nikodym theorem; uniqueness is up to sets of probability zero.

Let $T\le N$ for a deterministic integer $N$. The stopped martingale has the finite-sum representationEach summand is integrable, and $\{T\ge k\}=\{T>k-1\}\in {\mathcal{F}}_{k-1}$ by the stopping time property. Therefore the conditional expectation identity for a martingale givesTaking expectations in the finite sum proves the bounded optional stopping theorem:No limiting argument or uniform-integrability assumption is needed for a bounded stopping time.

For the real-valued variables here, take a $\mathcal{G}$-measurable version of $X=\mathbb{E}[Y\mid \mathcal{G}]$. Since the variables are bounded, the conditional expectation defining identity extends to the bounded $\mathcal{G}$-measurable multiplier $X$, givingThe equality case for conditional second moments now givesA nonnegative random variable with zero expectation vanishes with probability one. Thus $X=Y$ as an almost sure equality. The same proof works for square-integrable variables; boundedness is more than is needed.

Fix $0<t<1$. The event $\{\tau \le t\}$ means that the future path through time one never exceeds the maximum already attained by time $t$. Given ${\mathcal{F}}_t$, future increments form an independent Brownian motion $W$, soHere the Brownian running maximum distribution is the one derived above. On $\{B_t<0\}$, an event of probability $1/2$, the finite quantity $M_t-B_t$ is strictly positive, so this conditional probability lies strictly between zero and one.

If $\tau$ were a stopping time, $\{\tau \le t\}$ would be ${\mathcal{F}}_t$-measurable, and the displayed conditional expectation of its indicator would equal that indicator, taking only zero and one. The contradiction proves $\tau$ is not an $({\mathcal{F}}_t)$-stopping time. Thus the Strong Markov property cannot be invoked at $\tau$ to assert a Brownian shifted process; the direct argument here does not assume that property at this random time.