布朗运动的样本路径为什么几乎必然处处连续，却又几乎必然处处不可微？
Why are the sample paths of Brownian motion almost surely continuous everywhere yet almost surely differentiable nowhere?
布朗运动增量 B(t+h)−B(t) 服从均值0、方差h的正态分布，典型位移按 √h 而非 h 收缩：h→0 时差商量级约为 1/√h 而发散，可微性几乎必然处处失效。柯尔莫哥洛夫连续性判据借助矩估计保证存在连续修正。“连续又处处不可微”正是标度指数为1/2、低于光滑所需的1的直接后果。
Brownian increments B(t+h)−B(t) are Gaussian with mean 0 and variance h, so typical displacements shrink like √h rather than h. As h→0 the difference quotient is of order 1/√h and blows up, so differentiability fails almost surely at every point. Meanwhile Kolmogorov's continuity criterion, built on moment estimates, guarantees a pathwise-continuous modification. Continuous-everywhere-yet-differentiable-nowhere is the direct consequence of the increment scaling exponent 1/2 lying below the value 1 required for smoothness.
难度 专业 · 适合 高中(15-18)
CC BY 4.0 · 已核验 · Verified 2026-11-03
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I. Karatzas 与 S. E. Shreve《布朗运动与随机分析》第2版 (Brownian Motion and Stochastic Calculus)，Springer GTM 113，1991。
I. Karatzas & S. E. Shreve, Brownian Motion and Stochastic Calculus, 2nd edition, Springer GTM 113, 1991.
独立核验依据 · Independent verification: 核验依据 1 OPENLIBRARY 标识 /works/OL1858792W：Brownian motion and sto
#随机过程 #布朗运动 #不可微性 #标度律
在 Galton–Watson 分支过程中，家族灭绝概率怎样确定？为什么后代平均数不超过 1 时灭绝几乎必然发生？
How is the extinction probability of a Galton–Watson branching process determined, and why is extinction almost certain when the mean number of offspring is at most one?
设一个个体留下 j 个后代的概率为 p_j，母函数为 f(s)=Σ p_j s^j，则灭绝概率 q 是方程 q=f(q) 的最小非负根。当后代期望 m=f′(1)≤1 且分布非退化时，q=1，种群几乎必然灭绝：随机波动得不到平均意义上的补偿，家系迟早中断。当 m>1 时 q<1，种群以正概率 1−q 无限延续，而初始祖先为 k 个时灭绝概率降为 q^k。
For a Galton–Watson process with offspring probabilities p_j and generating function f(s), the extinction probability q is the smallest nonnegative solution of q = f(q). If the mean offspring number f′(1) is at most one and the law is nondegenerate, then q=1: fluctuations are never compensated on average, so the family line eventually vanishes almost surely. If the mean exceeds one, then q<1 and the population survives forever with positive probability; starting from k initial ancestors, the extinction probability drops to q^k.
难度 专业 · 适合 高中(15-18)
CC BY 4.0 · 已核验 · Verified 2026-11-02
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T.E. Harris《分支过程理论》，Springer-Verlag，1963 年
T.E. Harris, The Theory of Branching Processes, Springer-Verlag, 1963
独立核验依据 · Independent verification: 核验依据 1 OPENLIBRARY 标识 /works/OL4649472W：The theory of branching
#分支过程 #灭绝概率 #母函数
强度为 λ 的泊松过程经过“独立稀疏化”——每个到达事件以概率 p 被保留，是否保留彼此独立——之后，保留下来的事件流为什么仍是泊松过程，其强度为何变成 pλ？
If a Poisson process of rate λ undergoes independent thinning — each arrival is kept with probability p, independently of everything else — why is the stream of retained events still a Poisson process, and why does its rate become pλ?
关键在于独立筛选与指数间隔时间的相互作用。相邻两次被保留事件之间的等待时间，是若干个独立的 Exp(λ) 间隔按几何分布加权求和，这样的几何混合仍服从 Exp(pλ)；由指数分布的无记忆性也能直接看出。不同区间的独立性继承自原过程，故新流的间隔独立同分布且无记忆，这正是强度 pλ 的泊松过程的刻画。
The mechanism lies in how independent filtering interacts with exponential interarrival times. The gap between successive retained events is a geometrically weighted sum of independent Exp(λ) gaps, and such a mixture is again Exp(pλ) — which also follows straight from memorylessness. Independence across disjoint intervals is inherited from the original process, so the thinned stream has i.i.d. memoryless gaps, precisely the characterizing property of a Poisson process of rate pλ.
难度 进阶 · 适合 初中(12-15)
CC BY 4.0 · 来源待核验 · Source pending review
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出处备注：作者 / 书名 / 版次 / 出版社 / 年份齐备，仅缺独立比对证据；已在可信度报告单列披露，不再重复核验。
Ross, S. M.，《随机过程》(Stochastic Processes)，第2版，Wiley，1996年。
Ross, S. M., Stochastic Processes, 2nd ed., Wiley, 1996.
#泊松过程 #稀疏化 #指数分布