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第273行： − + − − − − Let <math>X \sim \operatorname{Pois}(\lambda)</math> and <math>Y \sim \operatorname{Pois}(\mu)</math> be independent random variables, with <math> \lambda < \mu </math>, then we have that − − Let <math>X \sim \operatorname{Pois}(\lambda)</math> and <math>Y \sim \operatorname{Pois}(\mu)</math> be independent random variables, with <math> \lambda < \mu </math>, then we have that − − 设 x sim 操作者名称{ Pois }(lambda) </math > 和 < math > y sim 操作者名称{ Pois }(mu) </math > 是独立的随机变量，并且带有 < math > lambda < mu </math > ，那么我们就有了 − + − − <math> − − 《数学》 − − \frac{e^{-(\sqrt{\mu} -\sqrt{\lambda})^2 }}{(\lambda + \mu)^2} - \frac{e^{-(\lambda + \mu)}}{2\sqrt{\lambda \mu}} - \frac{e^{-(\lambda + \mu)}}{4\lambda \mu} \leq P(X - Y \geq 0) \leq e^{- (\sqrt{\mu} -\sqrt{\lambda})^2} − − Frac { e ^ {-(sqrt { mu }-sqrt { lambda }) ^ 2}{(lambda + mu) ^ 2}-frac { e ^ {-(lambda + mu)}}{2 sqrt { lambda mu }-frac { e ^ { e ^ {-(lambda + mu)}-(lambda + mu)}}}{4 lambda } leq p (x-y geq 0) leq ^ e ^ {-(sqrt { mu }-sqrt { lambda }) ^ 2} − </math> − − 数学 − − − − The upper bound is proved using a standard Chernoff bound. − − The upper bound is proved using a standard Chernoff bound. − − 利用标准的''' 切诺夫界Chernoff bound'''证明了上界的存在性。 − − − − The lower bound can be proved by noting that <math> P(X-Y\geq0\mid X+Y=i)</math> is the probability that <math>Z \geq \frac{i}{2}</math>, where <math>Z \sim \operatorname{Bin}\left(i, \frac{\lambda}{\lambda+\mu}\right)</math>, which is bounded below by <math> \frac{1}{(i+1)^2} e^{\left(-iD\left(0.5 \| \frac{\lambda}{\lambda+\mu}\right)\right)} </math>, where <math>D</math> is [[Kullback–Leibler divergence|relative entropy]] (See the entry on [[Binomial distribution#Tail Bounds|bounds on tails of binomial distributions]] for details). Further noting that <math> X+Y \sim \operatorname{Pois}(\lambda+\mu)</math>, and computing a lower bound on the unconditional probability gives the result. More details can be found in the appendix of Kamath ''et al.''.{{r|Kamath2015}} − The lower bound can be proved by noting that <math> P(X-Y\geq0\mid X+Y=i)</math> is the probability that <math>Z \geq \frac{i}{2}</math>, where <math>Z \sim \operatorname{Bin}\left(i, \frac{\lambda}{\lambda+\mu}\right)</math>, which is bounded below by <math> \frac{1}{(i+1)^2} e^{\left(-iD\left(0.5 \| \frac{\lambda}{\lambda+\mu}\right)\right)} </math>, where <math>D</math> is relative entropy (See the entry on bounds on tails of binomial distributions for details). Further noting that <math> X+Y \sim \operatorname{Pois}(\lambda+\mu)</math>, and computing a lower bound on the unconditional probability gives the result. More details can be found in the appendix of Kamath et al..+ − 下限可以通过下面的例子来证明: < math > p (X-Y geq0 mid x + y = i) </math > 是 < math > z geq frac { i }{2} </math > ，其中 < math > z sim 操作员名{ Bin }左(i，frac { lambda } + mu }右) </math > ,下面由 < math > frac {1}{(i + 1) ^ 2} e ^ { left (- iD left (0.5 | frac { lambda }(- iD left (0.5 | frac { lambda + mu } right)}}} </math > 限定，其中 < math > d </math > 是相对熵。进一步注意到 < math > x + y sim 操作者名{ Pois }(lambda + mu) </math > ，并计算无条件概率的下限得到结果。更多的细节可以在卡马斯等人的附录中找到。 + +
→Poisson races 泊松族群
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=== Poisson races 泊松族群===
=== 泊松族群===
设<math>X \sim \operatorname{Pois}(\lambda)</math>和<math>Y \sim \operatorname{Pois}(\mu)</math>是独立的随机变量，并且带有 <math> \lambda < \mu </math>，那么我们就有了
:<math>
:<math>
\frac{e^{-(\sqrt{\mu} -\sqrt{\lambda})^2 }}{(\lambda + \mu)^2} - \frac{e^{-(\lambda + \mu)}}{2\sqrt{\lambda \mu}} - \frac{e^{-(\lambda + \mu)}}{4\lambda \mu} \leq P(X - Y \geq 0) \leq e^{- (\sqrt{\mu} -\sqrt{\lambda})^2}
\frac{e^{-(\sqrt{\mu} -\sqrt{\lambda})^2 }}{(\lambda + \mu)^2} - \frac{e^{-(\lambda + \mu)}}{2\sqrt{\lambda \mu}} - \frac{e^{-(\lambda + \mu)}}{4\lambda \mu} \leq P(X - Y \geq 0) \leq e^{- (\sqrt{\mu} -\sqrt{\lambda})^2}
</math>
</math>
利用标准的切诺夫界证明了上界的存在性。
下限可以通过下面的例子来证明: <math> P(X-Y\geq0\mid X+Y=i)</math>是<math>Z \geq \frac{i}{2}</math> ，其中<math>Z \sim \operatorname{Bin}\left(i, \frac{\lambda}{\lambda+\mu}\right)</math>,下面由<math> \frac{1}{(i+1)^2} e^{\left(-iD\left(0.5 \| \frac{\lambda}{\lambda+\mu}\right)\right)} </math>限定，其中<math>D</math>是相对熵。进一步注意到<math> X+Y \sim \operatorname{Pois}(\lambda+\mu)</math>，并计算无条件概率的下限得到结果。更多的细节可以在Kamath等人的附录中找到。
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== ''' 相关分布'Related distributions''==
== ''' 相关分布'Related distributions''==