概率 泊松过程零到达
Prob Poisson Zero Arrivals
题目详情
订单到达交易台遵循泊松过程，速率为每分钟 4 个订单。一分钟内恰好收到 0 个订单的概率是多少？
任务：泊松分布 P(X=k) = λ^k × e^(-λ) / k!。此处 λ=4，k=0：P(X=0) = 4^0 × e^(-4) / 0! = e^(-4) ≈ 0.0183。一分钟无订单的概率约 1.83%。
英文原题
Orders arrive at a trading desk according to a Poisson process with a rate of 4 orders per minute. What is the probability of receiving exactly 0 orders in a given minute?
解析
问题分析
Orders arrive at a trading desk according to a Poisson process with a rate of 4 orders per minute. What is the probability of receiving exactly 0 orders in a given minute?
解法
根据题目要求实现相应功能。核心逻辑需要：
// 核心数据结构和方法——根据题目 API 约定实现
// 1. 确定状态表示——选择支持所需操作的数据结构
// 2. 实现核心算法——确保 O(·) 时间复杂度和正确性
// 3. 处理边界条件——空输入、极值参数、并发访问验证
用具体输入验证：构造已知输入的测试用例，确认输出匹配预期结果。
复杂度与边界
- 时间复杂度：取决于选用的算法
- 空间复杂度：取决于数据规模
- 关键边界条件：空输入、极值参数、并发场景下的正确性保证
英文解析
Analysis
Orders arrive at a trading desk according to a Poisson process with a rate of 4 orders per minute. Determine the probability of receiving exactly 0 orders in a given minute? The Poisson distribution gives P(X=k) = (lambda^k * e^(-lambda)) / k!, where lambda = 4 (average orders per minute) and k = 0.
Solution
double poissonZeroProbability(double lambda = 4.0) {
// P(X=0) = lambda^0 * e^(-lambda) / 0! = e^(-lambda)
return std::exp(-lambda); // e^(-4) ≈ 0.01832
}Complexity & Edge Cases
- Time complexity: O(1)
- Space complexity: O(1)
- Edge cases: (1) Very high lambda makes P(X=0) essentially zero (2) Very low lambda (sparse events) makes P(X=0) close to 1 (3) Poisson assumes independent arrivals
Verification
Compute e^(-4) = 0.01832. Verify by simulation: generate 1M Poisson(4) random variables, count fraction with X=0. Should approximate 0.0183.
Key Considerations
The Poisson distribution is the standard model for order arrival in market microstructure. P(X=0) = e^(-lambda) is the probability of a quiet minute - no orders arriving. In low-frequency markets, this probability can be significant, affecting strategy design: if most minutes have zero arrivals, the strategy must handle long idle periods efficiently.