模拟泊松等待时间
我需要模拟泊松等待时间。我发现了许多模拟到达数量的示例，但我需要在给定平均等待时间的情况下模拟一次到达的等待时间。
我一直在寻找这样的代码：
public int getPoisson(double lambda)
{
double L = Math.exp(-lambda);
double p = 1.0;
int k = 0;
do
{
k++;
p *= rand.nextDouble();
p *= Math.random();
} while (p > L);
return k - 1;
}
但那是到达次数，而不是到达时间。
效率优先于准确性，更多的是因为功耗而不是时间。我使用的语言是 Java，如果算法只使用 Random 类中可用的方法，那就最好了，但这不是必需的。
I need to simulate Poisson wait times. I've found many examples of simulating the number of arrivals, but I need to simulate the wait time for one arrival, given an average wait time.
I keep finding code like this:
public int getPoisson(double lambda)
{
double L = Math.exp(-lambda);
double p = 1.0;
int k = 0;
do
{
k++;
p *= rand.nextDouble();
p *= Math.random();
} while (p > L);
return k - 1;
}
but that is for number of arrivals, not arrival times.
Efficieny is preferred to accuracy, more because of power consumption than time. The language I am working in is Java, and it would be best if the algorithm only used methods available in the Random class, but this is not required.
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到达之间的时间呈指数分布，您可以使用以下公式生成随机变量
X~exp(lambda)：更多信息请参阅 生成指数变量。
请注意，到达之间的时间也与首次到达之间的时间相匹配，因为指数分布是无记忆。
Time between arrivals is an exponential distribution, and you can generate a random variable
X~exp(lambda)with the formula:More info on generating exponential variable.
Note that time between arrival also matches time until first arrival, because exponential distribution is memoryless.
如果要模拟屏幕上出现的地震、闪电或生物，通常的方法是假设平均到达率为 λ 的泊松分布。
更容易做的事情是模拟到达间隔：
使用泊松分布，随着时间的推移，到达的可能性更大。它对应于该概率密度函数的累积分布。泊松分布随机变量的期望值等于 λ，其方差也等于 λ。
最简单的方法是对具有指数形式 (e)^-λt 的累积分布进行“采样”，得出 t = -ln(U)/λ。您选择一个统一的随机数 U 并代入公式以获得下一个事件之前应该经过的时间。
不幸的是，由于 U 通常属于 [0,1[ ，这可能会导致日志出现问题，因此使用 t= -ln(1-U)/λ 更容易避免它。
示例代码可以在下面的链接中找到。
https://stackoverflow.com/a/5615564/1650437
If you want to simulate earthquakes, or lightning or critters appearing on a screen, the usual method is to assume a Poisson Distribution with an average arrival rate λ.
The easier thing to do is to simulate inter-arrivals:
With a Poisson distribution, the arrivals get more likely as time passes. It corresponds to the cumulative distribution for that probability density function. The expected value of a Poisson-distributed random variable is equal to λ and so is its variance.
The simplest way is to 'sample' the cumulative distribution which has an exponential form (e)^-λt which gives t = -ln(U)/λ. You choose a uniform random number U and plug in the formula to get the time that should pass before the next event.
Unfortunately, because U usually belongs to [0,1[ that could cause issues with the log, so it's easier to avoid it by using t= -ln(1-U)/λ.
Sample code can be found at the link below.
https://stackoverflow.com/a/5615564/1650437