blocks|key|1815393|text|由于您可以访问统一的随机数生成器，因此使用inversion+method可以很容易地生成与您知道其CDF的其他分布一起分布的随机数。|type|unstyled|depth|inlineStyleRanges|entityRanges|offset|length|data|1815394|因此，在[0,1)中生成一个均匀随机数u，然后通过以下方法计算x：|style|CODE|1815395|x+=+log(1-u)/(-λ)，|1815396|其中λ是指数分布的速率参数。现在，x是一个具有指数分布的随机数。请注意，上面的log是自然对数ln。|1815397|entityMap|0|LINK|mutability|MUTABLE|url|http://en.wikipedia.org/wiki/Inverse_transform_sampling^0|L|G|0|0|4|5|J|1|V|1|0|0|H|0|2|1|H|1|13|3|1B|2|0^^$0|@$1|2|3|4|5|6|7|T|8|@]|9|@$A|U|B|V|1|W]]|C|$]]|$1|D|3|E|5|6|7|X|8|@$A|Y|B|Z|F|G]|$A|10|B|11|F|G]|$A|12|B|13|F|G]]|9|@]|C|$]]|$1|H|3|I|5|6|7|14|8|@$A|15|B|16|F|G]]|9|@]|C|$]]|$1|J|3|K|5|6|7|17|8|@$A|18|B|19|F|G]|$A|1A|B|1B|F|G]|$A|1C|B|1D|F|G]|$A|1E|B|1F|F|G]]|9|@]|C|$]]|$1|L|3|-4|5|6|7|1G|8|@]|9|@]|C|$]]]|M|$N|$5|O|P|Q|C|$R|S]]]]

Since you have access to a uniform random number generator, generating a random number distributed with other distribution whose CDF you know is easy using the <a href="http://en.wikipedia.org/wiki/Inverse_transform_sampling" rel="noreferrer">inversion method</a>.

So, generate a uniform random number, <code>u</code>, in <code>[0,1)</code>, then calculate <code>x</code> by:

<code>x = log(1-u)/(-λ)</code>,

where <code>λ</code> is the rate parameter of the exponential distribution. Now, <code>x</code> is a random number with an exponential distribution. Note that <code>log</code> above is <code>ln</code>, the natural logarithm.

blocks|key|2028853|text|抽样的基本定理认为，如果你能对期望的分布进行归一化、积分和倒置，你就可以自由地回家了。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|2028854|如果您在[a,b]上标准化了所需的分布F(x)。你可以计算|offset|length|style|CODE|2028855|C(y)+=+\int_a%5Ey+F(x)+dx|code-block|syntax|javascript|2028856|将其反转以获得C%5E{-1}，在[0，1]上均匀抛出z并查找|2028857|x_i+=+C%5E{-1}(z_i)|2028858|它将具有所需的分布。|2028859|2028860|在你的例子中：F(x)+=+ke%5E{-kx}和我将假设你想要[0,infinity]。我们得到：|2028861|C(y)+=+1+-+e%5E{-ky}|2028862|它是可逆的，给出|2028863|x+=+-1/k++ln(1+-+z)|2028864|对于在[0,1)上均匀抛出的z。|2028865|2028866|但是，坦率地说，使用经过良好调试的库更聪明，除非您是为了自己的启迪而这样做。|2028867|entityMap^0|0|4|5|J|4|0|0|7|6|P|1|0|0|0|0|7|F|U|C|0|0|0|0|3|5|0|0|0^^$0|@$1|2|3|4|5|6|7|18|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|19|8|@$D|1A|E|1B|F|G]|$D|1C|E|1D|F|G]]|9|@]|A|$]]|$1|H|3|I|5|J|7|1E|8|@]|9|@]|A|$K|L]]|$1|M|3|N|5|6|7|1F|8|@$D|1G|E|1H|F|G]|$D|1I|E|1J|F|G]]|9|@]|A|$]]|$1|O|3|P|5|J|7|1K|8|@]|9|@]|A|$K|L]]|$1|Q|3|R|5|6|7|1L|8|@]|9|@]|A|$]]|$1|S|3|-4|5|6|7|1M|8|@]|9|@]|A|$]]|$1|T|3|U|5|6|7|1N|8|@$D|1O|E|1P|F|G]|$D|1Q|E|1R|F|G]]|9|@]|A|$]]|$1|V|3|W|5|J|7|1S|8|@]|9|@]|A|$K|L]]|$1|X|3|Y|5|6|7|1T|8|@]|9|@]|A|$]]|$1|Z|3|10|5|J|7|1U|8|@]|9|@]|A|$K|L]]|$1|11|3|12|5|6|7|1V|8|@$D|1W|E|1X|F|G]]|9|@]|A|$]]|$1|13|3|-4|5|6|7|1Y|8|@]|9|@]|A|$]]|$1|14|3|15|5|6|7|1Z|8|@]|9|@]|A|$]]|$1|16|3|-4|5|6|7|20|8|@]|9|@]|A|$]]]|17|$]]

The Fundamental Theorem of Sampling holds that if you can normalize, integrate and invert the desired distribution you are home free.

If you have a desired distribution <code>F(x)</code> normalized on <code>[a,b]</code>. You compute

<pre><code>C(y) = \int_a^y F(x) dx
</code></pre>

invert that to get <code>C^{-1}</code>, throw <code>z</code> uniformly on [0,1) and find 

<pre><code>x_i = C^{-1}(z_i)
</code></pre>

which will have the desired distribution.

<hr>

In your case: <code>F(x) = ke^{-kx}</code> and I will assume that you want <code>[0,infinity]</code>. We get :

<pre><code>C(y) = 1 - e^{-ky}
</code></pre>

which is invertable to give

<pre><code>x = -1/k ln(1 - z)
</code></pre>

for z thrown uniformly on <code>[0,1)</code>.

<hr>

But, frankly, using a well debugged library is smarter unless you're doing this for your own edification.

blocks|key|1815654|text|这是我在维基百科上找到的公式：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1815655|T=+-Ln(u)+/λ|1815656|我们创建一个在0，1中具有均匀分布(u)的随机数，我们得到x：|1815657|Random+R=+new+Random()；|1815658|double+u=+R.+NextDouble()；|1815659|双x=+-Math.Log(u)/(λ)；|1815660|entityMap^0|0|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|N|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|O|8|@]|9|@]|A|$]]|$1|D|3|E|5|6|7|P|8|@]|9|@]|A|$]]|$1|F|3|G|5|6|7|Q|8|@]|9|@]|A|$]]|$1|H|3|I|5|6|7|R|8|@]|9|@]|A|$]]|$1|J|3|K|5|6|7|S|8|@]|9|@]|A|$]]|$1|L|3|-4|5|6|7|T|8|@]|9|@]|A|$]]]|M|$]]

This is the formula I found on Wikipedia :
T = -Ln(u) / λ
We create a random number with a uniform distribution (u) in [0,1]and we get x :
Random R = new Random();
double u = R. NextDouble();
double x = -Math.Log(u)/(λ);

blocks|key|1815356|text|如果您想要好的随机数，可以考虑链接到gsl例程：http://www.gnu.org/software/gsl/。他们有常规的gsl_ran_exponential。如果你想使用在[0，1)上均匀分布的内置生成器(例如，对于一些大的N，u=Random.Next(0，N-1)/N+)来生成随机数，那么只需使用：|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|1815357|-mu+*+log+(1-u)|code-block|syntax|javascript|1815358|请参阅gsl源代码中的randist/expential.c。|1815359|编辑:只是为了与后面的一些答案进行比较-这等同于u=+1/lambda。µ是分布的平均值，也称为OP链接到的wikipedia页面上的scale参数，lambda是rate参数。|1815360|entityMap|0|LINK|mutability|MUTABLE|url|http://www.gnu.org/software/gsl/^0|1R|J|O|W|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|W|8|@$9|X|A|Y|B|C]]|D|@$9|Z|A|10|1|11]]|E|$]]|$1|F|3|G|5|H|7|12|8|@]|D|@]|E|$I|J]]|$1|K|3|L|5|6|7|13|8|@]|D|@]|E|$]]|$1|M|3|N|5|6|7|14|8|@]|D|@]|E|$]]|$1|O|3|-4|5|6|7|15|8|@]|D|@]|E|$]]]|P|$Q|$5|R|S|T|E|$U|V]]]]

If you want good random numbers, consider linking to the gsl routines: <a href="http://www.gnu.org/software/gsl/" rel="noreferrer">http://www.gnu.org/software/gsl/</a>. They have the routine <code>gsl_ran_exponential</code>. If you want to generate random numbers using a built-in generator with a uniform distribution on [0, 1) (e.g. u=Random.Next(0, N-1)/N, for some large N), then just use:

<pre><code>-mu * log (1-u)
</code></pre>

See randist/exponential.c in the gsl source.

EDIT: just for comparison with some later answers - this is equivalent with mu = 1/lambda. mu here is the mean of the distribution, also called the scale parameter on the wikipedia page the OP linked to, and lambda is the rate parameter.

blocks|key|2028912|text|指数分布的一个有趣的性质:考虑一个具有指数到达间隔时间的到达过程。取任意时间段(t1、t2)和该时间段内的到达量。这些到达的数据在t1和t2之间均匀分布。(Sheldon+Ross，随机过程)。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|2028913|如果我有一个伪随机数生成器，并且出于某种原因(例如，我的软件不能计算对数)，你不想做上面的转换，而是想要一个指数r.v。平均值为1.0。|2028914|您可以：|2028915|1)创建1001U(0，1)个随机变量。|2028916|2)按顺序排序|2028917|3)第一个减去第二个，第二个减去第三个，……才能得到1000个差异。|2028918|4)这些差异是指数RVs，来自均值=+1.0的分布。|2028919|我认为效率较低，但却是达到同样目的的一种手段。|2028920|entityMap^0|0|0|0|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|R|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|S|8|@]|9|@]|A|$]]|$1|D|3|E|5|6|7|T|8|@]|9|@]|A|$]]|$1|F|3|G|5|6|7|U|8|@]|9|@]|A|$]]|$1|H|3|I|5|6|7|V|8|@]|9|@]|A|$]]|$1|J|3|K|5|6|7|W|8|@]|9|@]|A|$]]|$1|L|3|M|5|6|7|X|8|@]|9|@]|A|$]]|$1|N|3|O|5|6|7|Y|8|@]|9|@]|A|$]]|$1|P|3|-4|5|6|7|Z|8|@]|9|@]|A|$]]]|Q|$]]

One interesting property of the exponential distribution: Consider an arrival process with exponential interarrival times. Take any period of time (t1, t2) and the arrivals in that period. Those arrivals are UNIFORMLY distributed between t1 and t2. (Sheldon Ross, Stochastic Processes).

If I have a pseudo-random number generator and, for some reason (e.g. my software can't calculate logs) you don't want to do the above transformation, but want an exponential r.v. with mean of 1.0. 

You can :

1) Create 1001 U(0,1) random variables.

2) Sort in order 

3) Subtract the second from the first, third from the second,... to get 1000 differences.

4) Those differences are exponential RVs with from a distribution with mean = 1.0. 

Less efficient, I think, but a means to the same end.

blocks|key|1815034|text|这是我在面对类似需求时使用的：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1815035|//+sorry..+pseudocode,+mine+was+in+Tcl:

int+weighted_random+(int+max)+{
++++float+random_number+=+rand();
++++return+floor(max+-+ceil(+max+*+random_number+*+random_number))
}|code-block|syntax|javascript|1815036|当然，这是随机数平方的公式，所以你是沿着二次曲线生成一个随机数。|1815037|entityMap^0|0|0|0^^$0|@$1|2|3|4|5|6|7|K|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|L|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|M|8|@]|9|@]|A|$]]|$1|I|3|-4|5|6|7|N|8|@]|9|@]|A|$]]]|J|$]]

This was what I used when faced with similar requirements:

<pre><code>// sorry.. pseudocode, mine was in Tcl:

int weighted_random (int max) {
 float random_number = rand();
 return floor(max - ceil( max * random_number * random_number))
}
</code></pre>

Of course this is the formula of squaring the random number so you're generating a random number along a quadratic curve.

I would like to generate some pseudorandom numbers and up until now I've been very content with the .Net library's <code>Random.Next(int min, int max)</code> function. PRNGs of this variety are supposed to be using a <a href="http://en.wikipedia.org/wiki/Uniform_distribution_(continuous)" rel="noreferrer">Uniform distribution</a>, but I would very much like to generate some numbers using an <a href="http://en.wikipedia.org/wiki/Exponential_distribution" rel="noreferrer">Exponential Distribution</a>.

I'm programming in C#, although I'll accept pseudocode or C++, Java or the like. 

Any suggestions / code snippets / algorithms / thoughts?

Pseudorandom Number Generator - Exponential Distribution

Java

我想生成一些伪随机数，到目前为止，我对.Net库的Random.Next(int min, int max)函数非常满意。这种类型的应该使用，但我非常希望使用PRNG生成一些数字。我正在用C#编程，尽管我会接受伪代码或C++、Java或类似的语言。有什么建议/代码片段/算法/想法吗？

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