结合两个正常随机变量 [英] Combining two normal random variables

问题描述

假设我有以下2个随机变量:

suppose I have the following 2 random variables :

X，其中均值= 6，标准差= 3.5
Y，平均值= -42，stdev = 5

X where mean = 6 and stdev = 3.5
Y where mean = -42 and stdev = 5

我想基于前两个变量创建一个新的随机变量Z，并且知道:X发生90％的时间，Y发生10％的时间.

I would like to create a new random variable Z based on the first two and knowing that : X happens 90% of the time and Y happens 10% of the time.

计算Z的平均值很容易:0.9 * 6 + 0.1 * -42 = 1.2

It is easy to calculate the mean for Z : 0.9 * 6 + 0.1 * -42 = 1.2

但是有可能在单个函数中为Z生成随机值吗? 当然，我可以按照以下方式做一些事情:

But is it possible to generate random values for Z in a single function? Of course, I could do something along those lines :

if (randIntBetween(1,10) > 1)
    GenerateRandomNormalValue(6, 3.5);
else
    GenerateRandomNormalValue(-42, 5);

但是我真的很想拥有一个函数，该函数可以用作这样的随机变量(Z)的概率密度函数，而该变量不一定是正常的.

But I would really like to have a single function that would act as a probability density function for such a random variable (Z) that is not necessary normal.

对不起的伪代码

感谢您的帮助！

编辑:这是一个具体的询问:

Edit : here would be one concrete interrogation :

假设我们将Z的5个连续值的结果相加.以大于10的数字结束的概率是多少?

Let's say we add the result of 5 consecutives values from Z. What would be the probability of ending with a number higher than 10?

推荐答案

但是我真的很想有一个 单一功能 这样的概率密度函数 随机变量(Z)不是 必不可少的.

But I would really like to have a single function that would act as a probability density function for such a random variable (Z) that is not necessary normal.

好吧，如果您想要密度，这里是:

Okay, if you want the density, here it is:

rho = 0.9 * density_of_x + 0.1 * density_of_y

但是，如果不这样做，您将无法从该密度采样:1)计算其CDF(繁琐，但不可行)2)将其求逆(为此您将需要一个数值求解器).或者，您可以拒绝抽样(或其他形式，例如重要性抽样).这是昂贵的，而且要正确设置很麻烦.

But you cannot sample from this density if you don't 1) compute its CDF (cumbersome, but not infeasible) 2) invert it (you will need a numerical solver for this). Or you can do rejection sampling (or variants, eg. importance sampling). This is costly, and cumbersome to get right.

因此，您应该使用"if"语句(即，调用生成器3次)，除非您有很强的理由不这样做(例如，使用准随机序列).

So you should go for the "if" statement (ie. call the generator 3 times), except if you have a very strong reason not to (using quasi-random sequences for instance).

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