互斥事件与独立事件

互斥事件(disjoint event):A∩B= ?,即P(A∩B)=0。在venn graph上可以清晰的看出不相交的两个区域就是两个互斥事件。
独立事件(independent event):P(A∩B)=P(A)P(B)。独立事件与互斥事件不同,无法在venn graph上看出。独立事件更像是多维空间中的概念,事件A对应于一个子空间,而事件B对应于另一个子空间,A∩B是两个空间的相交叠的子空间。而P(x)则是一个特殊的函数,满足P(A∩B)=P(A)P(B),因此独立性不仅取决于空间,还取决于函数本身。独立事件要么通过假设得到,要么通过验证P(A∩B)=P(A)P(B)得到。

下面一个小例子可以检验你是否真正理解了独立事件。
You have probably heard it before. Now you can solve it rigorously. It is called the “Monty Hall Problem.” A prize is placed at random behind one of three doors. You pick a door. To be concrete, let’s suppose you always pick door 1. Now Monty Hall chooses one of the other two doors, opens it and shows you that it is empty. He then gives you the opportunity to keep your door or switch to the other unopened door. Should you stay or switch? Intuition suggests it doesn’t matter. The correct answer is that you should switch. Prove it. It will help to specify the sample space and the relevant events carefully. Thus writeΩ={(w1,w2),wi∈{1,2,3}},where w1 is where the prize is and w2 is the door Monty opens.
翻译:
三门问题(Monty Hall problem)亦称为蒙提霍尔问题、蒙特霍问题或蒙提霍尔悖论,大致出自美国的电视游戏节目Let’s Make a Deal。问题名字来自该节目的主持人蒙提·霍尔(Monty Hall)。参赛者会看见三扇关闭了的门,其中一扇的后面有一辆汽车,选中后面有车的那扇门可赢得该汽车,另外两扇门后面则各藏有一只山羊。当参赛者选定了一扇门,但未去开启它的时候,节目主持人开启剩下两扇门的其中一扇,露出其中一只山羊。主持人其后会问参赛者要不要换另一扇仍然关上的门。问题是:换另一扇门会否增加参赛者赢得汽车的机率?如果严格按照上述的条件,即主持人清楚地知道,哪扇门后是羊,那么答案是会。不换门的话,赢得汽车的几率是1/3。换门的话,赢得汽车的几率是2/3。

参考文献:
【1】https://en.wikipedia.org/wiki/Monty_Hall_problem
【2】Wasserman L. All of Statistics: A Concise Course in Statistical Inference[M]. Springer, 2010.

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