作者sky428 (人)
看板Math
标题[机统] 关於期望值
时间Sun Jan 23 12:49:23 2011
Question:
A fair die is rolled repeatedly.
Let X be the number of rolls needed to obtaina 5
and Y the number of rolls needed to obtain a 6.
Calculate E(X|Y = 2).
下面是他给的解答
Solution:
X follows a geometric distribution with p = 1/6.
Y = 2 implies the first roll is not a 6 and the second roll is a 6.
This means a 5 is obtained for the first time on the first roll
(probability = 20%)
or a 5 is obtained for the first time on the third or later roll
(probability = 80%).
E(X|X>=3) = 1/p + 2 = 6+2 = 8, so E(X|Y=2) = .2(1) + .8(8) = 6.6
我机率没学好，英文也很糟，所以看不懂>"<
我的疑问是...
0.2是怎麽来的??他所代表的意思是P(X=1|Y=2)??
(是因为第一次掷到的可能只剩下点数1~5，所以是1/5?)
为什麽不是(1/5)*(1/6).......我知道我哪里想错了
P(X=1|Y=2)=(1/6)*(1/6)/(5/6)*(1/6)=1/5
E(X|X>=3) = 1/p + 2 = 6+2 = 8
﹋﹋﹋﹌﹌﹌﹋﹋﹋﹋这地方不懂
E(X|Y=2) = .2(1) + .8(8) = 6.6
拜托解答一下，谢谢!!!!
--
※ 发信站: 批踢踢实业坊(ptt.cc)
◆ From: 61.227.138.102
1F：→ yhliu :"为什麽不是(1/5)*(1/6)?" 01/23 13:06
2F：→ yhliu :我倒想问: 为甚麽是 (1/5)*(1/6)? 01/23 13:06
3F：→ yhliu :又: E[X|X>=3]=1/p+2 = 6+2 01/23 13:07
4F：→ yhliu :整个计算的想法是: 01/23 13:08
5F：→ yhliu :E[X|Y=2] = P[X=1|Y=2]*1+P[X>1|Y=2]E[X|X>1,Y=2] 01/23 13:09
※ 编辑: sky428 来自: 61.227.138.102 (01/23 14:25)
6F：→ sky428 :谢谢~ 我修改好了~ 不过我不懂E[X|X>=3]=1/p+2 01/23 14:30
7F：→ yhliu :E[X|X>=3] = 2+E[X], 前两次已知不是所要点数, 从第 01/23 18:33
8F：→ yhliu :3次开始算平均需要再 E[X] 次. 01/23 18:34
9F：→ sky428 :懂了~ 谢谢!!! 01/23 20:15