作者sky428 (人)
看板Math
標題[機統] 關於期望值
時間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).
下面是他給的解答
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
我機率沒學好，英文也很糟，所以看不懂>"<
我的疑問是...
0.2是怎麼來的??他所代表的意思是P(X=1|Y=2)??
(是因為第一次擲到的可能只剩下點數1~5，所以是1/5?)
為什麼不是(1/5)*(1/6).......我知道我哪裡想錯了
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
﹋﹋﹋﹌﹌﹌﹋﹋﹋﹋這地方不懂
E(X|Y=2) = .2(1) + .8(8) = 6.6
拜託解答一下，謝謝!!!!
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◆ From: 61.227.138.102
1F：→ yhliu :"為什麼不是(1/5)*(1/6)?" 01/23 13:06
2F：→ yhliu :我倒想問: 為甚麼是 (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 :整個計算的想法是: 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
※ 編輯: sky428 來自: 61.227.138.102 (01/23 14:25)
6F：→ sky428 :謝謝~ 我修改好了~ 不過我不懂E[X|X>=3]=1/p+2 01/23 14:30
7F：→ yhliu :E[X|X>=3] = 2+E[X], 前兩次已知不是所要點數, 從第 01/23 18:33
8F：→ yhliu :3次開始算平均需要再 E[X] 次. 01/23 18:34
9F：→ sky428 :懂了~ 謝謝!!! 01/23 20:15