h*****g
发帖数: 21 | 1 1，X,Y are standard distribution. X+Y=Z, Z is a constant, what is the
probability P(X|X+Y=Z)?
2. There are 100 seats, 100 people with number 1 to 100. each people can sit
on any seat. what is the expectation where id(seat)=id(people)? | i******t
发帖数: 370 | 2 老题目了
1. P(X|Z) ~ N(Z/2, sqrt(2)). Use Bayesian formula.
2. sum_{i=1}^{100}1/i. Use induction.
sit
【在 h*****g 的大作中提到】
: 1，X,Y are standard distribution. X+Y=Z, Z is a constant, what is the
: probability P(X|X+Y=Z)?
: 2. There are 100 seats, 100 people with number 1 to 100. each people can sit
: on any seat. what is the expectation where id(seat)=id(people)?
| h*****g
发帖数: 21 | 3 第一题还不太明白。
第二题,如果换成两个人，那么可能的排序是1 2 或者2 1， 各，占1/2概率，对号入座
的个数的期望值为： 2*1/2+0*1/2=1， 但按照你的公式是1.5？
【在 i******t 的大作中提到】
: 老题目了
: 1. P(X|Z) ~ N(Z/2, sqrt(2)). Use Bayesian formula.
: 2. sum_{i=1}^{100}1/i. Use induction.
:
: sit
| P****d
发帖数: 369 | | i******t
发帖数: 370 | 5 Possible to make some mistakes if not careful :P
【在 P****d 的大作中提到】
: 老题目也可以两道都做错lol
| i******t
发帖数: 370 | 6 Corrected for both problems.
For Problem 2, E(n)=1/n*(E(n-1)+1)+(n-1)/n*(E(n-1)-1/(n-1))=E(n-1)
【在 h*****g 的大作中提到】
: 第一题还不太明白。
: 第二题,如果换成两个人，那么可能的排序是1 2 或者2 1， 各，占1/2概率，对号入座
: 的个数的期望值为： 2*1/2+0*1/2=1， 但按照你的公式是1.5？
| h*****g
发帖数: 21 | 7 这就对了， 第二题第二项第二个因式(E(n-1)-1/(n-1))想了好久才明白。
多谢大侠了。
对了，这个老题目是哪里的？ MarkJoshi 和Zhou的书里都没有见过。 | M****e
发帖数: 3715 | 8 use indicator for problem 2
Xi=1 when person i seats on seat i, 0 otherwise
P(Xi=1)=1/100, then E(Xi)=1/100
E(sum{Xi})= sum{E(Xi)}=1
sit
【在 h*****g 的大作中提到】
: 1，X,Y are standard distribution. X+Y=Z, Z is a constant, what is the
: probability P(X|X+Y=Z)?
: 2. There are 100 seats, 100 people with number 1 to 100. each people can sit
: on any seat. what is the expectation where id(seat)=id(people)?
| h*****g
发帖数: 21 | 9 第一题还不明白，汗。 能详细解释下吗？
【在 i******t 的大作中提到】
: Corrected for both problems.
: For Problem 2, E(n)=1/n*(E(n-1)+1)+(n-1)/n*(E(n-1)-1/(n-1))=E(n-1)
| f*******y
发帖数: 267 | 10 X+Y ~ N(0, 2)
use bayesian theorem, P(X|X+Y=Z) = P(X)P(Z-X)/P(X+Y=Z)
and you will get normal dist with mean Z/2, std^2 = 1/2
for the second problem, can you explain the term
(E(n-1)-1/(n-1)) ??
thanks. | | | h*****g
发帖数: 21 | 11 the second term (n-1)/n*(E(n-1)-1/(n-1)) descirbes the expected number of
seats( and the person with the same number as the seats) when the nth guy
didn't seat in the nth seat( the probability is (n-1)/n).
Then there are totaly n-1 seat and n-1 person left. Assume that the nth guy
occupy the seat with the number of m, we can temporarily link the nth seat
and the mth person, then the expected number of expected number of seats is
E(n-1) which is the sum of the expected number for the mth perosn and the
expeced number for the rest(n-2) person.
E(rest)+1/(n-1)*1, then E(rest)=E(n-1)-1/(n-1), here is the scheme:
seat: 1,2,3,...,m,...n
guys: a,b,c,....n,...m
【在 f*******y 的大作中提到】
: X+Y ~ N(0, 2)
: use bayesian theorem, P(X|X+Y=Z) = P(X)P(Z-X)/P(X+Y=Z)
: and you will get normal dist with mean Z/2, std^2 = 1/2
: for the second problem, can you explain the term
: (E(n-1)-1/(n-1)) ??
: thanks.
| h*****g
发帖数: 21 | 12 厚颜再求教Prob. 1, 话说压根就没看懂题目，大牛不要喷饭。
X 和Y 是标准正态分布， 那么P（X|X+Y=0)是什么意思？记号P（A|B）的意思不是说
event B 成立的条件下A 成立的概率吗？那么P（X|Y）是什么意思？Y 和X 是个在无穷
区间的分布密度，放在这里是什么意思。Bayesian公式说的是P(A|B)*P（B）=P（B|A）
*P（A）里A和B都是事件，P（A）表示A事件发生的概率P(A|B)表示事件B发生的条件下A
发生的概率，把分布密度X，Y弄到这里面来是什么意思? | h*****g
发帖数: 21 | 13 厚颜再求教Prob. 1, 话说压根就没看懂题目，大牛不要喷饭。
X 和Y 是标准正态分布， 那么P（X|X+Y=0)是什么意思？记号P（A|B）的意思不是说
event B 成立的条件下A 成立的概率吗？那么P（X|Y）是什么意思？Y 和X 是个在无穷
区间的分布密度，放在这里是什么意思。Bayesian公式说的是P(A|B)*P（B）=P（B|A）
*P（A）里A和B都是事件，P（A）表示A事件发生的概率P(A|B)表示事件B发生的条件下A
发生的概率，把分布密度X，Y弄到这里面来是什么意思?
木人帮忙吗？
【在 f*******y 的大作中提到】
: X+Y ~ N(0, 2)
: use bayesian theorem, P(X|X+Y=Z) = P(X)P(Z-X)/P(X+Y=Z)
: and you will get normal dist with mean Z/2, std^2 = 1/2
: for the second problem, can you explain the term
: (E(n-1)-1/(n-1)) ??
: thanks.
| h****y
发帖数: 49 | 14 1.
define random variable W1=X+Y; W2=X-Y;
You can prove that W1 and W2 are independent given that X and Y are
independent.
Then X=(W1+W2)/2
Given that W1=Z, we have X=Z/2+W2/2
W2/2~N(0,1/sqrt(2)), so X~N(Z/2, 1/sqrt(2))
2. use define random variables xi_i = 1 when match and 0 otherwise.
Expectation is a linear operator. Xi_i are not independent, but it doesn't
matter here.
sit
【在 h*****g 的大作中提到】
: 1，X,Y are standard distribution. X+Y=Z, Z is a constant, what is the
: probability P(X|X+Y=Z)?
: 2. There are 100 seats, 100 people with number 1 to 100. each people can sit
: on any seat. what is the expectation where id(seat)=id(people)?
| h*****g
发帖数: 21 | 15 Smart idea to introduce random virables, but can you explain why it doesn't
matter when the two virable don't independent?
【在 h****y 的大作中提到】
: 1.
: define random variable W1=X+Y; W2=X-Y;
: You can prove that W1 and W2 are independent given that X and Y are
: independent.
: Then X=(W1+W2)/2
: Given that W1=Z, we have X=Z/2+W2/2
: W2/2~N(0,1/sqrt(2)), so X~N(Z/2, 1/sqrt(2))
: 2. use define random variables xi_i = 1 when match and 0 otherwise.
: Expectation is a linear operator. Xi_i are not independent, but it doesn't
: matter here.
| h****y
发帖数: 49 | 16 E(X+Y)=EX+EY no matter X Y are independent or not.
e.g. E(X+X)=EX+EX,
This is a result of the linearity of Expectation.
t
【在 h*****g 的大作中提到】
: Smart idea to introduce random virables, but can you explain why it doesn't
: matter when the two virable don't independent?
|
|
