s****n
发帖数: 41 | 1 sorry, 很老的一道题，我被卡住了，怎么都想不通为什么我的方法不对，能帮我解释
一下嘛
“What is the average of the lesser of two random numbers from 0 to 1? ”
我用的方法为：
E[lesser of the two]=E[X|Y]*P(X
其中E[X|Y]=int(0_1)int(0_y)x*dx*dy=1/6，
为什么答案是1/3呢？ |
A****s
发帖数: 129 | 2 the integration is only on half of the prob space.
so you should have the integration on the whole prob space as 1/6*2=1/3, I
guess
geometrically you have a tetrahedron whose volume is 1/6
but the base area is only 1/2
【在 s****n 的大作中提到】
: sorry, 很老的一道题，我被卡住了，怎么都想不通为什么我的方法不对，能帮我解释
: 一下嘛
: “What is the average of the lesser of two random numbers from 0 to 1? ”
: 我用的方法为：
: E[lesser of the two]=E[X|Y]*P(X: 其中E[X|Y]=int(0_1)int(0_y)x*dx*dy=1/6，
: 为什么答案是1/3呢？
|
p*****k
发帖数: 318 | 3 swchen, for your approach, when calculating E[X|Y],
you need to normalize the probability:
E[X|Y]=int(0_1)dy[int(0_y)dx*x] / int(0_1)dy[int(0_y)dx*1] =1/3
Allens has also given a very nice geometrical pic above. |
b********u
发帖数: 63 | 4 you confused E[X|Y] and E[X|Y=y]. the former is still a random variable.
【在 s****n 的大作中提到】
: sorry, 很老的一道题，我被卡住了，怎么都想不通为什么我的方法不对，能帮我解释
: 一下嘛
: “What is the average of the lesser of two random numbers from 0 to 1? ”
: 我用的方法为：
: E[lesser of the two]=E[X|Y]*P(X: 其中E[X|Y]=int(0_1)int(0_y)x*dx*dy=1/6，
: 为什么答案是1/3呢？
|
h**6
发帖数: 4160 | 5 My solution:
E[lesser of the two] = int(0_1)dy{[int(0_y)dx*x] + int(y_1)dx*y]} =1/3 |
k*******d
发帖数: 1340 | 6 This is how I want to do it. Essentially the same as han6
We actually want to compute :
E[XI_{X<=Y} + YI_{Y<=X}], where I_{condition} is the indicator function
Then, simply use the definition
\int_0^1 (\int_0^y x dx + y \int_y^1 dx) dy |
