【 以下文字转载自 Mathematics 讨论区 】 发信人: jinanddong (辐射), 信区: Mathematics 标 题: 知道random process (X+Y) 和 X 的distribution, 如何得到 Y? 发信站: BBS 未名空间站 (Sun Nov 16 00:46:53 2008) 现在我有两组实验数据, 可以分别画出两组数据的cummulative distribution function (CDF). 假设一组是X, 另一组是X+Y, X 是不要的那部分, 请问如何得到y 的distribution? Y最终是要用于simulation的, 所以不一定要精确的distibution, 只要知道每个P(Y=y) , 就可以了.
d*2
3 楼
...I think they are very easy to solve.....
l*r
4 楼
EM algorithm
y y)
【在 j********g 的大作中提到】 : 【 以下文字转载自 Mathematics 讨论区 】 : 发信人: jinanddong (辐射), 信区: Mathematics : 标 题: 知道random process (X+Y) 和 X 的distribution, 如何得到 Y? : 发信站: BBS 未名空间站 (Sun Nov 16 00:46:53 2008) : 现在我有两组实验数据, 可以分别画出两组数据的cummulative distribution : function (CDF). 假设一组是X, 另一组是X+Y, X 是不要的那部分, 请问如何得到y : 的distribution? : Y最终是要用于simulation的, 所以不一定要精确的distibution, 只要知道每个P(Y=y) : , 就可以了.
l*t
5 楼
我都作了,题目比较简单
L*a
6 楼
This is a basic quesiton of probability. Y = (X+Y) - X = (X+Y) + (-X). the pdf of the sum of two random variable is the convolution of the pdf of the two random variable.
y y)
【在 j********g 的大作中提到】 : 【 以下文字转载自 Mathematics 讨论区 】 : 发信人: jinanddong (辐射), 信区: Mathematics : 标 题: 知道random process (X+Y) 和 X 的distribution, 如何得到 Y? : 发信站: BBS 未名空间站 (Sun Nov 16 00:46:53 2008) : 现在我有两组实验数据, 可以分别画出两组数据的cummulative distribution : function (CDF). 假设一组是X, 另一组是X+Y, X 是不要的那部分, 请问如何得到y : 的distribution? : Y最终是要用于simulation的, 所以不一定要精确的distibution, 只要知道每个P(Y=y) : , 就可以了.
x*i
7 楼
i did them to make sure i understood the LOS. they're easier than real questions on the exam I think. 60% right is kinda low, but what truly measures your readiness should be the mock exams
Let Z=X+Y. Under the assumption that X and Y are independent, then PDF of Z is the convolution of PDF of X and PDF of Y. So knowing PDF(Z) and PDF(X), one can obtain the PDF(Y) by de-convolution. Fourier transform can be useful here. When X and Y are not independent, the solution will not be unique. Laputa's solution is not correct, because (X+Y) and (-X) are in general dependent, unless X is constant.
j*c
9 楼
同意。但是虽然不是简单的convolution,也不难求的Y的分布
Z useful
【在 p******h 的大作中提到】 : Let Z=X+Y. Under the assumption that X and Y are independent, then PDF of Z : is the convolution of PDF of X and PDF of Y. So knowing PDF(Z) and PDF(X), : one can obtain the PDF(Y) by de-convolution. Fourier transform can be useful : here. : When X and Y are not independent, the solution will not be unique. : Laputa's solution is not correct, because (X+Y) and (-X) are in general : dependent, unless X is constant.