Recent content by gimmytang

Thank you for your answer. It's correct since the distribution function is integrated to 1. I always have some confusion about the right domain of the variable since it's a function of the other variable. gim

Hi, There are two i.i.d uniform random variables X and Y. Now I need to know the density of Y/X. My method is like this: Let U=Y/X, V=X. Then the marginal density of U is what I need. f_{U}(u)={\int_{-\infty}^{\infty}f_{U,V}(u,v)dv}={\int_{0}^{1}f_{X,Y}(u,uv)|v|dv}={\int_{0}^{1}vdv}=1/2...

Actually the marginal distribution of U, namely the distribution of the ratio of two uniform variables, is the only thing that I am interested. To be more clear: f_{U}(u)={\int_{-\infty}^{\infty}f_{U,V}(u,v)dv}={\int_{0}^{1}f_{X,Y}(u,uv)|v|dv}={\int_{0}^{1}vdv}=1/2 Now the question is my...

Hello, Let X ~ U(0,1), Y ~U(0,1), and independent from each other. To calculate the density of U=Y/X, let V=X, then: f_{U,V}(u,v)=f_{X,Y}(v,uv)|v| by change of variables. Then: f_{U}(u)=\int_{0}^{1}{f_{X,Y}(v,uv)|v|dv}=\int_{0}^{1}{vdv}={1\over 2}, 0<u<\infty, which is not integrated to 1...

yep, you are right. thanks!

Hi, I need to calculate the density function of Z=X+Y, where X and Y are independent uniform distributed on [0,1]. The calculation is in the following: f_{Z}(z)=\int_{A}dx a. If 0<z<1, A={x:0<x<z} then f(z) = z; b. If 1<z<2, A={x:0<x<1} then f(z) = 1; Step b is wrong, but I don't know where...

I am still wondering the joint distribution of X and Y. There must be a solution to that. If it is not too difficult, please give me some hints. Thanks! gim

Thank you for your useful hint! The result following your method is E(XY)=0, then cor(X,Y)=0. In this sense X and Y are uncorrelated, but they are fully associated. gim

Thank you, snoble!

Suppose there are 10 nodes in a graph, I need to generate edges between nodes, but there are two conditions to be satisfied: 1) each node can have maximize of two edges. 2) no loop in the graph. The question is how to run a program which gives an algorithm to generate such a graph randomly...

X~N(0,1), Y=X^2~\chi^2(1), find E(XY). My thoughts are in the following: To calculate E(XY), I need to know f(x,y), since E(XY)=\int{xyf(x,y)dxdy}. To calculate f(x,y), I need to know F(x,y), since f(x,y)=d(F(x,y)/dxdy. F(x,y)=P(X\leq x, Y\leq y) \\ =P(X\leq x, X^{2} \leq...

Is it simply a definition which reflects the linear relationship between two variables?

I got it. It is easy to see after drawing a Venn diagram. There are four partitions on the diagram. So the possible sigma field is any combination of the four partitions, which has the number of 16 in total. Thank you all!

maybe aliens tried to contact us through microwave 200 years ago, but at that time nobody could sense that. :bugeye:

natural constant e=2.718...