November 22nd, 2010, 10:01 am
thank you, outrun!your solution is very inspired, accutally, the next question of this problem is 'how to simulate the correlated random variables', the common practice is Cholesky decomposition and eigenvalue decomposition.can you clarify this point more clearly?Quote You can see that "b" dosen't show up in the first two constrains Quoteappearently, 'b' indeed shows up in the second constrains.do you mean that whatever b is, the a*d = 0.6 and b*d + c*e =0.8 can be held simultaneously?here's my question:since here x,y,z are standard normals, cov(x,x) = cov(y,y)=cov(z,z) = 1the entire equations should be:a*d = 0.6b*d + c*e =0.8a^2 = 1b^2 + c^2 = 1d^2 + e^2 + f^2 =1a*b = rho_xyassume b = 1,(here rho_xy = cov(x,y) = a*b = 1)then a = 1 or -1 (come from a^2 = 1)c = 0 (come from b^2 + c^2 = 1)d = 0.6 or -0.6 (come from a*d = 0.6)but b*d + c*e =0.8 can never be satisfiedso can I suppose b cannot be equal to 1?
Last edited by
fogsnow on November 21st, 2010, 11:00 pm, edited 1 time in total.