This is a pretty simple-minded question, but here goes.Suppose I have the following set up, where x and y are gaussian random variables, m, s1, s2 are constants (m is a mean, s1 and s2 are variances):x|y ~ N(y,s1)y ~ N(m,s2)From this, I can determine thatx ~ N(m,s1+s2)Now I would like to invert this to find y|x. It seems as though the answer is not unique! I could have y|x ~ N(k1*x,k2) where 0 < k1^2 < s1/(s1+s2) and k2 = s1 - (s1+s2)*k1^2.Have I completely messed up here? Or, if I'm right, how does one choose the optimal values of k1 and k2? I have seen in passing one author who uses k1 = k2 = s1/(1+s1) when s2=1, but no explanation was given (http://www.cs.toronto.edu/~roweis/csc25 ... aradox.pdf).Thanks!-- TMK --212-460-5430 home917-656-5351 cellt o p k a t z @ m s n . c o m

y/x ~ N( (xs2+ms1)/(s1+s2),s1s2/s1+s2)

Thank you, sanjay, this is very helpful. But I still think that this formula is not necessarily unique. I see that I neglected a constant term in the mean of y|x, it should be y|x = N(k1*x+b,k2). Then this gives us two equations with three parameters to solve: k1*m + b = m, (s1+s2)*k1^2 + k2 = s2. Your values of k1 = s2/(s1+s2), k2 = s1*s2/(s1+s2), b = m*s1/(s1+s2) solve these two equations, but so does the unconditional distribution of y, k1 = 0, k2 = s2, b = m. What makes your solution the "correct" one?Thanks!-- TMK --212-460-5430 home917-656-5351 cellt o p k a t z @ m s n . c o m

i didnt understand why you are equating the conditional(y/x) mean and variance with the unconditional Y mean and variance.Unconditional Y will take all the possible values for X and the Conditional one depends on a given X=x.For a given X=x the solution is unique.

you have x/y, y and x distribution function. f(y/x) = f(x,y)/f(x) = f(x/y)*f(y)/f(x)