Postby **cyanyoshi** » Sun Apr 30, 2017 11:54 pm UTC

My intuition is that with random variables X, Y and Z, then you can get everything you need from the pdf f(x,y,z). By normalizing f(x,y_{0},z_{0}), it seems that you should get f(x|y_{0},z_{0}), the conditional probability distribution of X given that Y=y_{0} and Z=z_{0}. Clearly, f(a,b|x_{0}) should be constant over the square (a,b)∈[0,x_{0}]^{2}, so f(a,b|x_{0})=1/(x_{0})^{2}. I am visualizing the domain of f(a,b,x) as an inverted pyramid with its vertex at (0,0,0) and its base is a square with vertices (0,0,1), (1,0,1), (1,1,1), and (0,1,1). Each horizontal slice should be a square with the same infinitesimal "mass" of dx. So then f(a,b,x) = 1/x^{2} within this pyramid and is zero elsewhere.

OK, now what about f(δ,x)? Fixing X=x_{0} again, f(a,b|x_{0}) is just a uniform distribution over a square. My gut tells me that f(δ|x_{0}) is proportional to the length of the level curve of δ(a,b) = |a-b| in this domain. This is just C*2*sqrt(2)*(x_{0}-δ). Normalizing over the domain [0,x] gives f(δ,x)=2*(x-δ)/x^{2}.

Now then, f(x|δ)=C*2*(x-δ)/x^{2}. x can take any value in [δ,1], so this must integrate to 1 over that interval. Doing the integral makes C=2*δ-2-2*ln(δ). Now we got f(x|δ) = 1/(δ-1-ln(δ))*(x-δ)/x^{2}. The expected value of x under this pdf is the integral of x*f(x|δ) on the interval of all possible x values, namely [δ,1]. Thus the expected value of x given δ is (1-δ+δ*ln(δ))(δ-1-ln(δ)).