Let X and Y be two gaussian random variables N(0, x) and N(0, y). X andY are correlated with a correlation .What is the law of E (X − Y |2X + Y )It seems very tedious to calculate the mean and variance of E(x-y|2x+y) directly from f(x,y).Any better ideas?Thanks

derive joint distribution of (X-Y) and (2X+Y) first. It's very easy since we know it should be bivariate normal. Then just use standard formulathus E(X-Y|2X+Y) = cov(X-Y, 2X+Y)/Var(2X+Y)*(2X+Y) + blablabla UPD. blablabla = 0 since E(X) = E(Y) = 0

xsg, I am not sure your formula is correctU := X-YV := 2X + YYou can write U/ sqrt(V(U)) = rho V / sqrt(V(V)) + (1- rho^2) Wwhere, U/ sqrt(V(U)), V / sqrt(V(V)) and W are N(0,1) and V independent from WSo E(U / V ) = rho sqrt(V(U))/ sqrt(V(V)) V, since E W = 0= Cov(U,V) sqrt(V(V)) / sqrt(V(U)) V = .. = bla as well -_-Thx

> So E(U / V ) = rho sqrt(V(U))/ sqrt(V(V)) V, since E W = 0this is correct, but this formula coincides with my solution (see update, I initially wrote it for arbitrary normal X and Y with nonzero mean)> = Cov(U,V) sqrt(V(V)) / sqrt(V(U)) VI think this is wrong

suppose X, Y are two random variables with joint density function f(x,y). Suppose U=aX+bY and V=cX+dY are linear combinations with ad-bc non zero. By definition, E(U|V) is a function g(V) so that E(h(V)U)=E(h(V)g(V)) for any function h. Lhs=\int_R \int_R h(V)Uf(U,V)dxdy=int_R\int_Rh(V)Uf(U,V)|(ad-bc)^{-1}|dUdVrhs=int_R\int_Rh(V)g(V)f(U,V)dxdy=int_R\int_Rh(V)g(V)f(U,V)|ad-bc|^{-1}dUdVBy comparison,g(V)=\int_R Uf(U,V)dU/int_Rf(U,V)dUwhere f(U,V) is obtained from f(x,y) by the inverse transformation fo (X,Y)-->(U,V).In this problem, since f(x,y) is gaussian, it should not be hard.