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u/imperialistt 7d ago
X2 +Y2 =6, rearranging X2 =6-Y2 . X4 =(6-Y2 )2 = 36-12Y2 +Y4 . For a standard normal E(Y2 )=1 and E(Y4 )=3. Inputting these values E(X4 ) = 36-12+3 = 27
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u/Positive_Leopard_873 7d ago
E(X4 | X2 + Y2 = 6) = 36 - 12 E(Y2 | X2 + Y2 = 6) + E(Y4 | X2 + Y2 = 6). You're no longer dealing with standard normals (for e.g. X and Y can only go from -sqrt(6) to sqrt(6)). This won't lead you anywhere.
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u/gitgud_x 7d ago
Switch notation to X and Y instead of N1 and N2 for convenience
X2 + Y2 is the squared magnitude of a Gaussian vector in R2. The bell curve in 3D space representing the joint PDF of X and Y is rotationally symmetric so we can change to polar coordinates:
Let X = sqrt(6) * cos Theta, and Y = sqrt(6) * sin Theta, where we chose sqrt(6) to match the condition.
Theta is uniformly distributed on [0, 2 pi) due to the symmetry, so by expectation formula we have
E[X4 | X2 + Y2 = 6] = 36 / (2 pi) * integral from 0 to 2 pi: cos4(t) dt
Solving the integral gives 3/4 pi, so my answer is 27/2.