X is normally distributed with mean 0, variance 3. Y is normally distributed with mean 0, variance 1. Cov (X,Y) =1. How do I find the expectation of X, given that 2X+3Y=1

Answer 1

The variance-covariance matrix is

3 1/2

1/2 1

The inverse is

4/11 -2/11

-2/11 12/11

Then, f(x,y) = 2/(11π) exp(-1/2 (4/11 x^2 + 4/11 xy +12/11y^2))

If 2x +3y = 1, y = (1 - 2x)/3

Then, f(x,y) = f(x, (1 - 2x)/3)

We substitute (1 - 2x)/3 for y. Then, we will be able to make use of symmetry to get the expected value.

Looking at the exponential term,

4/11 x^2 + 4/11 xy +12/11y^2 = 4/11 x^2 - 4/11 x(1 - 2x)/3 + 12/11 [(1 - 2x)/3]^2 =

4/11 x^2 - 4/33x + 8/33x^2 + 4/33 - 16/33x +16/33x^2 =

12/11x^2 -20/33x +4/33

We complete the square, but the constant is irrelevant, all we care about is the symmetry.

12/11(x-5/18)^2 + ...

The mean is 5/18