Question about inverse-variance weighting

Suppose we want to make inference on an unobserved realization x of a random variable ˜x, which is normally distributed with mean μx and variance σ2x. Suppose there is another random variable ˜y (whose unobserved realization we’ll similarly call y) that is normally distributed with mean μy and variance σ2y. Let σxy be the covariance of ˜x and ˜y.

Now suppose we observe a signal on x,
a=x+˜u,
where ˜uN(0,ϕ2x), and a signal on y,
b=y+˜v,
where ˜vN(0,ϕ2y). Assume that ˜u and ˜v are independent.

What is the distribution of x conditional on a and b?

What I know so far:
Using inverse-variance weighting,
E(x|a)=1σ2xμx+1ϕ2xa1σ2x+1ϕ2x,
and
Var(x|a)=11σ2x+1ϕ2x.

Since x and y are jointly drawn, b should carry some information about x. Other than realizing this, I’m stuck. Any help is appreciated!

Answer

I’m not sure whether the inverse-variance weighting formulas apply here. However I think you might compute the conditional distribution of x given a and b by assuming that x, y, a and b follow a joint multivariate normal distribution.

Specifically, if you assume (compatibly with what specified in the question) that
[xyuv]N([μxμy00],[σ2xσxy00σxyσ2y0000ϕ2x0000ϕ2y])
then, letting a=x+u and b=y+v, you can find that
[xab]N([μxμxμy],[σ2xσ2xσxyσ2xσ2x+ϕ2xσxyσxyσxyσ2y+ϕ2y]).
(Note that in the above it is implicitly assumed that u and v are independent between each other and also with x and y.)

From this you could find the conditional distribution of x given a and b using standard properties of the multivariate normal distribution (see here for example: http://en.wikipedia.org/wiki/Multivariate_normal_distribution#Conditional_distributions).

Attribution
Source : Link , Question Author : bad_at_math , Answer Author : a.arfe

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