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I am relatively new to using Bayes rules for continuous variables and am having trouble setting up part of the formula and am looking for help. The example I am trying to work through is the following:

Suppose there are two numbers x and y that were chosen at random. We have a prior belief that x and y were chosen from a standard multivariate normal distribution. We are then given the following piece of information: x+y=1.0.

The question that I am interested in is how this should change our posterior belief about the values of x and y. From Bayes rule:

posterior distribution is proportional to P(x+y=1 | prior) * P(prior)

What I am stuck on is figuring out what to enter in the formula for P(x+y=1 | prior), would someone be willing to point me in the right direction?

My first thought was that it should be 0, because the volume of a line is zero but then the problem doesn't make much sense. But then I got to thinking that if I had a second such piece of information, say x+2y=5, then I could figure out x and y exactly and it wouldn't make sense for Bayes rule not to be able to make use of this kind of information. Thank you for your help.

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  • $\begingroup$ If x and y are mutlivariate, then x+y cannot be one. At the very least, it has to be a multivariate vector of ones, in which case the answer may depend on the dimension. The probability that x+y equals any given value is zero, and that may be another technical difficulty that you may be encountering. $\endgroup$
    – StasK
    Nov 12 '14 at 4:19
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You should write the problem as: $$ x \sim {\cal N}(0,1) \\ y \sim {\cal N}(0,1) \\ z = x+y $$ Now condition on the event $z=1$. The conditional distribution of $z$ is $p(z | x, y) = \delta(z - (x+y))$ where $\delta$ is the Dirac delta function. In this problem it helps to think of the Dirac delta function as the limit of a Gaussian whose variance is going to zero: $$ p(z | x,y) = \lim_{v \rightarrow 0} {\cal N}(z; x+y,v) $$

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