What is $\text{Var}[x | x+y+z = 3]$ where $x, y$ and $z$ are independent standard normal random variables? There seems to be a way to get the result without any heavy lifting calculation. It seems related to some sort of symmetry related trick like $\text{E}[x| x+y+z = 3] = 1$, but not sure how to perform it.
What is $\text{Var} [x | x+y+z = 3]$ where$ x, y$ and $z$ are independent standard normal random variables?
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1 Answers
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$w = x+y+z$ has a normal distribution, and $x = u + w/3$ where $u = (2x-y-z)/3$ is independent of $w$. Thus the conditional distribution of $x$ given $w=3$ is the distribution of $u+1$, namely a normal distribution with mean $1$ and variance $(4+1+1)/9 = 2/3$.
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0Can you explain how you got the transformation w and v from x,y,z? – 2017-01-31
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0$w = x+y+z$ because that's what you're conditioning on. $u = x - c w$ where constant $c$ is chosen so that $u$ and $w$ are uncorrelated (and therefore independent, since they are jointly normal). – 2017-01-31