Mixing
How to compute variance of conditional expectation?
$begingroup$
Suppose that $(xi, eta)$ is a random vector with absolute continuous distribution: $$f_{(xi,eta)}(x,y)= left{ begin{array}{ll}
e^{-x} & textrm{when $x>0, yin (0,1)$}\
0 & textrm{in other cases}.\
end{array} right.
$$
Let $zeta=2xi - eta.$ Find variance of random variable $E(xi|zeta).$
I have an idea to use a conditional variance formula: $$operatorname{Var}(xi)=E(operatorname{Var}(xi|zeta))+operatorname{Var}(E(xi|zeta))$$ and then compute $operatorname{Var}(E(xi|zeta))$ from here, but I don't know how to get $operatorname{Var}(xi)$ and $E(operatorname{Var}(xi|zeta))$?
conditional-expectation variance
edited Jan 8 at 9:23
Davide Giraudo
126k16150261
asked Jan 7 at 13:40
A.FueA.Fue
65
$endgroup$
add a comment |
$begingroup$
Suppose that $(xi, eta)$ is a random vector with absolute continuous distribution: $$f_{(xi,eta)}(x,y)= left{ begin{array}{ll}
e^{-x} & textrm{when $x>0, yin (0,1)$}\
0 & textrm{in other cases}.\
end{array} right.
$$
Let $zeta=2xi - eta.$ Find variance of random variable $E(xi|zeta).$
I have an idea to use a conditional variance formula: $$operatorname{Var}(xi)=E(operatorname{Var}(xi|zeta))+operatorname{Var}(E(xi|zeta))$$ and then compute $operatorname{Var}(E(xi|zeta))$ from here, but I don't know how to get $operatorname{Var}(xi)$ and $E(operatorname{Var}(xi|zeta))$?
conditional-expectation variance
edited Jan 8 at 9:23
Davide Giraudo
126k16150261
asked Jan 7 at 13:40
A.FueA.Fue
65
$endgroup$
$begingroup$
The variance of $xi$ is easy to obtain because $xi$ is independent of $eta$ so it's just the variance of an exponential(1) variable. But now all you've really done is reduced the problem to computing $mathbb{E}[mathrm{Var}(xi mid zeta)]$ and it's not obvious to me that this is easier than computing $mathrm{Var}(mathbb{E}[xi mid zeta])$ directly.
$endgroup$
– Ian
Jan 7 at 14:02
add a comment |
$begingroup$
Suppose that $(xi, eta)$ is a random vector with absolute continuous distribution: $$f_{(xi,eta)}(x,y)= left{ begin{array}{ll}
e^{-x} & textrm{when $x>0, yin (0,1)$}\
0 & textrm{in other cases}.\
end{array} right.
$$
Let $zeta=2xi - eta.$ Find variance of random variable $E(xi|zeta).$
I have an idea to use a conditional variance formula: $$operatorname{Var}(xi)=E(operatorname{Var}(xi|zeta))+operatorname{Var}(E(xi|zeta))$$ and then compute $operatorname{Var}(E(xi|zeta))$ from here, but I don't know how to get $operatorname{Var}(xi)$ and $E(operatorname{Var}(xi|zeta))$?
conditional-expectation variance
edited Jan 8 at 9:23
Davide Giraudo
126k16150261
asked Jan 7 at 13:40
A.FueA.Fue
65
$endgroup$
Suppose that $(xi, eta)$ is a random vector with absolute continuous distribution: $$f_{(xi,eta)}(x,y)= left{ begin{array}{ll}
e^{-x} & textrm{when $x>0, yin (0,1)$}\
0 & textrm{in other cases}.\
end{array} right.
$$
Let $zeta=2xi - eta.$ Find variance of random variable $E(xi|zeta).$
I have an idea to use a conditional variance formula: $$operatorname{Var}(xi)=E(operatorname{Var}(xi|zeta))+operatorname{Var}(E(xi|zeta))$$ and then compute $operatorname{Var}(E(xi|zeta))$ from here, but I don't know how to get $operatorname{Var}(xi)$ and $E(operatorname{Var}(xi|zeta))$?
conditional-expectation variance
conditional-expectation variance
edited Jan 8 at 9:23
Davide Giraudo
126k16150261
asked Jan 7 at 13:40
A.FueA.Fue
65
edited Jan 8 at 9:23
Davide Giraudo
126k16150261
asked Jan 7 at 13:40
A.FueA.Fue
65
edited Jan 8 at 9:23
Davide Giraudo
126k16150261
edited Jan 8 at 9:23
Davide Giraudo
126k16150261
edited Jan 8 at 9:23
Davide Giraudo
126k16150261
126k16150261
asked Jan 7 at 13:40
A.FueA.Fue
65
asked Jan 7 at 13:40
A.FueA.Fue
65
asked Jan 7 at 13:40
A.FueA.Fue
65
65
$begingroup$
The variance of $xi$ is easy to obtain because $xi$ is independent of $eta$ so it's just the variance of an exponential(1) variable. But now all you've really done is reduced the problem to computing $mathbb{E}[mathrm{Var}(xi mid zeta)]$ and it's not obvious to me that this is easier than computing $mathrm{Var}(mathbb{E}[xi mid zeta])$ directly.
$endgroup$
– Ian
Jan 7 at 14:02
add a comment |
$begingroup$
The variance of $xi$ is easy to obtain because $xi$ is independent of $eta$ so it's just the variance of an exponential(1) variable. But now all you've really done is reduced the problem to computing $mathbb{E}[mathrm{Var}(xi mid zeta)]$ and it's not obvious to me that this is easier than computing $mathrm{Var}(mathbb{E}[xi mid zeta])$ directly.
$endgroup$
– Ian
Jan 7 at 14:02
$begingroup$
The variance of $xi$ is easy to obtain because $xi$ is independent of $eta$ so it's just the variance of an exponential(1) variable. But now all you've really done is reduced the problem to computing $mathbb{E}[mathrm{Var}(xi mid zeta)]$ and it's not obvious to me that this is easier than computing $mathrm{Var}(mathbb{E}[xi mid zeta])$ directly.
$endgroup$
– Ian
Jan 7 at 14:02
$begingroup$
The variance of $xi$ is easy to obtain because $xi$ is independent of $eta$ so it's just the variance of an exponential(1) variable. But now all you've really done is reduced the problem to computing $mathbb{E}[mathrm{Var}(xi mid zeta)]$ and it's not obvious to me that this is easier than computing $mathrm{Var}(mathbb{E}[xi mid zeta])$ directly.
$endgroup$
– Ian
Jan 7 at 14:02
add a comment |
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$begingroup$
The variance of $xi$ is easy to obtain because $xi$ is independent of $eta$ so it's just the variance of an exponential(1) variable. But now all you've really done is reduced the problem to computing $mathbb{E}[mathrm{Var}(xi mid zeta)]$ and it's not obvious to me that this is easier than computing $mathrm{Var}(mathbb{E}[xi mid zeta])$ directly.
$endgroup$
– Ian
Jan 7 at 14:02