Multivariate Normal : expectation of X given Y is doubly-truncated












2














Let $(X, Y)$ be distributed as a multivariate normal with parameters



$$
mu =
begin{bmatrix}
mu_X \ mu_Y
end{bmatrix}
qquad
Sigma =
begin{bmatrix}
sigma_X^2 & sigma_{XY} \ sigma_{XY} & sigma_Y^2
end{bmatrix}.
$$



I would like to calculate $E(X | y_1 < Y < y_2)$, where $y_1$ and $y_2$ are constants.



From Wikipedia, I have managed to work out that
$$
E(X | Y > y_1) = mu_X + frac{sigma_{XY}}{sigma_Y} left[frac{phileft(frac{y_1 - mu_y}{sigma_Y}right)}{1 - Phileft(frac{y_1 - mu_y}{sigma_Y}right)}right]\
E(X | Y < y_2) = mu_X - frac{sigma_{XY}}{sigma_Y} left[frac{phileft(frac{y_2 - mu_y}{sigma_Y}right)}{Phileft(frac{y_2 - mu_y}{sigma_Y}right)}right]
,$$
where $phi(cdot)$ and $Phi(cdot)$ are, respectively, the p.d.f. and the c.d.f. of the Normal distribution. I could not figure out how to calculate $E(X | y_1 < Y < y_2)$, though.



I have searched for an answer in similar posts from the Stack Exchange network, but I couldn't get a clue from them that would solve this issue. For the record, here are some of them:




  1. Conditional expectation in the multivariate normal distribution

  2. Expectation of conditional normal distribution

  3. Conditional expectation of bivariate normal

  4. https://math.stackexchange.com/q/2807096/83294






conditional-expectation multivariate-normal







share|cite|improve this question


















  • 1




    See stats.stackexchange.com/questions/356023/….
    – StubbornAtom
    4 hours ago










  • @StubbornAtom, thank you. I am studying the answers there and will post one here (since my question is different and the solution to it is but a step used in the answers posted there) ASAP.
    – Waldir Leoncio
    3 hours ago
















2














Let $(X, Y)$ be distributed as a multivariate normal with parameters



$$
mu =
begin{bmatrix}
mu_X \ mu_Y
end{bmatrix}
qquad
Sigma =
begin{bmatrix}
sigma_X^2 & sigma_{XY} \ sigma_{XY} & sigma_Y^2
end{bmatrix}.
$$



I would like to calculate $E(X | y_1 < Y < y_2)$, where $y_1$ and $y_2$ are constants.



From Wikipedia, I have managed to work out that
$$
E(X | Y > y_1) = mu_X + frac{sigma_{XY}}{sigma_Y} left[frac{phileft(frac{y_1 - mu_y}{sigma_Y}right)}{1 - Phileft(frac{y_1 - mu_y}{sigma_Y}right)}right]\
E(X | Y < y_2) = mu_X - frac{sigma_{XY}}{sigma_Y} left[frac{phileft(frac{y_2 - mu_y}{sigma_Y}right)}{Phileft(frac{y_2 - mu_y}{sigma_Y}right)}right]
,$$
where $phi(cdot)$ and $Phi(cdot)$ are, respectively, the p.d.f. and the c.d.f. of the Normal distribution. I could not figure out how to calculate $E(X | y_1 < Y < y_2)$, though.



I have searched for an answer in similar posts from the Stack Exchange network, but I couldn't get a clue from them that would solve this issue. For the record, here are some of them:




  1. Conditional expectation in the multivariate normal distribution

  2. Expectation of conditional normal distribution

  3. Conditional expectation of bivariate normal

  4. https://math.stackexchange.com/q/2807096/83294






conditional-expectation multivariate-normal







share|cite|improve this question


















  • 1




    See stats.stackexchange.com/questions/356023/….
    – StubbornAtom
    4 hours ago










  • @StubbornAtom, thank you. I am studying the answers there and will post one here (since my question is different and the solution to it is but a step used in the answers posted there) ASAP.
    – Waldir Leoncio
    3 hours ago














2












2








2







Let $(X, Y)$ be distributed as a multivariate normal with parameters



$$
mu =
begin{bmatrix}
mu_X \ mu_Y
end{bmatrix}
qquad
Sigma =
begin{bmatrix}
sigma_X^2 & sigma_{XY} \ sigma_{XY} & sigma_Y^2
end{bmatrix}.
$$



I would like to calculate $E(X | y_1 < Y < y_2)$, where $y_1$ and $y_2$ are constants.



From Wikipedia, I have managed to work out that
$$
E(X | Y > y_1) = mu_X + frac{sigma_{XY}}{sigma_Y} left[frac{phileft(frac{y_1 - mu_y}{sigma_Y}right)}{1 - Phileft(frac{y_1 - mu_y}{sigma_Y}right)}right]\
E(X | Y < y_2) = mu_X - frac{sigma_{XY}}{sigma_Y} left[frac{phileft(frac{y_2 - mu_y}{sigma_Y}right)}{Phileft(frac{y_2 - mu_y}{sigma_Y}right)}right]
,$$
where $phi(cdot)$ and $Phi(cdot)$ are, respectively, the p.d.f. and the c.d.f. of the Normal distribution. I could not figure out how to calculate $E(X | y_1 < Y < y_2)$, though.



I have searched for an answer in similar posts from the Stack Exchange network, but I couldn't get a clue from them that would solve this issue. For the record, here are some of them:




  1. Conditional expectation in the multivariate normal distribution

  2. Expectation of conditional normal distribution

  3. Conditional expectation of bivariate normal

  4. https://math.stackexchange.com/q/2807096/83294






conditional-expectation multivariate-normal







share|cite|improve this question













Let $(X, Y)$ be distributed as a multivariate normal with parameters



$$
mu =
begin{bmatrix}
mu_X \ mu_Y
end{bmatrix}
qquad
Sigma =
begin{bmatrix}
sigma_X^2 & sigma_{XY} \ sigma_{XY} & sigma_Y^2
end{bmatrix}.
$$



I would like to calculate $E(X | y_1 < Y < y_2)$, where $y_1$ and $y_2$ are constants.



From Wikipedia, I have managed to work out that
$$
E(X | Y > y_1) = mu_X + frac{sigma_{XY}}{sigma_Y} left[frac{phileft(frac{y_1 - mu_y}{sigma_Y}right)}{1 - Phileft(frac{y_1 - mu_y}{sigma_Y}right)}right]\
E(X | Y < y_2) = mu_X - frac{sigma_{XY}}{sigma_Y} left[frac{phileft(frac{y_2 - mu_y}{sigma_Y}right)}{Phileft(frac{y_2 - mu_y}{sigma_Y}right)}right]
,$$
where $phi(cdot)$ and $Phi(cdot)$ are, respectively, the p.d.f. and the c.d.f. of the Normal distribution. I could not figure out how to calculate $E(X | y_1 < Y < y_2)$, though.



I have searched for an answer in similar posts from the Stack Exchange network, but I couldn't get a clue from them that would solve this issue. For the record, here are some of them:




  1. Conditional expectation in the multivariate normal distribution

  2. Expectation of conditional normal distribution

  3. Conditional expectation of bivariate normal

  4. https://math.stackexchange.com/q/2807096/83294






conditional-expectation multivariate-normal




conditional-expectation multivariate-normal






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked 4 hours ago









Waldir Leoncio

1,39252234




1,39252234








  • 1




    See stats.stackexchange.com/questions/356023/….
    – StubbornAtom
    4 hours ago










  • @StubbornAtom, thank you. I am studying the answers there and will post one here (since my question is different and the solution to it is but a step used in the answers posted there) ASAP.
    – Waldir Leoncio
    3 hours ago














  • 1




    See stats.stackexchange.com/questions/356023/….
    – StubbornAtom
    4 hours ago










  • @StubbornAtom, thank you. I am studying the answers there and will post one here (since my question is different and the solution to it is but a step used in the answers posted there) ASAP.
    – Waldir Leoncio
    3 hours ago








1




1




See stats.stackexchange.com/questions/356023/….
– StubbornAtom
4 hours ago




See stats.stackexchange.com/questions/356023/….
– StubbornAtom
4 hours ago












@StubbornAtom, thank you. I am studying the answers there and will post one here (since my question is different and the solution to it is but a step used in the answers posted there) ASAP.
– Waldir Leoncio
3 hours ago




@StubbornAtom, thank you. I am studying the answers there and will post one here (since my question is different and the solution to it is but a step used in the answers posted there) ASAP.
– Waldir Leoncio
3 hours ago










1 Answer
1






active

oldest

votes


















3














you should use the same approach in point 3. the conditional expectation
$$E(X|Y)=mu_X+sigma_{XY}frac{Y-mu_Y}{sigma_Y^2}$$
Then take the expectation of the RHS of this expression given $y_1<Y<y_2$. The only random variable is $Y$ and this has conditional expectation of
$$E(Y|y_1<Y<y_2)=mu_Y+sigma_Yleft[frac{phileft(frac{y_1-mu_Y}{sigma_Y}right)-phileft(frac{y_2-mu_Y}{sigma_Y}right)}{Phileft(frac{y_2-mu_Y}{sigma_Y}right)-Phileft(frac{y_1-mu_Y}{sigma_Y}right)}right]
$$
Plugging this in gives you
$$E(X|y_1<Y<y_2)=mu_X+frac{sigma_{XY}}{sigma_Y}left[frac{phileft(frac{y_1-mu_Y}{sigma_Y}right)-phileft(frac{y_2-mu_Y}{sigma_Y}right)}{Phileft(frac{y_2-mu_Y}{sigma_Y}right)-Phileft(frac{y_1-mu_Y}{sigma_Y}right)}right]
$$



this also contains your two answers as special cases $y_1to-infty$ and $y_2toinfty$






share|cite|improve this answer





















  • Thank you for your insightful answer, it helped me with a subsequential issue I was having over here.
    – Waldir Leoncio
    3 hours ago











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1 Answer
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1 Answer
1






active

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active

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active

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3














you should use the same approach in point 3. the conditional expectation
$$E(X|Y)=mu_X+sigma_{XY}frac{Y-mu_Y}{sigma_Y^2}$$
Then take the expectation of the RHS of this expression given $y_1<Y<y_2$. The only random variable is $Y$ and this has conditional expectation of
$$E(Y|y_1<Y<y_2)=mu_Y+sigma_Yleft[frac{phileft(frac{y_1-mu_Y}{sigma_Y}right)-phileft(frac{y_2-mu_Y}{sigma_Y}right)}{Phileft(frac{y_2-mu_Y}{sigma_Y}right)-Phileft(frac{y_1-mu_Y}{sigma_Y}right)}right]
$$
Plugging this in gives you
$$E(X|y_1<Y<y_2)=mu_X+frac{sigma_{XY}}{sigma_Y}left[frac{phileft(frac{y_1-mu_Y}{sigma_Y}right)-phileft(frac{y_2-mu_Y}{sigma_Y}right)}{Phileft(frac{y_2-mu_Y}{sigma_Y}right)-Phileft(frac{y_1-mu_Y}{sigma_Y}right)}right]
$$



this also contains your two answers as special cases $y_1to-infty$ and $y_2toinfty$






share|cite|improve this answer





















  • Thank you for your insightful answer, it helped me with a subsequential issue I was having over here.
    – Waldir Leoncio
    3 hours ago
















3














you should use the same approach in point 3. the conditional expectation
$$E(X|Y)=mu_X+sigma_{XY}frac{Y-mu_Y}{sigma_Y^2}$$
Then take the expectation of the RHS of this expression given $y_1<Y<y_2$. The only random variable is $Y$ and this has conditional expectation of
$$E(Y|y_1<Y<y_2)=mu_Y+sigma_Yleft[frac{phileft(frac{y_1-mu_Y}{sigma_Y}right)-phileft(frac{y_2-mu_Y}{sigma_Y}right)}{Phileft(frac{y_2-mu_Y}{sigma_Y}right)-Phileft(frac{y_1-mu_Y}{sigma_Y}right)}right]
$$
Plugging this in gives you
$$E(X|y_1<Y<y_2)=mu_X+frac{sigma_{XY}}{sigma_Y}left[frac{phileft(frac{y_1-mu_Y}{sigma_Y}right)-phileft(frac{y_2-mu_Y}{sigma_Y}right)}{Phileft(frac{y_2-mu_Y}{sigma_Y}right)-Phileft(frac{y_1-mu_Y}{sigma_Y}right)}right]
$$



this also contains your two answers as special cases $y_1to-infty$ and $y_2toinfty$






share|cite|improve this answer





















  • Thank you for your insightful answer, it helped me with a subsequential issue I was having over here.
    – Waldir Leoncio
    3 hours ago














3












3








3






you should use the same approach in point 3. the conditional expectation
$$E(X|Y)=mu_X+sigma_{XY}frac{Y-mu_Y}{sigma_Y^2}$$
Then take the expectation of the RHS of this expression given $y_1<Y<y_2$. The only random variable is $Y$ and this has conditional expectation of
$$E(Y|y_1<Y<y_2)=mu_Y+sigma_Yleft[frac{phileft(frac{y_1-mu_Y}{sigma_Y}right)-phileft(frac{y_2-mu_Y}{sigma_Y}right)}{Phileft(frac{y_2-mu_Y}{sigma_Y}right)-Phileft(frac{y_1-mu_Y}{sigma_Y}right)}right]
$$
Plugging this in gives you
$$E(X|y_1<Y<y_2)=mu_X+frac{sigma_{XY}}{sigma_Y}left[frac{phileft(frac{y_1-mu_Y}{sigma_Y}right)-phileft(frac{y_2-mu_Y}{sigma_Y}right)}{Phileft(frac{y_2-mu_Y}{sigma_Y}right)-Phileft(frac{y_1-mu_Y}{sigma_Y}right)}right]
$$



this also contains your two answers as special cases $y_1to-infty$ and $y_2toinfty$






share|cite|improve this answer












you should use the same approach in point 3. the conditional expectation
$$E(X|Y)=mu_X+sigma_{XY}frac{Y-mu_Y}{sigma_Y^2}$$
Then take the expectation of the RHS of this expression given $y_1<Y<y_2$. The only random variable is $Y$ and this has conditional expectation of
$$E(Y|y_1<Y<y_2)=mu_Y+sigma_Yleft[frac{phileft(frac{y_1-mu_Y}{sigma_Y}right)-phileft(frac{y_2-mu_Y}{sigma_Y}right)}{Phileft(frac{y_2-mu_Y}{sigma_Y}right)-Phileft(frac{y_1-mu_Y}{sigma_Y}right)}right]
$$
Plugging this in gives you
$$E(X|y_1<Y<y_2)=mu_X+frac{sigma_{XY}}{sigma_Y}left[frac{phileft(frac{y_1-mu_Y}{sigma_Y}right)-phileft(frac{y_2-mu_Y}{sigma_Y}right)}{Phileft(frac{y_2-mu_Y}{sigma_Y}right)-Phileft(frac{y_1-mu_Y}{sigma_Y}right)}right]
$$



this also contains your two answers as special cases $y_1to-infty$ and $y_2toinfty$







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered 3 hours ago









probabilityislogic

18.9k36384




18.9k36384












  • Thank you for your insightful answer, it helped me with a subsequential issue I was having over here.
    – Waldir Leoncio
    3 hours ago


















  • Thank you for your insightful answer, it helped me with a subsequential issue I was having over here.
    – Waldir Leoncio
    3 hours ago
















Thank you for your insightful answer, it helped me with a subsequential issue I was having over here.
– Waldir Leoncio
3 hours ago




Thank you for your insightful answer, it helped me with a subsequential issue I was having over here.
– Waldir Leoncio
3 hours ago


















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