Mixing
How to find E[|Product of Two Gaussians|]
Clash Royale CLAN TAG#URR8PPP
up vote
2
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The Problem
Let $Y sim N(0,B)$ and $Z sim N(0,A)$. I would like to find
$$boxedYZ$$ This seems like a natural question that would arise in probability theory, but I have not been able to find any results online or in textbooks. (Note: Here $Y$,$Z$ are NOT necessarily independent. However, they are jointly normal with known pdf $f_Y,Z(y,z)$; nevertheless simply evaluating the double integral to directly find $mathbbE[|YZ|]$ directly seems to not work.)
The Attempt
By the law of total expectation, we have that $mathbbE left[|YZ| right] = mathbbE left[mathbbE[(mid YZ mid )|Z] right]$. So, first we will try to find $mathbbE[(mid YZ mid )|Z]$. Well,
$$mathbbE[(mid YZ mid )|Z]=|Z| mathbbE[(mid Ymid) |Z]$$
We note that $|Y|$ follows a half-normal distribution. This is where I am stuck. I'm not sure where to go from here, perhaps I need to use some results pertaining to the folded normal dist. Or maybe try a different approach entirely to find $mathbbE[|YZ|]$.
Edit: We have that $Y|Z sim N left(0+fracsqrtBsqrtArho(z-0),(1-rho^2 )B right)$ where $rho$ is the correlation coefficient. Then $|Y|Z|$ follows a folded normal distribution, so we can get $mathbbE[|Y|Z|]$ from that. The issue here is that we want $mathbbE[(|Y|)|Z]$.
Any help with this problem is immensely appreciated. Thank you!
probability probability-theory probability-distributions expectation conditional-expectation
asked Jul 17 at 19:29
David_Shmij
398116
add a comment |Â
up vote
2
down vote
favorite
The Problem
Let $Y sim N(0,B)$ and $Z sim N(0,A)$. I would like to find
$$boxedYZ$$ This seems like a natural question that would arise in probability theory, but I have not been able to find any results online or in textbooks. (Note: Here $Y$,$Z$ are NOT necessarily independent. However, they are jointly normal with known pdf $f_Y,Z(y,z)$; nevertheless simply evaluating the double integral to directly find $mathbbE[|YZ|]$ directly seems to not work.)
The Attempt
By the law of total expectation, we have that $mathbbE left[|YZ| right] = mathbbE left[mathbbE[(mid YZ mid )|Z] right]$. So, first we will try to find $mathbbE[(mid YZ mid )|Z]$. Well,
$$mathbbE[(mid YZ mid )|Z]=|Z| mathbbE[(mid Ymid) |Z]$$
We note that $|Y|$ follows a half-normal distribution. This is where I am stuck. I'm not sure where to go from here, perhaps I need to use some results pertaining to the folded normal dist. Or maybe try a different approach entirely to find $mathbbE[|YZ|]$.
Edit: We have that $Y|Z sim N left(0+fracsqrtBsqrtArho(z-0),(1-rho^2 )B right)$ where $rho$ is the correlation coefficient. Then $|Y|Z|$ follows a folded normal distribution, so we can get $mathbbE[|Y|Z|]$ from that. The issue here is that we want $mathbbE[(|Y|)|Z]$.
Any help with this problem is immensely appreciated. Thank you!
probability probability-theory probability-distributions expectation conditional-expectation
asked Jul 17 at 19:29
David_Shmij
398116
1
Are they at least jointly normal?
– E-A
Jul 17 at 20:30
1
I don't think finding $E[XY]$ is easy without having the joint distribution of $(X,Y)$ or the covariance, but who knows...
– Shashi
Jul 17 at 20:32
@E-A Yes, $Y$ and $Z$ are jointly normal, so i definitely have the joint distribution $f_Y,Z(y,z)$.
– David_Shmij
Jul 17 at 20:34
2
If $(X, Y) sim mathcalN(0, Sigma)$ with $Sigma = beginpmatrix a & b \ b & c endpmatrix$, then $$ mathbbE[|XY|] = frac2pileft( sqrtdetSigma + b arctanleft( fracbsqrtdetSigma right) right). $$
– Sangchul Lee
Jul 17 at 22:44
@SangchulLee !!! Nice! Where did you get this from/ how to derive it ? I(f you write this as an answer with some justification I'll accept it . Thanks!
– David_Shmij
Jul 18 at 15:22
add a comment |Â
up vote
2
down vote
favorite
up vote
2
down vote
favorite
The Problem
Let $Y sim N(0,B)$ and $Z sim N(0,A)$. I would like to find
$$boxedYZ$$ This seems like a natural question that would arise in probability theory, but I have not been able to find any results online or in textbooks. (Note: Here $Y$,$Z$ are NOT necessarily independent. However, they are jointly normal with known pdf $f_Y,Z(y,z)$; nevertheless simply evaluating the double integral to directly find $mathbbE[|YZ|]$ directly seems to not work.)
The Attempt
By the law of total expectation, we have that $mathbbE left[|YZ| right] = mathbbE left[mathbbE[(mid YZ mid )|Z] right]$. So, first we will try to find $mathbbE[(mid YZ mid )|Z]$. Well,
$$mathbbE[(mid YZ mid )|Z]=|Z| mathbbE[(mid Ymid) |Z]$$
We note that $|Y|$ follows a half-normal distribution. This is where I am stuck. I'm not sure where to go from here, perhaps I need to use some results pertaining to the folded normal dist. Or maybe try a different approach entirely to find $mathbbE[|YZ|]$.
Edit: We have that $Y|Z sim N left(0+fracsqrtBsqrtArho(z-0),(1-rho^2 )B right)$ where $rho$ is the correlation coefficient. Then $|Y|Z|$ follows a folded normal distribution, so we can get $mathbbE[|Y|Z|]$ from that. The issue here is that we want $mathbbE[(|Y|)|Z]$.
Any help with this problem is immensely appreciated. Thank you!
probability probability-theory probability-distributions expectation conditional-expectation
asked Jul 17 at 19:29
David_Shmij
398116
The Problem
Let $Y sim N(0,B)$ and $Z sim N(0,A)$. I would like to find
$$boxedYZ$$ This seems like a natural question that would arise in probability theory, but I have not been able to find any results online or in textbooks. (Note: Here $Y$,$Z$ are NOT necessarily independent. However, they are jointly normal with known pdf $f_Y,Z(y,z)$; nevertheless simply evaluating the double integral to directly find $mathbbE[|YZ|]$ directly seems to not work.)
The Attempt
By the law of total expectation, we have that $mathbbE left[|YZ| right] = mathbbE left[mathbbE[(mid YZ mid )|Z] right]$. So, first we will try to find $mathbbE[(mid YZ mid )|Z]$. Well,
$$mathbbE[(mid YZ mid )|Z]=|Z| mathbbE[(mid Ymid) |Z]$$
We note that $|Y|$ follows a half-normal distribution. This is where I am stuck. I'm not sure where to go from here, perhaps I need to use some results pertaining to the folded normal dist. Or maybe try a different approach entirely to find $mathbbE[|YZ|]$.
Edit: We have that $Y|Z sim N left(0+fracsqrtBsqrtArho(z-0),(1-rho^2 )B right)$ where $rho$ is the correlation coefficient. Then $|Y|Z|$ follows a folded normal distribution, so we can get $mathbbE[|Y|Z|]$ from that. The issue here is that we want $mathbbE[(|Y|)|Z]$.
Any help with this problem is immensely appreciated. Thank you!
probability probability-theory probability-distributions expectation conditional-expectation
asked Jul 17 at 19:29
David_Shmij
398116
edited Jul 17 at 22:05
asked Jul 17 at 19:29
David_Shmij
398116
asked Jul 17 at 19:29
David_Shmij
398116
asked Jul 17 at 19:29
David_Shmij
398116
398116
1
Are they at least jointly normal?
– E-A
Jul 17 at 20:30
1
I don't think finding $E[XY]$ is easy without having the joint distribution of $(X,Y)$ or the covariance, but who knows...
– Shashi
Jul 17 at 20:32
@E-A Yes, $Y$ and $Z$ are jointly normal, so i definitely have the joint distribution $f_Y,Z(y,z)$.
– David_Shmij
Jul 17 at 20:34
2
If $(X, Y) sim mathcalN(0, Sigma)$ with $Sigma = beginpmatrix a & b \ b & c endpmatrix$, then $$ mathbbE[|XY|] = frac2pileft( sqrtdetSigma + b arctanleft( fracbsqrtdetSigma right) right). $$
– Sangchul Lee
Jul 17 at 22:44
@SangchulLee !!! Nice! Where did you get this from/ how to derive it ? I(f you write this as an answer with some justification I'll accept it . Thanks!
– David_Shmij
Jul 18 at 15:22
add a comment |Â
1
Are they at least jointly normal?
– E-A
Jul 17 at 20:30
1
I don't think finding $E[XY]$ is easy without having the joint distribution of $(X,Y)$ or the covariance, but who knows...
– Shashi
Jul 17 at 20:32
@E-A Yes, $Y$ and $Z$ are jointly normal, so i definitely have the joint distribution $f_Y,Z(y,z)$.
– David_Shmij
Jul 17 at 20:34
2
If $(X, Y) sim mathcalN(0, Sigma)$ with $Sigma = beginpmatrix a & b \ b & c endpmatrix$, then $$ mathbbE[|XY|] = frac2pileft( sqrtdetSigma + b arctanleft( fracbsqrtdetSigma right) right). $$
– Sangchul Lee
Jul 17 at 22:44
@SangchulLee !!! Nice! Where did you get this from/ how to derive it ? I(f you write this as an answer with some justification I'll accept it . Thanks!
– David_Shmij
Jul 18 at 15:22
1
1
Are they at least jointly normal?
– E-A
Jul 17 at 20:30
Are they at least jointly normal?
– E-A
Jul 17 at 20:30
1
1
I don't think finding $E[XY]$ is easy without having the joint distribution of $(X,Y)$ or the covariance, but who knows...
– Shashi
Jul 17 at 20:32
I don't think finding $E[XY]$ is easy without having the joint distribution of $(X,Y)$ or the covariance, but who knows...
– Shashi
Jul 17 at 20:32
@E-A Yes, $Y$ and $Z$ are jointly normal, so i definitely have the joint distribution $f_Y,Z(y,z)$.
– David_Shmij
Jul 17 at 20:34
@E-A Yes, $Y$ and $Z$ are jointly normal, so i definitely have the joint distribution $f_Y,Z(y,z)$.
– David_Shmij
Jul 17 at 20:34
2
2
If $(X, Y) sim mathcalN(0, Sigma)$ with $Sigma = beginpmatrix a & b \ b & c endpmatrix$, then $$ mathbbE[|XY|] = frac2pileft( sqrtdetSigma + b arctanleft( fracbsqrtdetSigma right) right). $$
– Sangchul Lee
Jul 17 at 22:44
If $(X, Y) sim mathcalN(0, Sigma)$ with $Sigma = beginpmatrix a & b \ b & c endpmatrix$, then $$ mathbbE[|XY|] = frac2pileft( sqrtdetSigma + b arctanleft( fracbsqrtdetSigma right) right). $$
– Sangchul Lee
Jul 17 at 22:44
@SangchulLee !!! Nice! Where did you get this from/ how to derive it ? I(f you write this as an answer with some justification I'll accept it . Thanks!
– David_Shmij
Jul 18 at 15:22
@SangchulLee !!! Nice! Where did you get this from/ how to derive it ? I(f you write this as an answer with some justification I'll accept it . Thanks!
– David_Shmij
Jul 18 at 15:22
add a comment |Â
1 Answer
1
active
oldest
votes
up vote
0
down vote
accepted
The answer is found in Corollary 3.1 in this - Guassian Integrals Involving Absolute Value Functions
answered Jul 26 at 18:28
David_Shmij
398116
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
accepted
The answer is found in Corollary 3.1 in this - Guassian Integrals Involving Absolute Value Functions
answered Jul 26 at 18:28
David_Shmij
398116
add a comment |Â
up vote
0
down vote
accepted
The answer is found in Corollary 3.1 in this - Guassian Integrals Involving Absolute Value Functions
answered Jul 26 at 18:28
David_Shmij
398116
add a comment |Â
up vote
0
down vote
accepted
up vote
0
down vote
accepted
The answer is found in Corollary 3.1 in this - Guassian Integrals Involving Absolute Value Functions
answered Jul 26 at 18:28
David_Shmij
398116
The answer is found in Corollary 3.1 in this - Guassian Integrals Involving Absolute Value Functions
answered Jul 26 at 18:28
David_Shmij
398116
answered Jul 26 at 18:28
David_Shmij
398116
answered Jul 26 at 18:28
David_Shmij
398116
answered Jul 26 at 18:28
David_Shmij
398116
398116
add a comment |Â
add a comment |Â
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1
Are they at least jointly normal?
– E-A
Jul 17 at 20:30
1
I don't think finding $E[XY]$ is easy without having the joint distribution of $(X,Y)$ or the covariance, but who knows...
– Shashi
Jul 17 at 20:32
@E-A Yes, $Y$ and $Z$ are jointly normal, so i definitely have the joint distribution $f_Y,Z(y,z)$.
– David_Shmij
Jul 17 at 20:34
2
If $(X, Y) sim mathcalN(0, Sigma)$ with $Sigma = beginpmatrix a & b \ b & c endpmatrix$, then $$ mathbbE[|XY|] = frac2pileft( sqrtdetSigma + b arctanleft( fracbsqrtdetSigma right) right). $$
– Sangchul Lee
Jul 17 at 22:44
@SangchulLee !!! Nice! Where did you get this from/ how to derive it ? I(f you write this as an answer with some justification I'll accept it . Thanks!
– David_Shmij
Jul 18 at 15:22