What is the joint density of 2 random variables that Linear combinations of the same random variables?
$begingroup$
Suppose we have random variable $W$ and $M$ that are independent standard normal random variables. If we were to define $X$ and $Y$ as:
$X=aW +bM$ and $Y=cW+bM$
How do we find the joint density of $X$ and $Y$? ie $f_{X,Y}$.
I found the pdf of $X$ and the pdf of $Y$ (linear combination of Normals are normal with new mean and variance) but im not sure where to go from there. I believe you can't just multiply $f_U$ and $f_V$ together because they share $W$ and $M$ which make them dependent. However, when you find $f_X$ and $f_Y$ those $M$ and $W$ terms just disappear leaving you with only $a$'s, $b$'s etc.
statistics normal-distribution
asked Aug 24 '17 at 18:17
RibbonSannyRibbonSanny
1
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add a comment |
$begingroup$
Suppose we have random variable $W$ and $M$ that are independent standard normal random variables. If we were to define $X$ and $Y$ as:
$X=aW +bM$ and $Y=cW+bM$
How do we find the joint density of $X$ and $Y$? ie $f_{X,Y}$.
I found the pdf of $X$ and the pdf of $Y$ (linear combination of Normals are normal with new mean and variance) but im not sure where to go from there. I believe you can't just multiply $f_U$ and $f_V$ together because they share $W$ and $M$ which make them dependent. However, when you find $f_X$ and $f_Y$ those $M$ and $W$ terms just disappear leaving you with only $a$'s, $b$'s etc.
statistics normal-distribution
asked Aug 24 '17 at 18:17
RibbonSannyRibbonSanny
1
$endgroup$
add a comment |
$begingroup$
Suppose we have random variable $W$ and $M$ that are independent standard normal random variables. If we were to define $X$ and $Y$ as:
$X=aW +bM$ and $Y=cW+bM$
How do we find the joint density of $X$ and $Y$? ie $f_{X,Y}$.
I found the pdf of $X$ and the pdf of $Y$ (linear combination of Normals are normal with new mean and variance) but im not sure where to go from there. I believe you can't just multiply $f_U$ and $f_V$ together because they share $W$ and $M$ which make them dependent. However, when you find $f_X$ and $f_Y$ those $M$ and $W$ terms just disappear leaving you with only $a$'s, $b$'s etc.
statistics normal-distribution
asked Aug 24 '17 at 18:17
RibbonSannyRibbonSanny
1
$endgroup$
Suppose we have random variable $W$ and $M$ that are independent standard normal random variables. If we were to define $X$ and $Y$ as:
$X=aW +bM$ and $Y=cW+bM$
How do we find the joint density of $X$ and $Y$? ie $f_{X,Y}$.
I found the pdf of $X$ and the pdf of $Y$ (linear combination of Normals are normal with new mean and variance) but im not sure where to go from there. I believe you can't just multiply $f_U$ and $f_V$ together because they share $W$ and $M$ which make them dependent. However, when you find $f_X$ and $f_Y$ those $M$ and $W$ terms just disappear leaving you with only $a$'s, $b$'s etc.
statistics normal-distribution
statistics normal-distribution
asked Aug 24 '17 at 18:17
RibbonSannyRibbonSanny
1
asked Aug 24 '17 at 18:17
RibbonSannyRibbonSanny
1
asked Aug 24 '17 at 18:17
RibbonSannyRibbonSanny
1
asked Aug 24 '17 at 18:17
RibbonSannyRibbonSanny
1
asked Aug 24 '17 at 18:17
RibbonSannyRibbonSanny
1
1
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add a comment |
2 Answers
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$begingroup$
Linear combinations of jointly Gaussian random variables are also jointly Gaussian.
Independent Gaussians are jointly Gaussian, so $(X,Y)$ follow a joint Gaussian distribution. This is specified by the $E[X],E[Y], sigma_X^2, sigma_Y^2, sigma_{XY}$.
$E[X] = E[aW+bM] = a E[W] + b E[M] =0$ and similarly $E[Y]=0$.
$sigma_X^2 = var(aW+bM) = a^2 var(W) + b^2 var(M) = a^2+b^2$ and similarly $sigma_Y^2 = c^2+b^2$.
$sigma_{XY} = E[XY] - E[X]E[Y] = E[XY] - 0 = E[(aW+bM)(cW+bM)] = E[acW^2+b^2M + (ab+bc) MW]= ac+b^2 + (ab+bc)E[MW]=ac+b^2$ since $E[MW]=E[M]E[W]=0$.
Thus, $(X,Y)$ follows a normal distribution with mean zero (vector) and covariance matrix $begin{bmatrix} sigma_X^2 & sigma_{XY} \ sigma_{XY} & sigma_Y^2 end{bmatrix}$.
answered Aug 24 '17 at 18:49
BatmanBatman
16.4k11735
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$begingroup$
Summarizing @Batman succinctly, let $A=begin{pmatrix}a&b\c&bend{pmatrix}$. From $begin{pmatrix}W\M end{pmatrix} sim N(0, I)$, we have
begin{align*}
begin{pmatrix}X\Y end{pmatrix} &= Abegin{pmatrix}W\M end{pmatrix}\
&sim N(A0, AIA^T)\
&= Nleft(0, begin{pmatrix}a^2+b^2 & ac+b^2 \ ac+b^2 & b^2+c^2end{pmatrix}right)
end{align*}
Note there is implicit assumption that $A$ is full rank, i.e. $aneq c$. To prove that independent linear combination of normal is normal, you could use the moment generating function. Linear combination of normal distribution
answered Aug 24 '17 at 19:17
Jirapat SamranvedhyaJirapat Samranvedhya
27118
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2 Answers
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2 Answers
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$begingroup$
Linear combinations of jointly Gaussian random variables are also jointly Gaussian.
Independent Gaussians are jointly Gaussian, so $(X,Y)$ follow a joint Gaussian distribution. This is specified by the $E[X],E[Y], sigma_X^2, sigma_Y^2, sigma_{XY}$.
$E[X] = E[aW+bM] = a E[W] + b E[M] =0$ and similarly $E[Y]=0$.
$sigma_X^2 = var(aW+bM) = a^2 var(W) + b^2 var(M) = a^2+b^2$ and similarly $sigma_Y^2 = c^2+b^2$.
$sigma_{XY} = E[XY] - E[X]E[Y] = E[XY] - 0 = E[(aW+bM)(cW+bM)] = E[acW^2+b^2M + (ab+bc) MW]= ac+b^2 + (ab+bc)E[MW]=ac+b^2$ since $E[MW]=E[M]E[W]=0$.
Thus, $(X,Y)$ follows a normal distribution with mean zero (vector) and covariance matrix $begin{bmatrix} sigma_X^2 & sigma_{XY} \ sigma_{XY} & sigma_Y^2 end{bmatrix}$.
answered Aug 24 '17 at 18:49
BatmanBatman
16.4k11735
$endgroup$
add a comment |
$begingroup$
Linear combinations of jointly Gaussian random variables are also jointly Gaussian.
Independent Gaussians are jointly Gaussian, so $(X,Y)$ follow a joint Gaussian distribution. This is specified by the $E[X],E[Y], sigma_X^2, sigma_Y^2, sigma_{XY}$.
$E[X] = E[aW+bM] = a E[W] + b E[M] =0$ and similarly $E[Y]=0$.
$sigma_X^2 = var(aW+bM) = a^2 var(W) + b^2 var(M) = a^2+b^2$ and similarly $sigma_Y^2 = c^2+b^2$.
$sigma_{XY} = E[XY] - E[X]E[Y] = E[XY] - 0 = E[(aW+bM)(cW+bM)] = E[acW^2+b^2M + (ab+bc) MW]= ac+b^2 + (ab+bc)E[MW]=ac+b^2$ since $E[MW]=E[M]E[W]=0$.
Thus, $(X,Y)$ follows a normal distribution with mean zero (vector) and covariance matrix $begin{bmatrix} sigma_X^2 & sigma_{XY} \ sigma_{XY} & sigma_Y^2 end{bmatrix}$.
answered Aug 24 '17 at 18:49
BatmanBatman
16.4k11735
$endgroup$
add a comment |
$begingroup$
Linear combinations of jointly Gaussian random variables are also jointly Gaussian.
Independent Gaussians are jointly Gaussian, so $(X,Y)$ follow a joint Gaussian distribution. This is specified by the $E[X],E[Y], sigma_X^2, sigma_Y^2, sigma_{XY}$.
$E[X] = E[aW+bM] = a E[W] + b E[M] =0$ and similarly $E[Y]=0$.
$sigma_X^2 = var(aW+bM) = a^2 var(W) + b^2 var(M) = a^2+b^2$ and similarly $sigma_Y^2 = c^2+b^2$.
$sigma_{XY} = E[XY] - E[X]E[Y] = E[XY] - 0 = E[(aW+bM)(cW+bM)] = E[acW^2+b^2M + (ab+bc) MW]= ac+b^2 + (ab+bc)E[MW]=ac+b^2$ since $E[MW]=E[M]E[W]=0$.
Thus, $(X,Y)$ follows a normal distribution with mean zero (vector) and covariance matrix $begin{bmatrix} sigma_X^2 & sigma_{XY} \ sigma_{XY} & sigma_Y^2 end{bmatrix}$.
answered Aug 24 '17 at 18:49
BatmanBatman
16.4k11735
$endgroup$
Linear combinations of jointly Gaussian random variables are also jointly Gaussian.
Independent Gaussians are jointly Gaussian, so $(X,Y)$ follow a joint Gaussian distribution. This is specified by the $E[X],E[Y], sigma_X^2, sigma_Y^2, sigma_{XY}$.
$E[X] = E[aW+bM] = a E[W] + b E[M] =0$ and similarly $E[Y]=0$.
$sigma_X^2 = var(aW+bM) = a^2 var(W) + b^2 var(M) = a^2+b^2$ and similarly $sigma_Y^2 = c^2+b^2$.
$sigma_{XY} = E[XY] - E[X]E[Y] = E[XY] - 0 = E[(aW+bM)(cW+bM)] = E[acW^2+b^2M + (ab+bc) MW]= ac+b^2 + (ab+bc)E[MW]=ac+b^2$ since $E[MW]=E[M]E[W]=0$.
Thus, $(X,Y)$ follows a normal distribution with mean zero (vector) and covariance matrix $begin{bmatrix} sigma_X^2 & sigma_{XY} \ sigma_{XY} & sigma_Y^2 end{bmatrix}$.
answered Aug 24 '17 at 18:49
BatmanBatman
16.4k11735
answered Aug 24 '17 at 18:49
BatmanBatman
16.4k11735
answered Aug 24 '17 at 18:49
BatmanBatman
16.4k11735
answered Aug 24 '17 at 18:49
BatmanBatman
16.4k11735
16.4k11735
add a comment |
add a comment |
$begingroup$
Summarizing @Batman succinctly, let $A=begin{pmatrix}a&b\c&bend{pmatrix}$. From $begin{pmatrix}W\M end{pmatrix} sim N(0, I)$, we have
begin{align*}
begin{pmatrix}X\Y end{pmatrix} &= Abegin{pmatrix}W\M end{pmatrix}\
&sim N(A0, AIA^T)\
&= Nleft(0, begin{pmatrix}a^2+b^2 & ac+b^2 \ ac+b^2 & b^2+c^2end{pmatrix}right)
end{align*}
Note there is implicit assumption that $A$ is full rank, i.e. $aneq c$. To prove that independent linear combination of normal is normal, you could use the moment generating function. Linear combination of normal distribution
answered Aug 24 '17 at 19:17
Jirapat SamranvedhyaJirapat Samranvedhya
27118
$endgroup$
add a comment |
$begingroup$
Summarizing @Batman succinctly, let $A=begin{pmatrix}a&b\c&bend{pmatrix}$. From $begin{pmatrix}W\M end{pmatrix} sim N(0, I)$, we have
begin{align*}
begin{pmatrix}X\Y end{pmatrix} &= Abegin{pmatrix}W\M end{pmatrix}\
&sim N(A0, AIA^T)\
&= Nleft(0, begin{pmatrix}a^2+b^2 & ac+b^2 \ ac+b^2 & b^2+c^2end{pmatrix}right)
end{align*}
Note there is implicit assumption that $A$ is full rank, i.e. $aneq c$. To prove that independent linear combination of normal is normal, you could use the moment generating function. Linear combination of normal distribution
answered Aug 24 '17 at 19:17
Jirapat SamranvedhyaJirapat Samranvedhya
27118
$endgroup$
add a comment |
$begingroup$
Summarizing @Batman succinctly, let $A=begin{pmatrix}a&b\c&bend{pmatrix}$. From $begin{pmatrix}W\M end{pmatrix} sim N(0, I)$, we have
begin{align*}
begin{pmatrix}X\Y end{pmatrix} &= Abegin{pmatrix}W\M end{pmatrix}\
&sim N(A0, AIA^T)\
&= Nleft(0, begin{pmatrix}a^2+b^2 & ac+b^2 \ ac+b^2 & b^2+c^2end{pmatrix}right)
end{align*}
Note there is implicit assumption that $A$ is full rank, i.e. $aneq c$. To prove that independent linear combination of normal is normal, you could use the moment generating function. Linear combination of normal distribution
answered Aug 24 '17 at 19:17
Jirapat SamranvedhyaJirapat Samranvedhya
27118
$endgroup$
Summarizing @Batman succinctly, let $A=begin{pmatrix}a&b\c&bend{pmatrix}$. From $begin{pmatrix}W\M end{pmatrix} sim N(0, I)$, we have
begin{align*}
begin{pmatrix}X\Y end{pmatrix} &= Abegin{pmatrix}W\M end{pmatrix}\
&sim N(A0, AIA^T)\
&= Nleft(0, begin{pmatrix}a^2+b^2 & ac+b^2 \ ac+b^2 & b^2+c^2end{pmatrix}right)
end{align*}
Note there is implicit assumption that $A$ is full rank, i.e. $aneq c$. To prove that independent linear combination of normal is normal, you could use the moment generating function. Linear combination of normal distribution
answered Aug 24 '17 at 19:17
Jirapat SamranvedhyaJirapat Samranvedhya
27118
answered Aug 24 '17 at 19:17
Jirapat SamranvedhyaJirapat Samranvedhya
27118
answered Aug 24 '17 at 19:17
Jirapat SamranvedhyaJirapat Samranvedhya
27118
answered Aug 24 '17 at 19:17
Jirapat SamranvedhyaJirapat Samranvedhya
27118
27118
add a comment |
add a comment |
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