Proving $(X,Y)$ is a normal vector when $Xsim N(1,1)$ and $Ymid Xsim N(3X,4)$
Suppose I have a random vector $(X,Y)$ with $Xsimmathcal{N}(1,1)$ and $Y|X = x simmathcal{N}(3x,4)$.
I need to prove that $(X,Y)$ is a normal vector as well.
To do that I want to explicitly write the vector of expected values and the $2x2$ matrix of variance and covariance.
I know that the first entry of the vector of expected values is 1 and the entry $C_{1,1}$ of the matrix is 1 as well. However I am struggling to see how I can derive the density function of Y from the conditional one.
Any suggestions?
probability normal-distribution conditional-probability bivariate-distributions
edited Jan 4 at 19:02
StubbornAtom
5,39411138
asked Jan 4 at 14:48
qcc101qcc101
477113
add a comment |
Suppose I have a random vector $(X,Y)$ with $Xsimmathcal{N}(1,1)$ and $Y|X = x simmathcal{N}(3x,4)$.
I need to prove that $(X,Y)$ is a normal vector as well.
To do that I want to explicitly write the vector of expected values and the $2x2$ matrix of variance and covariance.
I know that the first entry of the vector of expected values is 1 and the entry $C_{1,1}$ of the matrix is 1 as well. However I am struggling to see how I can derive the density function of Y from the conditional one.
Any suggestions?
probability normal-distribution conditional-probability bivariate-distributions
edited Jan 4 at 19:02
StubbornAtom
5,39411138
asked Jan 4 at 14:48
qcc101qcc101
477113
Joint density of $(X,Y)$ is $f_{X,Y}(x,y)=f_{Ymid X}(y)f_X(x)=...$
– StubbornAtom
Jan 4 at 15:07
Yes, but I am trying to calculate E[Y], V[Y] and Cov[X,Y] to "fill" the matrices. I am struggling with Cov[X,Y], specifically with E[XY]. Do you have an hint for that?
– qcc101
Jan 4 at 15:10
Comment made by @StubbornAtom below his/her answer needs to be emphasized. Just finding the mean vector and the variance-covariance matrix do not show the joint-normality of $X$ and $Y$.
– Just_to_Answer
Jan 4 at 20:15
add a comment |
Suppose I have a random vector $(X,Y)$ with $Xsimmathcal{N}(1,1)$ and $Y|X = x simmathcal{N}(3x,4)$.
I need to prove that $(X,Y)$ is a normal vector as well.
To do that I want to explicitly write the vector of expected values and the $2x2$ matrix of variance and covariance.
I know that the first entry of the vector of expected values is 1 and the entry $C_{1,1}$ of the matrix is 1 as well. However I am struggling to see how I can derive the density function of Y from the conditional one.
Any suggestions?
probability normal-distribution conditional-probability bivariate-distributions
edited Jan 4 at 19:02
StubbornAtom
5,39411138
asked Jan 4 at 14:48
qcc101qcc101
477113
Suppose I have a random vector $(X,Y)$ with $Xsimmathcal{N}(1,1)$ and $Y|X = x simmathcal{N}(3x,4)$.
I need to prove that $(X,Y)$ is a normal vector as well.
To do that I want to explicitly write the vector of expected values and the $2x2$ matrix of variance and covariance.
I know that the first entry of the vector of expected values is 1 and the entry $C_{1,1}$ of the matrix is 1 as well. However I am struggling to see how I can derive the density function of Y from the conditional one.
Any suggestions?
probability normal-distribution conditional-probability bivariate-distributions
probability normal-distribution conditional-probability bivariate-distributions
edited Jan 4 at 19:02
StubbornAtom
5,39411138
asked Jan 4 at 14:48
qcc101qcc101
477113
edited Jan 4 at 19:02
StubbornAtom
5,39411138
asked Jan 4 at 14:48
qcc101qcc101
477113
edited Jan 4 at 19:02
StubbornAtom
5,39411138
edited Jan 4 at 19:02
StubbornAtom
5,39411138
edited Jan 4 at 19:02
StubbornAtom
5,39411138
5,39411138
asked Jan 4 at 14:48
qcc101qcc101
477113
asked Jan 4 at 14:48
qcc101qcc101
477113
asked Jan 4 at 14:48
qcc101qcc101
477113
477113
Joint density of $(X,Y)$ is $f_{X,Y}(x,y)=f_{Ymid X}(y)f_X(x)=...$
– StubbornAtom
Jan 4 at 15:07
Yes, but I am trying to calculate E[Y], V[Y] and Cov[X,Y] to "fill" the matrices. I am struggling with Cov[X,Y], specifically with E[XY]. Do you have an hint for that?
– qcc101
Jan 4 at 15:10
Comment made by @StubbornAtom below his/her answer needs to be emphasized. Just finding the mean vector and the variance-covariance matrix do not show the joint-normality of $X$ and $Y$.
– Just_to_Answer
Jan 4 at 20:15
add a comment |
Joint density of $(X,Y)$ is $f_{X,Y}(x,y)=f_{Ymid X}(y)f_X(x)=...$
– StubbornAtom
Jan 4 at 15:07
Yes, but I am trying to calculate E[Y], V[Y] and Cov[X,Y] to "fill" the matrices. I am struggling with Cov[X,Y], specifically with E[XY]. Do you have an hint for that?
– qcc101
Jan 4 at 15:10
Comment made by @StubbornAtom below his/her answer needs to be emphasized. Just finding the mean vector and the variance-covariance matrix do not show the joint-normality of $X$ and $Y$.
– Just_to_Answer
Jan 4 at 20:15
Joint density of $(X,Y)$ is $f_{X,Y}(x,y)=f_{Ymid X}(y)f_X(x)=...$
– StubbornAtom
Jan 4 at 15:07
Joint density of $(X,Y)$ is $f_{X,Y}(x,y)=f_{Ymid X}(y)f_X(x)=...$
– StubbornAtom
Jan 4 at 15:07
Yes, but I am trying to calculate E[Y], V[Y] and Cov[X,Y] to "fill" the matrices. I am struggling with Cov[X,Y], specifically with E[XY]. Do you have an hint for that?
– qcc101
Jan 4 at 15:10
Yes, but I am trying to calculate E[Y], V[Y] and Cov[X,Y] to "fill" the matrices. I am struggling with Cov[X,Y], specifically with E[XY]. Do you have an hint for that?
– qcc101
Jan 4 at 15:10
Comment made by @StubbornAtom below his/her answer needs to be emphasized. Just finding the mean vector and the variance-covariance matrix do not show the joint-normality of $X$ and $Y$.
– Just_to_Answer
Jan 4 at 20:15
Comment made by @StubbornAtom below his/her answer needs to be emphasized. Just finding the mean vector and the variance-covariance matrix do not show the joint-normality of $X$ and $Y$.
– Just_to_Answer
Jan 4 at 20:15
add a comment |
2 Answers
2
active
oldest
votes
You could say $Y=3X+Z$ where $Zsim N(0,4)$ independent of $X$.
It is then an easy step to say that $Ysim N(3times 1+0,3^2times 1+4)$ and
- $E[X]=1$
- $E[Y]=3$
- $text{Var}(X)=1$
- $text{Var}(Y)=13$
- $text{Cov}(X,Y)=text{Cov}(X,3X)=3text{Var}(X)=3$
answered Jan 4 at 15:26
HenryHenry
98.4k475162
This is the direction I took as well. However I did not use an additional Z. Therefore I am struggling with calculating the covariance between X and Y, could you take a look at my comment to the answer below?
– qcc101
Jan 4 at 15:28
add a comment |
You can proceed with moment generating functions.
Joint MGF of $(X,Y)$ is
begin{align}
M(s,t)&=Eleft[e^{sX+tY}right]
\&=Eleft[E(e^{sX+tY}mid X)right]
\&=Eleft[e^{sX}E(e^{tY}mid X)right]
end{align}
From the given information, you should be able to show that this MGF corresponds to the MGF of a bivariate normal distribution. That would complete your proof using the uniqueness property of MGF.
answered Jan 4 at 15:17
StubbornAtomStubbornAtom
5,39411138
I am not sure I am following you. In my problem I did find Variance and Expected value of Y, using the tower property of conditioning. I stil need to find the two symmetric entries of the matrix of variance and covariance to complete what I want to do. That is I need to find Covariance[X,Y]. Calculating, I get to this: $$Cov[X,Y] = E[XY] - 3E[X^2] -E[Y] + 3E[X]$$ Above, I know everything except E[XY]. I did the following: $ E[XY] = E[ E[XY|X]] = E[YE[X|X]] = E[Y E[X]]$ but I get the wrong result ( covariance should be 3 but I get -3). Could you point out what I did wrong?
– qcc101
Jan 4 at 15:26
@qcc101 $E(XY)=E(E(XYmid X))=E(XE(Ymid X))=...$
– StubbornAtom
Jan 4 at 15:29
Right, silly mistake. I will investigate the part about moment generating functions as well. Thank you.
– qcc101
Jan 4 at 15:30
1
@qcc101 Without the argument that $X$ and $Y$ are linear combinations of independent normal variables and hence $(X,Y)$ is bivariate normal (as seen in the answer by @Henry), I don't think finding the variance-covariance matrix (only) as you are trying to do is sufficient for the given question.
– StubbornAtom
Jan 4 at 15:37
add a comment |
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2 Answers
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2 Answers
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You could say $Y=3X+Z$ where $Zsim N(0,4)$ independent of $X$.
It is then an easy step to say that $Ysim N(3times 1+0,3^2times 1+4)$ and
- $E[X]=1$
- $E[Y]=3$
- $text{Var}(X)=1$
- $text{Var}(Y)=13$
- $text{Cov}(X,Y)=text{Cov}(X,3X)=3text{Var}(X)=3$
answered Jan 4 at 15:26
HenryHenry
98.4k475162
This is the direction I took as well. However I did not use an additional Z. Therefore I am struggling with calculating the covariance between X and Y, could you take a look at my comment to the answer below?
– qcc101
Jan 4 at 15:28
add a comment |
You could say $Y=3X+Z$ where $Zsim N(0,4)$ independent of $X$.
It is then an easy step to say that $Ysim N(3times 1+0,3^2times 1+4)$ and
- $E[X]=1$
- $E[Y]=3$
- $text{Var}(X)=1$
- $text{Var}(Y)=13$
- $text{Cov}(X,Y)=text{Cov}(X,3X)=3text{Var}(X)=3$
answered Jan 4 at 15:26
HenryHenry
98.4k475162
This is the direction I took as well. However I did not use an additional Z. Therefore I am struggling with calculating the covariance between X and Y, could you take a look at my comment to the answer below?
– qcc101
Jan 4 at 15:28
add a comment |
You could say $Y=3X+Z$ where $Zsim N(0,4)$ independent of $X$.
It is then an easy step to say that $Ysim N(3times 1+0,3^2times 1+4)$ and
- $E[X]=1$
- $E[Y]=3$
- $text{Var}(X)=1$
- $text{Var}(Y)=13$
- $text{Cov}(X,Y)=text{Cov}(X,3X)=3text{Var}(X)=3$
answered Jan 4 at 15:26
HenryHenry
98.4k475162
You could say $Y=3X+Z$ where $Zsim N(0,4)$ independent of $X$.
It is then an easy step to say that $Ysim N(3times 1+0,3^2times 1+4)$ and
- $E[X]=1$
- $E[Y]=3$
- $text{Var}(X)=1$
- $text{Var}(Y)=13$
- $text{Cov}(X,Y)=text{Cov}(X,3X)=3text{Var}(X)=3$
answered Jan 4 at 15:26
HenryHenry
98.4k475162
answered Jan 4 at 15:26
HenryHenry
98.4k475162
answered Jan 4 at 15:26
HenryHenry
98.4k475162
answered Jan 4 at 15:26
HenryHenry
98.4k475162
98.4k475162
This is the direction I took as well. However I did not use an additional Z. Therefore I am struggling with calculating the covariance between X and Y, could you take a look at my comment to the answer below?
– qcc101
Jan 4 at 15:28
add a comment |
This is the direction I took as well. However I did not use an additional Z. Therefore I am struggling with calculating the covariance between X and Y, could you take a look at my comment to the answer below?
– qcc101
Jan 4 at 15:28
This is the direction I took as well. However I did not use an additional Z. Therefore I am struggling with calculating the covariance between X and Y, could you take a look at my comment to the answer below?
– qcc101
Jan 4 at 15:28
This is the direction I took as well. However I did not use an additional Z. Therefore I am struggling with calculating the covariance between X and Y, could you take a look at my comment to the answer below?
– qcc101
Jan 4 at 15:28
add a comment |
You can proceed with moment generating functions.
Joint MGF of $(X,Y)$ is
begin{align}
M(s,t)&=Eleft[e^{sX+tY}right]
\&=Eleft[E(e^{sX+tY}mid X)right]
\&=Eleft[e^{sX}E(e^{tY}mid X)right]
end{align}
From the given information, you should be able to show that this MGF corresponds to the MGF of a bivariate normal distribution. That would complete your proof using the uniqueness property of MGF.
answered Jan 4 at 15:17
StubbornAtomStubbornAtom
5,39411138
I am not sure I am following you. In my problem I did find Variance and Expected value of Y, using the tower property of conditioning. I stil need to find the two symmetric entries of the matrix of variance and covariance to complete what I want to do. That is I need to find Covariance[X,Y]. Calculating, I get to this: $$Cov[X,Y] = E[XY] - 3E[X^2] -E[Y] + 3E[X]$$ Above, I know everything except E[XY]. I did the following: $ E[XY] = E[ E[XY|X]] = E[YE[X|X]] = E[Y E[X]]$ but I get the wrong result ( covariance should be 3 but I get -3). Could you point out what I did wrong?
– qcc101
Jan 4 at 15:26
@qcc101 $E(XY)=E(E(XYmid X))=E(XE(Ymid X))=...$
– StubbornAtom
Jan 4 at 15:29
Right, silly mistake. I will investigate the part about moment generating functions as well. Thank you.
– qcc101
Jan 4 at 15:30
1
@qcc101 Without the argument that $X$ and $Y$ are linear combinations of independent normal variables and hence $(X,Y)$ is bivariate normal (as seen in the answer by @Henry), I don't think finding the variance-covariance matrix (only) as you are trying to do is sufficient for the given question.
– StubbornAtom
Jan 4 at 15:37
add a comment |
You can proceed with moment generating functions.
Joint MGF of $(X,Y)$ is
begin{align}
M(s,t)&=Eleft[e^{sX+tY}right]
\&=Eleft[E(e^{sX+tY}mid X)right]
\&=Eleft[e^{sX}E(e^{tY}mid X)right]
end{align}
From the given information, you should be able to show that this MGF corresponds to the MGF of a bivariate normal distribution. That would complete your proof using the uniqueness property of MGF.
answered Jan 4 at 15:17
StubbornAtomStubbornAtom
5,39411138
I am not sure I am following you. In my problem I did find Variance and Expected value of Y, using the tower property of conditioning. I stil need to find the two symmetric entries of the matrix of variance and covariance to complete what I want to do. That is I need to find Covariance[X,Y]. Calculating, I get to this: $$Cov[X,Y] = E[XY] - 3E[X^2] -E[Y] + 3E[X]$$ Above, I know everything except E[XY]. I did the following: $ E[XY] = E[ E[XY|X]] = E[YE[X|X]] = E[Y E[X]]$ but I get the wrong result ( covariance should be 3 but I get -3). Could you point out what I did wrong?
– qcc101
Jan 4 at 15:26
@qcc101 $E(XY)=E(E(XYmid X))=E(XE(Ymid X))=...$
– StubbornAtom
Jan 4 at 15:29
Right, silly mistake. I will investigate the part about moment generating functions as well. Thank you.
– qcc101
Jan 4 at 15:30
1
@qcc101 Without the argument that $X$ and $Y$ are linear combinations of independent normal variables and hence $(X,Y)$ is bivariate normal (as seen in the answer by @Henry), I don't think finding the variance-covariance matrix (only) as you are trying to do is sufficient for the given question.
– StubbornAtom
Jan 4 at 15:37
add a comment |
You can proceed with moment generating functions.
Joint MGF of $(X,Y)$ is
begin{align}
M(s,t)&=Eleft[e^{sX+tY}right]
\&=Eleft[E(e^{sX+tY}mid X)right]
\&=Eleft[e^{sX}E(e^{tY}mid X)right]
end{align}
From the given information, you should be able to show that this MGF corresponds to the MGF of a bivariate normal distribution. That would complete your proof using the uniqueness property of MGF.
answered Jan 4 at 15:17
StubbornAtomStubbornAtom
5,39411138
You can proceed with moment generating functions.
Joint MGF of $(X,Y)$ is
begin{align}
M(s,t)&=Eleft[e^{sX+tY}right]
\&=Eleft[E(e^{sX+tY}mid X)right]
\&=Eleft[e^{sX}E(e^{tY}mid X)right]
end{align}
From the given information, you should be able to show that this MGF corresponds to the MGF of a bivariate normal distribution. That would complete your proof using the uniqueness property of MGF.
answered Jan 4 at 15:17
StubbornAtomStubbornAtom
5,39411138
answered Jan 4 at 15:17
StubbornAtomStubbornAtom
5,39411138
answered Jan 4 at 15:17
StubbornAtomStubbornAtom
5,39411138
answered Jan 4 at 15:17
StubbornAtomStubbornAtom
5,39411138
5,39411138
I am not sure I am following you. In my problem I did find Variance and Expected value of Y, using the tower property of conditioning. I stil need to find the two symmetric entries of the matrix of variance and covariance to complete what I want to do. That is I need to find Covariance[X,Y]. Calculating, I get to this: $$Cov[X,Y] = E[XY] - 3E[X^2] -E[Y] + 3E[X]$$ Above, I know everything except E[XY]. I did the following: $ E[XY] = E[ E[XY|X]] = E[YE[X|X]] = E[Y E[X]]$ but I get the wrong result ( covariance should be 3 but I get -3). Could you point out what I did wrong?
– qcc101
Jan 4 at 15:26
@qcc101 $E(XY)=E(E(XYmid X))=E(XE(Ymid X))=...$
– StubbornAtom
Jan 4 at 15:29
Right, silly mistake. I will investigate the part about moment generating functions as well. Thank you.
– qcc101
Jan 4 at 15:30
1
@qcc101 Without the argument that $X$ and $Y$ are linear combinations of independent normal variables and hence $(X,Y)$ is bivariate normal (as seen in the answer by @Henry), I don't think finding the variance-covariance matrix (only) as you are trying to do is sufficient for the given question.
– StubbornAtom
Jan 4 at 15:37
add a comment |
I am not sure I am following you. In my problem I did find Variance and Expected value of Y, using the tower property of conditioning. I stil need to find the two symmetric entries of the matrix of variance and covariance to complete what I want to do. That is I need to find Covariance[X,Y]. Calculating, I get to this: $$Cov[X,Y] = E[XY] - 3E[X^2] -E[Y] + 3E[X]$$ Above, I know everything except E[XY]. I did the following: $ E[XY] = E[ E[XY|X]] = E[YE[X|X]] = E[Y E[X]]$ but I get the wrong result ( covariance should be 3 but I get -3). Could you point out what I did wrong?
– qcc101
Jan 4 at 15:26
@qcc101 $E(XY)=E(E(XYmid X))=E(XE(Ymid X))=...$
– StubbornAtom
Jan 4 at 15:29
Right, silly mistake. I will investigate the part about moment generating functions as well. Thank you.
– qcc101
Jan 4 at 15:30
1
@qcc101 Without the argument that $X$ and $Y$ are linear combinations of independent normal variables and hence $(X,Y)$ is bivariate normal (as seen in the answer by @Henry), I don't think finding the variance-covariance matrix (only) as you are trying to do is sufficient for the given question.
– StubbornAtom
Jan 4 at 15:37
I am not sure I am following you. In my problem I did find Variance and Expected value of Y, using the tower property of conditioning. I stil need to find the two symmetric entries of the matrix of variance and covariance to complete what I want to do. That is I need to find Covariance[X,Y]. Calculating, I get to this: $$Cov[X,Y] = E[XY] - 3E[X^2] -E[Y] + 3E[X]$$ Above, I know everything except E[XY]. I did the following: $ E[XY] = E[ E[XY|X]] = E[YE[X|X]] = E[Y E[X]]$ but I get the wrong result ( covariance should be 3 but I get -3). Could you point out what I did wrong?
– qcc101
Jan 4 at 15:26
I am not sure I am following you. In my problem I did find Variance and Expected value of Y, using the tower property of conditioning. I stil need to find the two symmetric entries of the matrix of variance and covariance to complete what I want to do. That is I need to find Covariance[X,Y]. Calculating, I get to this: $$Cov[X,Y] = E[XY] - 3E[X^2] -E[Y] + 3E[X]$$ Above, I know everything except E[XY]. I did the following: $ E[XY] = E[ E[XY|X]] = E[YE[X|X]] = E[Y E[X]]$ but I get the wrong result ( covariance should be 3 but I get -3). Could you point out what I did wrong?
– qcc101
Jan 4 at 15:26
@qcc101 $E(XY)=E(E(XYmid X))=E(XE(Ymid X))=...$
– StubbornAtom
Jan 4 at 15:29
@qcc101 $E(XY)=E(E(XYmid X))=E(XE(Ymid X))=...$
– StubbornAtom
Jan 4 at 15:29
Right, silly mistake. I will investigate the part about moment generating functions as well. Thank you.
– qcc101
Jan 4 at 15:30
Right, silly mistake. I will investigate the part about moment generating functions as well. Thank you.
– qcc101
Jan 4 at 15:30
1
1
@qcc101 Without the argument that $X$ and $Y$ are linear combinations of independent normal variables and hence $(X,Y)$ is bivariate normal (as seen in the answer by @Henry), I don't think finding the variance-covariance matrix (only) as you are trying to do is sufficient for the given question.
– StubbornAtom
Jan 4 at 15:37
@qcc101 Without the argument that $X$ and $Y$ are linear combinations of independent normal variables and hence $(X,Y)$ is bivariate normal (as seen in the answer by @Henry), I don't think finding the variance-covariance matrix (only) as you are trying to do is sufficient for the given question.
– StubbornAtom
Jan 4 at 15:37
add a comment |
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Joint density of $(X,Y)$ is $f_{X,Y}(x,y)=f_{Ymid X}(y)f_X(x)=...$
– StubbornAtom
Jan 4 at 15:07
Yes, but I am trying to calculate E[Y], V[Y] and Cov[X,Y] to "fill" the matrices. I am struggling with Cov[X,Y], specifically with E[XY]. Do you have an hint for that?
– qcc101
Jan 4 at 15:10
Comment made by @StubbornAtom below his/her answer needs to be emphasized. Just finding the mean vector and the variance-covariance matrix do not show the joint-normality of $X$ and $Y$.
– Just_to_Answer
Jan 4 at 20:15