Proving $(X,Y)$ is a normal vector when $Xsim N(1,1)$ and $Ymid Xsim N(3X,4)$












1












$begingroup$


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







share|cite|improve this question











$endgroup$












  • $begingroup$
    Joint density of $(X,Y)$ is $f_{X,Y}(x,y)=f_{Ymid X}(y)f_X(x)=...$
    $endgroup$
    – StubbornAtom
    Jan 4 at 15:07










  • $begingroup$
    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?
    $endgroup$
    – qcc101
    Jan 4 at 15:10










  • $begingroup$
    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 $Y$.
    $endgroup$
    – Just_to_Answer
    Jan 4 at 20:15


















1












$begingroup$


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







share|cite|improve this question











$endgroup$












  • $begingroup$
    Joint density of $(X,Y)$ is $f_{X,Y}(x,y)=f_{Ymid X}(y)f_X(x)=...$
    $endgroup$
    – StubbornAtom
    Jan 4 at 15:07










  • $begingroup$
    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?
    $endgroup$
    – qcc101
    Jan 4 at 15:10










  • $begingroup$
    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 $Y$.
    $endgroup$
    – Just_to_Answer
    Jan 4 at 20:15
















1












1








1


1



$begingroup$


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







share|cite|improve this question











$endgroup$




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






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 4 at 19:02









StubbornAtom

6,29831440




6,29831440










asked Jan 4 at 14:48









qcc101qcc101

629213




629213












  • $begingroup$
    Joint density of $(X,Y)$ is $f_{X,Y}(x,y)=f_{Ymid X}(y)f_X(x)=...$
    $endgroup$
    – StubbornAtom
    Jan 4 at 15:07










  • $begingroup$
    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?
    $endgroup$
    – qcc101
    Jan 4 at 15:10










  • $begingroup$
    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 $Y$.
    $endgroup$
    – Just_to_Answer
    Jan 4 at 20:15




















  • $begingroup$
    Joint density of $(X,Y)$ is $f_{X,Y}(x,y)=f_{Ymid X}(y)f_X(x)=...$
    $endgroup$
    – StubbornAtom
    Jan 4 at 15:07










  • $begingroup$
    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?
    $endgroup$
    – qcc101
    Jan 4 at 15:10










  • $begingroup$
    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 $Y$.
    $endgroup$
    – Just_to_Answer
    Jan 4 at 20:15


















$begingroup$
Joint density of $(X,Y)$ is $f_{X,Y}(x,y)=f_{Ymid X}(y)f_X(x)=...$
$endgroup$
– StubbornAtom
Jan 4 at 15:07




$begingroup$
Joint density of $(X,Y)$ is $f_{X,Y}(x,y)=f_{Ymid X}(y)f_X(x)=...$
$endgroup$
– StubbornAtom
Jan 4 at 15:07












$begingroup$
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?
$endgroup$
– qcc101
Jan 4 at 15:10




$begingroup$
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?
$endgroup$
– qcc101
Jan 4 at 15:10












$begingroup$
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 $Y$.
$endgroup$
– Just_to_Answer
Jan 4 at 20:15






$begingroup$
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 $Y$.
$endgroup$
– Just_to_Answer
Jan 4 at 20:15












2 Answers
2






active

oldest

votes


















2












$begingroup$

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




  • $E[X]=1$

  • $E[Y]=3$

  • $text{Var}(X)=1$

  • $text{Var}(Y)=13$

  • $text{Cov}(X,Y)=text{Cov}(X,3X)=3text{Var}(X)=3$






share|cite|improve this answer









$endgroup$













  • $begingroup$
    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?
    $endgroup$
    – qcc101
    Jan 4 at 15:28



















2












$begingroup$

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.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    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?
    $endgroup$
    – qcc101
    Jan 4 at 15:26










  • $begingroup$
    @qcc101 $E(XY)=E(E(XYmid X))=E(XE(Ymid X))=...$
    $endgroup$
    – StubbornAtom
    Jan 4 at 15:29










  • $begingroup$
    Right, silly mistake. I will investigate the part about moment generating functions as well. Thank you.
    $endgroup$
    – qcc101
    Jan 4 at 15:30






  • 1




    $begingroup$
    @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.
    $endgroup$
    – StubbornAtom
    Jan 4 at 15:37














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2 Answers
2






active

oldest

votes








2 Answers
2






active

oldest

votes









active

oldest

votes






active

oldest

votes









2












$begingroup$

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




  • $E[X]=1$

  • $E[Y]=3$

  • $text{Var}(X)=1$

  • $text{Var}(Y)=13$

  • $text{Cov}(X,Y)=text{Cov}(X,3X)=3text{Var}(X)=3$






share|cite|improve this answer









$endgroup$













  • $begingroup$
    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?
    $endgroup$
    – qcc101
    Jan 4 at 15:28
















2












$begingroup$

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




  • $E[X]=1$

  • $E[Y]=3$

  • $text{Var}(X)=1$

  • $text{Var}(Y)=13$

  • $text{Cov}(X,Y)=text{Cov}(X,3X)=3text{Var}(X)=3$






share|cite|improve this answer









$endgroup$













  • $begingroup$
    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?
    $endgroup$
    – qcc101
    Jan 4 at 15:28














2












2








2





$begingroup$

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




  • $E[X]=1$

  • $E[Y]=3$

  • $text{Var}(X)=1$

  • $text{Var}(Y)=13$

  • $text{Cov}(X,Y)=text{Cov}(X,3X)=3text{Var}(X)=3$






share|cite|improve this answer









$endgroup$



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




  • $E[X]=1$

  • $E[Y]=3$

  • $text{Var}(X)=1$

  • $text{Var}(Y)=13$

  • $text{Cov}(X,Y)=text{Cov}(X,3X)=3text{Var}(X)=3$







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Jan 4 at 15:26









HenryHenry

101k482170




101k482170












  • $begingroup$
    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?
    $endgroup$
    – qcc101
    Jan 4 at 15:28


















  • $begingroup$
    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?
    $endgroup$
    – qcc101
    Jan 4 at 15:28
















$begingroup$
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?
$endgroup$
– qcc101
Jan 4 at 15:28




$begingroup$
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?
$endgroup$
– qcc101
Jan 4 at 15:28











2












$begingroup$

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.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    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?
    $endgroup$
    – qcc101
    Jan 4 at 15:26










  • $begingroup$
    @qcc101 $E(XY)=E(E(XYmid X))=E(XE(Ymid X))=...$
    $endgroup$
    – StubbornAtom
    Jan 4 at 15:29










  • $begingroup$
    Right, silly mistake. I will investigate the part about moment generating functions as well. Thank you.
    $endgroup$
    – qcc101
    Jan 4 at 15:30






  • 1




    $begingroup$
    @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.
    $endgroup$
    – StubbornAtom
    Jan 4 at 15:37


















2












$begingroup$

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.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    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?
    $endgroup$
    – qcc101
    Jan 4 at 15:26










  • $begingroup$
    @qcc101 $E(XY)=E(E(XYmid X))=E(XE(Ymid X))=...$
    $endgroup$
    – StubbornAtom
    Jan 4 at 15:29










  • $begingroup$
    Right, silly mistake. I will investigate the part about moment generating functions as well. Thank you.
    $endgroup$
    – qcc101
    Jan 4 at 15:30






  • 1




    $begingroup$
    @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.
    $endgroup$
    – StubbornAtom
    Jan 4 at 15:37
















2












2








2





$begingroup$

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.






share|cite|improve this answer









$endgroup$



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.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Jan 4 at 15:17









StubbornAtomStubbornAtom

6,29831440




6,29831440












  • $begingroup$
    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?
    $endgroup$
    – qcc101
    Jan 4 at 15:26










  • $begingroup$
    @qcc101 $E(XY)=E(E(XYmid X))=E(XE(Ymid X))=...$
    $endgroup$
    – StubbornAtom
    Jan 4 at 15:29










  • $begingroup$
    Right, silly mistake. I will investigate the part about moment generating functions as well. Thank you.
    $endgroup$
    – qcc101
    Jan 4 at 15:30






  • 1




    $begingroup$
    @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.
    $endgroup$
    – StubbornAtom
    Jan 4 at 15:37




















  • $begingroup$
    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?
    $endgroup$
    – qcc101
    Jan 4 at 15:26










  • $begingroup$
    @qcc101 $E(XY)=E(E(XYmid X))=E(XE(Ymid X))=...$
    $endgroup$
    – StubbornAtom
    Jan 4 at 15:29










  • $begingroup$
    Right, silly mistake. I will investigate the part about moment generating functions as well. Thank you.
    $endgroup$
    – qcc101
    Jan 4 at 15:30






  • 1




    $begingroup$
    @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.
    $endgroup$
    – StubbornAtom
    Jan 4 at 15:37


















$begingroup$
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?
$endgroup$
– qcc101
Jan 4 at 15:26




$begingroup$
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?
$endgroup$
– qcc101
Jan 4 at 15:26












$begingroup$
@qcc101 $E(XY)=E(E(XYmid X))=E(XE(Ymid X))=...$
$endgroup$
– StubbornAtom
Jan 4 at 15:29




$begingroup$
@qcc101 $E(XY)=E(E(XYmid X))=E(XE(Ymid X))=...$
$endgroup$
– StubbornAtom
Jan 4 at 15:29












$begingroup$
Right, silly mistake. I will investigate the part about moment generating functions as well. Thank you.
$endgroup$
– qcc101
Jan 4 at 15:30




$begingroup$
Right, silly mistake. I will investigate the part about moment generating functions as well. Thank you.
$endgroup$
– qcc101
Jan 4 at 15:30




1




1




$begingroup$
@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.
$endgroup$
– StubbornAtom
Jan 4 at 15:37






$begingroup$
@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.
$endgroup$
– StubbornAtom
Jan 4 at 15:37




















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