Finding the joint mgf of two random variables












0












$begingroup$



Let the joint pdf of $(X, Y)$ be given by



$$f(x, y) = frac{1}{sqrt{2pi}} text{exp}left(-y-frac{(x -
y)^{2}}{2}right) hspace{1cm} text{ for } y > 0, -infty < x <
infty$$



Find the joint mgf of $X$ and $Y$.




My attempt:



Recall that every joint probability distribution function can be decomposed into the product of a marginal density and a conditional density. That is, we can write



$$f(x, y) = f_{Xmid Y}(x mid y) cdot f_{Y}(y).$$



Defining $f_{X mid Y}(xmid y) = frac{1}{sqrt{2pi}} text{exp}(-(-x-y)^{2}/2)$ and $f_{Y}(y) = text{exp}(-y)$ works.



Now, note that $X | Y$ follows a normal distribution with parameters $y$ and $1$. Also, $Y$ follows an exponential distribution with parameter $1$. Therefore,



$$M_{X, Y}(s, t) = M_{X mid Y}(s) cdot M_{Y}(t) $$



$$= frac{text{exp}(ys + s^{2}/2)}{1 - t} $$



for $t < 1$.




Is my solution right? If not, what is the correct way to solve this problem?






probability probability-theory probability-distributions







share|cite|improve this question











$endgroup$












  • $begingroup$
    It's a 2D normal distribution. Try completing the squares inside the exponential function first, then directly apply the mgf for multivariate normal.
    $endgroup$
    – Vim
    Dec 13 '18 at 3:56










  • $begingroup$
    The joint MGF should be a function of two variables
    $endgroup$
    – angryavian
    Dec 13 '18 at 3:57










  • $begingroup$
    I edited my post
    $endgroup$
    – joseph
    Dec 13 '18 at 4:00
















0












$begingroup$



Let the joint pdf of $(X, Y)$ be given by



$$f(x, y) = frac{1}{sqrt{2pi}} text{exp}left(-y-frac{(x -
y)^{2}}{2}right) hspace{1cm} text{ for } y > 0, -infty < x <
infty$$



Find the joint mgf of $X$ and $Y$.




My attempt:



Recall that every joint probability distribution function can be decomposed into the product of a marginal density and a conditional density. That is, we can write



$$f(x, y) = f_{Xmid Y}(x mid y) cdot f_{Y}(y).$$



Defining $f_{X mid Y}(xmid y) = frac{1}{sqrt{2pi}} text{exp}(-(-x-y)^{2}/2)$ and $f_{Y}(y) = text{exp}(-y)$ works.



Now, note that $X | Y$ follows a normal distribution with parameters $y$ and $1$. Also, $Y$ follows an exponential distribution with parameter $1$. Therefore,



$$M_{X, Y}(s, t) = M_{X mid Y}(s) cdot M_{Y}(t) $$



$$= frac{text{exp}(ys + s^{2}/2)}{1 - t} $$



for $t < 1$.




Is my solution right? If not, what is the correct way to solve this problem?






probability probability-theory probability-distributions







share|cite|improve this question











$endgroup$












  • $begingroup$
    It's a 2D normal distribution. Try completing the squares inside the exponential function first, then directly apply the mgf for multivariate normal.
    $endgroup$
    – Vim
    Dec 13 '18 at 3:56










  • $begingroup$
    The joint MGF should be a function of two variables
    $endgroup$
    – angryavian
    Dec 13 '18 at 3:57










  • $begingroup$
    I edited my post
    $endgroup$
    – joseph
    Dec 13 '18 at 4:00














0












0








0





$begingroup$



Let the joint pdf of $(X, Y)$ be given by



$$f(x, y) = frac{1}{sqrt{2pi}} text{exp}left(-y-frac{(x -
y)^{2}}{2}right) hspace{1cm} text{ for } y > 0, -infty < x <
infty$$



Find the joint mgf of $X$ and $Y$.




My attempt:



Recall that every joint probability distribution function can be decomposed into the product of a marginal density and a conditional density. That is, we can write



$$f(x, y) = f_{Xmid Y}(x mid y) cdot f_{Y}(y).$$



Defining $f_{X mid Y}(xmid y) = frac{1}{sqrt{2pi}} text{exp}(-(-x-y)^{2}/2)$ and $f_{Y}(y) = text{exp}(-y)$ works.



Now, note that $X | Y$ follows a normal distribution with parameters $y$ and $1$. Also, $Y$ follows an exponential distribution with parameter $1$. Therefore,



$$M_{X, Y}(s, t) = M_{X mid Y}(s) cdot M_{Y}(t) $$



$$= frac{text{exp}(ys + s^{2}/2)}{1 - t} $$



for $t < 1$.




Is my solution right? If not, what is the correct way to solve this problem?






probability probability-theory probability-distributions







share|cite|improve this question











$endgroup$





Let the joint pdf of $(X, Y)$ be given by



$$f(x, y) = frac{1}{sqrt{2pi}} text{exp}left(-y-frac{(x -
y)^{2}}{2}right) hspace{1cm} text{ for } y > 0, -infty < x <
infty$$



Find the joint mgf of $X$ and $Y$.




My attempt:



Recall that every joint probability distribution function can be decomposed into the product of a marginal density and a conditional density. That is, we can write



$$f(x, y) = f_{Xmid Y}(x mid y) cdot f_{Y}(y).$$



Defining $f_{X mid Y}(xmid y) = frac{1}{sqrt{2pi}} text{exp}(-(-x-y)^{2}/2)$ and $f_{Y}(y) = text{exp}(-y)$ works.



Now, note that $X | Y$ follows a normal distribution with parameters $y$ and $1$. Also, $Y$ follows an exponential distribution with parameter $1$. Therefore,



$$M_{X, Y}(s, t) = M_{X mid Y}(s) cdot M_{Y}(t) $$



$$= frac{text{exp}(ys + s^{2}/2)}{1 - t} $$



for $t < 1$.




Is my solution right? If not, what is the correct way to solve this problem?






probability probability-theory probability-distributions




probability probability-theory probability-distributions






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 13 '18 at 3:59







joseph

















asked Dec 13 '18 at 3:52









josephjoseph

500111




500111












  • $begingroup$
    It's a 2D normal distribution. Try completing the squares inside the exponential function first, then directly apply the mgf for multivariate normal.
    $endgroup$
    – Vim
    Dec 13 '18 at 3:56










  • $begingroup$
    The joint MGF should be a function of two variables
    $endgroup$
    – angryavian
    Dec 13 '18 at 3:57










  • $begingroup$
    I edited my post
    $endgroup$
    – joseph
    Dec 13 '18 at 4:00


















  • $begingroup$
    It's a 2D normal distribution. Try completing the squares inside the exponential function first, then directly apply the mgf for multivariate normal.
    $endgroup$
    – Vim
    Dec 13 '18 at 3:56










  • $begingroup$
    The joint MGF should be a function of two variables
    $endgroup$
    – angryavian
    Dec 13 '18 at 3:57










  • $begingroup$
    I edited my post
    $endgroup$
    – joseph
    Dec 13 '18 at 4:00
















$begingroup$
It's a 2D normal distribution. Try completing the squares inside the exponential function first, then directly apply the mgf for multivariate normal.
$endgroup$
– Vim
Dec 13 '18 at 3:56




$begingroup$
It's a 2D normal distribution. Try completing the squares inside the exponential function first, then directly apply the mgf for multivariate normal.
$endgroup$
– Vim
Dec 13 '18 at 3:56












$begingroup$
The joint MGF should be a function of two variables
$endgroup$
– angryavian
Dec 13 '18 at 3:57




$begingroup$
The joint MGF should be a function of two variables
$endgroup$
– angryavian
Dec 13 '18 at 3:57












$begingroup$
I edited my post
$endgroup$
– joseph
Dec 13 '18 at 4:00




$begingroup$
I edited my post
$endgroup$
– joseph
Dec 13 '18 at 4:00










1 Answer
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oldest

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1












$begingroup$

You have correctly deduced that $$Xmid Ysimmathcal N(Y,1)quad,text{ where }Ysimtext{Exp}(1)$$



So the MGF of $(X,Y)$ should be



begin{align}
M(s,t)&=E,(e^{sX+tY})
\&=Eleft[E,(e^{sX+tY}mid Y)right]
\&=Eleft[e^{tY}E,(e^{sX}mid Y)right]
\&=Eleft[e^{tY}e^{sY+s^2/2}right]
\&=e^{s^2/2},Eleft[e^{(t+s)Y}right]
\&=frac{e^{s^2/2}}{1-(s+t)}quad,,s+t<1
end{align}



The formula that you used, namely $M_{X, Y}(s, t) = M_{X mid Y}(s) cdot M_{Y}(t)$, doesn't make much sense.






share|cite|improve this answer









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    $begingroup$

    You have correctly deduced that $$Xmid Ysimmathcal N(Y,1)quad,text{ where }Ysimtext{Exp}(1)$$



    So the MGF of $(X,Y)$ should be



    begin{align}
    M(s,t)&=E,(e^{sX+tY})
    \&=Eleft[E,(e^{sX+tY}mid Y)right]
    \&=Eleft[e^{tY}E,(e^{sX}mid Y)right]
    \&=Eleft[e^{tY}e^{sY+s^2/2}right]
    \&=e^{s^2/2},Eleft[e^{(t+s)Y}right]
    \&=frac{e^{s^2/2}}{1-(s+t)}quad,,s+t<1
    end{align}



    The formula that you used, namely $M_{X, Y}(s, t) = M_{X mid Y}(s) cdot M_{Y}(t)$, doesn't make much sense.






    share|cite|improve this answer









    $endgroup$


















      1












      $begingroup$

      You have correctly deduced that $$Xmid Ysimmathcal N(Y,1)quad,text{ where }Ysimtext{Exp}(1)$$



      So the MGF of $(X,Y)$ should be



      begin{align}
      M(s,t)&=E,(e^{sX+tY})
      \&=Eleft[E,(e^{sX+tY}mid Y)right]
      \&=Eleft[e^{tY}E,(e^{sX}mid Y)right]
      \&=Eleft[e^{tY}e^{sY+s^2/2}right]
      \&=e^{s^2/2},Eleft[e^{(t+s)Y}right]
      \&=frac{e^{s^2/2}}{1-(s+t)}quad,,s+t<1
      end{align}



      The formula that you used, namely $M_{X, Y}(s, t) = M_{X mid Y}(s) cdot M_{Y}(t)$, doesn't make much sense.






      share|cite|improve this answer









      $endgroup$
















        1












        1








        1





        $begingroup$

        You have correctly deduced that $$Xmid Ysimmathcal N(Y,1)quad,text{ where }Ysimtext{Exp}(1)$$



        So the MGF of $(X,Y)$ should be



        begin{align}
        M(s,t)&=E,(e^{sX+tY})
        \&=Eleft[E,(e^{sX+tY}mid Y)right]
        \&=Eleft[e^{tY}E,(e^{sX}mid Y)right]
        \&=Eleft[e^{tY}e^{sY+s^2/2}right]
        \&=e^{s^2/2},Eleft[e^{(t+s)Y}right]
        \&=frac{e^{s^2/2}}{1-(s+t)}quad,,s+t<1
        end{align}



        The formula that you used, namely $M_{X, Y}(s, t) = M_{X mid Y}(s) cdot M_{Y}(t)$, doesn't make much sense.






        share|cite|improve this answer









        $endgroup$



        You have correctly deduced that $$Xmid Ysimmathcal N(Y,1)quad,text{ where }Ysimtext{Exp}(1)$$



        So the MGF of $(X,Y)$ should be



        begin{align}
        M(s,t)&=E,(e^{sX+tY})
        \&=Eleft[E,(e^{sX+tY}mid Y)right]
        \&=Eleft[e^{tY}E,(e^{sX}mid Y)right]
        \&=Eleft[e^{tY}e^{sY+s^2/2}right]
        \&=e^{s^2/2},Eleft[e^{(t+s)Y}right]
        \&=frac{e^{s^2/2}}{1-(s+t)}quad,,s+t<1
        end{align}



        The formula that you used, namely $M_{X, Y}(s, t) = M_{X mid Y}(s) cdot M_{Y}(t)$, doesn't make much sense.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Dec 13 '18 at 13:06









        StubbornAtomStubbornAtom

        5,98811238




        5,98811238






























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