Finding joint pmfs from marginal pmfs
$begingroup$
Let a, b > 0. The random variables X and Y are independent and their densities are :
f(x) = 1/gamma(a)*x^(x-1)*e^-x, x>= 0
f(y) = 1/gamma(b)*y^(b-1)*e^-y, y>= 0
Let U=X+Y and V=X/X+Y
Find the joint density of U and V and show they are independent.
So far I have found J(u,v) to find the determinant is -u. I'm confused how to plug this into the marginal pdfs to find the joint pdfs.
probability-distributions
asked Nov 14 '13 at 4:16
fredfred
1615
$endgroup$
add a comment |
$begingroup$
Let a, b > 0. The random variables X and Y are independent and their densities are :
f(x) = 1/gamma(a)*x^(x-1)*e^-x, x>= 0
f(y) = 1/gamma(b)*y^(b-1)*e^-y, y>= 0
Let U=X+Y and V=X/X+Y
Find the joint density of U and V and show they are independent.
So far I have found J(u,v) to find the determinant is -u. I'm confused how to plug this into the marginal pdfs to find the joint pdfs.
probability-distributions
asked Nov 14 '13 at 4:16
fredfred
1615
$endgroup$
add a comment |
$begingroup$
Let a, b > 0. The random variables X and Y are independent and their densities are :
f(x) = 1/gamma(a)*x^(x-1)*e^-x, x>= 0
f(y) = 1/gamma(b)*y^(b-1)*e^-y, y>= 0
Let U=X+Y and V=X/X+Y
Find the joint density of U and V and show they are independent.
So far I have found J(u,v) to find the determinant is -u. I'm confused how to plug this into the marginal pdfs to find the joint pdfs.
probability-distributions
asked Nov 14 '13 at 4:16
fredfred
1615
$endgroup$
Let a, b > 0. The random variables X and Y are independent and their densities are :
f(x) = 1/gamma(a)*x^(x-1)*e^-x, x>= 0
f(y) = 1/gamma(b)*y^(b-1)*e^-y, y>= 0
Let U=X+Y and V=X/X+Y
Find the joint density of U and V and show they are independent.
So far I have found J(u,v) to find the determinant is -u. I'm confused how to plug this into the marginal pdfs to find the joint pdfs.
probability-distributions
probability-distributions
asked Nov 14 '13 at 4:16
fredfred
1615
asked Nov 14 '13 at 4:16
fredfred
1615
asked Nov 14 '13 at 4:16
fredfred
1615
asked Nov 14 '13 at 4:16
fredfred
1615
asked Nov 14 '13 at 4:16
fredfred
1615
1615
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Firstly, when working with absolutely continuous distributions (those with densities), you refer to the density as the pdf (probability density function), not the pmf (probability mass function [used when there are "masses" at singletons]).
The change of variables formula tells us that
$$
f_{(U,V)}(u,v)=f_{(X,Y)}(x,y)|det frac{d(x,y)}{d(u,v)}| = f_X(x)f_Y(y)cdot|-u|.
$$
where he we have used the fact that $f_{(X,Y)}=f_Xf_Y$ by independence of $X,Y$.
Write $x$ and $y$ in terms of $u,v$ and plug them into the $x,y$ in the above formula. You will find that $f_{(U,V)}$ will end up being the product of two familiar looking densities $h_1(u),h_2(v)$. It then immediately follows that $f_U = h_1, f_V = h_2$ and so $f_{(U,V)} = f_U f_V$ which tells us that $U,V$ are independent.
Can you run through the calculation and see what $h_1,h_2$ appear?
Edit: Note that $frac{d(x,y)}{d(u,v)}=J(u,v)$ is meant to be evaluated at $(u,v)$.
Hint 1: $x=uv, y=u(1-v)$, and you correctly calculated $J(u,v)=-u$. Plugging into the densities I am getting
$$
f_{(U,V)}(u,v) = left{begin{array}{cc}
&
begin{array}{cc}
frac{e^{-u} (u v)^{a-1} (u(1-v))^b}{(1-v) Gamma (a) Gamma (b)} & 0leq uv,0leq u(1-v) \
0 & text{else} \
end{array}
\
end{array}right.
$$
Do you see what $h_1,h_2$ are now?
edited Nov 14 '13 at 5:55
Did
248k23224463
answered Nov 14 '13 at 4:37
nullUsernullUser
16.7k442104
$endgroup$
$begingroup$
yes thank you for the help.
$endgroup$
– fred
Nov 14 '13 at 4:42
$begingroup$
Nevermind I'm not sure exactly what h1 and h2 are. Do they have something to do with the gamma function?
$endgroup$
– fred
Nov 14 '13 at 5:01
$begingroup$
Your PDFs are supposed to be $Gamma(a,1),Gamma(b,1)$, correct? You made a typo in the first one writing $x^{x-1}$ instead of $x^{a-1}$.
$endgroup$
– nullUser
Nov 14 '13 at 5:14
$begingroup$
yes that is correct
$endgroup$
– fred
Nov 14 '13 at 5:18
add a comment |
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1 Answer
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1 Answer
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$begingroup$
Firstly, when working with absolutely continuous distributions (those with densities), you refer to the density as the pdf (probability density function), not the pmf (probability mass function [used when there are "masses" at singletons]).
The change of variables formula tells us that
$$
f_{(U,V)}(u,v)=f_{(X,Y)}(x,y)|det frac{d(x,y)}{d(u,v)}| = f_X(x)f_Y(y)cdot|-u|.
$$
where he we have used the fact that $f_{(X,Y)}=f_Xf_Y$ by independence of $X,Y$.
Write $x$ and $y$ in terms of $u,v$ and plug them into the $x,y$ in the above formula. You will find that $f_{(U,V)}$ will end up being the product of two familiar looking densities $h_1(u),h_2(v)$. It then immediately follows that $f_U = h_1, f_V = h_2$ and so $f_{(U,V)} = f_U f_V$ which tells us that $U,V$ are independent.
Can you run through the calculation and see what $h_1,h_2$ appear?
Edit: Note that $frac{d(x,y)}{d(u,v)}=J(u,v)$ is meant to be evaluated at $(u,v)$.
Hint 1: $x=uv, y=u(1-v)$, and you correctly calculated $J(u,v)=-u$. Plugging into the densities I am getting
$$
f_{(U,V)}(u,v) = left{begin{array}{cc}
&
begin{array}{cc}
frac{e^{-u} (u v)^{a-1} (u(1-v))^b}{(1-v) Gamma (a) Gamma (b)} & 0leq uv,0leq u(1-v) \
0 & text{else} \
end{array}
\
end{array}right.
$$
Do you see what $h_1,h_2$ are now?
edited Nov 14 '13 at 5:55
Did
248k23224463
answered Nov 14 '13 at 4:37
nullUsernullUser
16.7k442104
$endgroup$
$begingroup$
yes thank you for the help.
$endgroup$
– fred
Nov 14 '13 at 4:42
$begingroup$
Nevermind I'm not sure exactly what h1 and h2 are. Do they have something to do with the gamma function?
$endgroup$
– fred
Nov 14 '13 at 5:01
$begingroup$
Your PDFs are supposed to be $Gamma(a,1),Gamma(b,1)$, correct? You made a typo in the first one writing $x^{x-1}$ instead of $x^{a-1}$.
$endgroup$
– nullUser
Nov 14 '13 at 5:14
$begingroup$
yes that is correct
$endgroup$
– fred
Nov 14 '13 at 5:18
add a comment |
$begingroup$
Firstly, when working with absolutely continuous distributions (those with densities), you refer to the density as the pdf (probability density function), not the pmf (probability mass function [used when there are "masses" at singletons]).
The change of variables formula tells us that
$$
f_{(U,V)}(u,v)=f_{(X,Y)}(x,y)|det frac{d(x,y)}{d(u,v)}| = f_X(x)f_Y(y)cdot|-u|.
$$
where he we have used the fact that $f_{(X,Y)}=f_Xf_Y$ by independence of $X,Y$.
Write $x$ and $y$ in terms of $u,v$ and plug them into the $x,y$ in the above formula. You will find that $f_{(U,V)}$ will end up being the product of two familiar looking densities $h_1(u),h_2(v)$. It then immediately follows that $f_U = h_1, f_V = h_2$ and so $f_{(U,V)} = f_U f_V$ which tells us that $U,V$ are independent.
Can you run through the calculation and see what $h_1,h_2$ appear?
Edit: Note that $frac{d(x,y)}{d(u,v)}=J(u,v)$ is meant to be evaluated at $(u,v)$.
Hint 1: $x=uv, y=u(1-v)$, and you correctly calculated $J(u,v)=-u$. Plugging into the densities I am getting
$$
f_{(U,V)}(u,v) = left{begin{array}{cc}
&
begin{array}{cc}
frac{e^{-u} (u v)^{a-1} (u(1-v))^b}{(1-v) Gamma (a) Gamma (b)} & 0leq uv,0leq u(1-v) \
0 & text{else} \
end{array}
\
end{array}right.
$$
Do you see what $h_1,h_2$ are now?
edited Nov 14 '13 at 5:55
Did
248k23224463
answered Nov 14 '13 at 4:37
nullUsernullUser
16.7k442104
$endgroup$
$begingroup$
yes thank you for the help.
$endgroup$
– fred
Nov 14 '13 at 4:42
$begingroup$
Nevermind I'm not sure exactly what h1 and h2 are. Do they have something to do with the gamma function?
$endgroup$
– fred
Nov 14 '13 at 5:01
$begingroup$
Your PDFs are supposed to be $Gamma(a,1),Gamma(b,1)$, correct? You made a typo in the first one writing $x^{x-1}$ instead of $x^{a-1}$.
$endgroup$
– nullUser
Nov 14 '13 at 5:14
$begingroup$
yes that is correct
$endgroup$
– fred
Nov 14 '13 at 5:18
add a comment |
$begingroup$
Firstly, when working with absolutely continuous distributions (those with densities), you refer to the density as the pdf (probability density function), not the pmf (probability mass function [used when there are "masses" at singletons]).
The change of variables formula tells us that
$$
f_{(U,V)}(u,v)=f_{(X,Y)}(x,y)|det frac{d(x,y)}{d(u,v)}| = f_X(x)f_Y(y)cdot|-u|.
$$
where he we have used the fact that $f_{(X,Y)}=f_Xf_Y$ by independence of $X,Y$.
Write $x$ and $y$ in terms of $u,v$ and plug them into the $x,y$ in the above formula. You will find that $f_{(U,V)}$ will end up being the product of two familiar looking densities $h_1(u),h_2(v)$. It then immediately follows that $f_U = h_1, f_V = h_2$ and so $f_{(U,V)} = f_U f_V$ which tells us that $U,V$ are independent.
Can you run through the calculation and see what $h_1,h_2$ appear?
Edit: Note that $frac{d(x,y)}{d(u,v)}=J(u,v)$ is meant to be evaluated at $(u,v)$.
Hint 1: $x=uv, y=u(1-v)$, and you correctly calculated $J(u,v)=-u$. Plugging into the densities I am getting
$$
f_{(U,V)}(u,v) = left{begin{array}{cc}
&
begin{array}{cc}
frac{e^{-u} (u v)^{a-1} (u(1-v))^b}{(1-v) Gamma (a) Gamma (b)} & 0leq uv,0leq u(1-v) \
0 & text{else} \
end{array}
\
end{array}right.
$$
Do you see what $h_1,h_2$ are now?
edited Nov 14 '13 at 5:55
Did
248k23224463
answered Nov 14 '13 at 4:37
nullUsernullUser
16.7k442104
$endgroup$
Firstly, when working with absolutely continuous distributions (those with densities), you refer to the density as the pdf (probability density function), not the pmf (probability mass function [used when there are "masses" at singletons]).
The change of variables formula tells us that
$$
f_{(U,V)}(u,v)=f_{(X,Y)}(x,y)|det frac{d(x,y)}{d(u,v)}| = f_X(x)f_Y(y)cdot|-u|.
$$
where he we have used the fact that $f_{(X,Y)}=f_Xf_Y$ by independence of $X,Y$.
Write $x$ and $y$ in terms of $u,v$ and plug them into the $x,y$ in the above formula. You will find that $f_{(U,V)}$ will end up being the product of two familiar looking densities $h_1(u),h_2(v)$. It then immediately follows that $f_U = h_1, f_V = h_2$ and so $f_{(U,V)} = f_U f_V$ which tells us that $U,V$ are independent.
Can you run through the calculation and see what $h_1,h_2$ appear?
Edit: Note that $frac{d(x,y)}{d(u,v)}=J(u,v)$ is meant to be evaluated at $(u,v)$.
Hint 1: $x=uv, y=u(1-v)$, and you correctly calculated $J(u,v)=-u$. Plugging into the densities I am getting
$$
f_{(U,V)}(u,v) = left{begin{array}{cc}
&
begin{array}{cc}
frac{e^{-u} (u v)^{a-1} (u(1-v))^b}{(1-v) Gamma (a) Gamma (b)} & 0leq uv,0leq u(1-v) \
0 & text{else} \
end{array}
\
end{array}right.
$$
Do you see what $h_1,h_2$ are now?
edited Nov 14 '13 at 5:55
Did
248k23224463
answered Nov 14 '13 at 4:37
nullUsernullUser
16.7k442104
edited Nov 14 '13 at 5:55
Did
248k23224463
edited Nov 14 '13 at 5:55
Did
248k23224463
edited Nov 14 '13 at 5:55
Did
248k23224463
248k23224463
answered Nov 14 '13 at 4:37
nullUsernullUser
16.7k442104
answered Nov 14 '13 at 4:37
nullUsernullUser
16.7k442104
answered Nov 14 '13 at 4:37
nullUsernullUser
16.7k442104
16.7k442104
$begingroup$
yes thank you for the help.
$endgroup$
– fred
Nov 14 '13 at 4:42
$begingroup$
Nevermind I'm not sure exactly what h1 and h2 are. Do they have something to do with the gamma function?
$endgroup$
– fred
Nov 14 '13 at 5:01
$begingroup$
Your PDFs are supposed to be $Gamma(a,1),Gamma(b,1)$, correct? You made a typo in the first one writing $x^{x-1}$ instead of $x^{a-1}$.
$endgroup$
– nullUser
Nov 14 '13 at 5:14
$begingroup$
yes that is correct
$endgroup$
– fred
Nov 14 '13 at 5:18
add a comment |
$begingroup$
yes thank you for the help.
$endgroup$
– fred
Nov 14 '13 at 4:42
$begingroup$
Nevermind I'm not sure exactly what h1 and h2 are. Do they have something to do with the gamma function?
$endgroup$
– fred
Nov 14 '13 at 5:01
$begingroup$
Your PDFs are supposed to be $Gamma(a,1),Gamma(b,1)$, correct? You made a typo in the first one writing $x^{x-1}$ instead of $x^{a-1}$.
$endgroup$
– nullUser
Nov 14 '13 at 5:14
$begingroup$
yes that is correct
$endgroup$
– fred
Nov 14 '13 at 5:18
$begingroup$
yes thank you for the help.
$endgroup$
– fred
Nov 14 '13 at 4:42
$begingroup$
yes thank you for the help.
$endgroup$
– fred
Nov 14 '13 at 4:42
$begingroup$
Nevermind I'm not sure exactly what h1 and h2 are. Do they have something to do with the gamma function?
$endgroup$
– fred
Nov 14 '13 at 5:01
$begingroup$
Nevermind I'm not sure exactly what h1 and h2 are. Do they have something to do with the gamma function?
$endgroup$
– fred
Nov 14 '13 at 5:01
$begingroup$
Your PDFs are supposed to be $Gamma(a,1),Gamma(b,1)$, correct? You made a typo in the first one writing $x^{x-1}$ instead of $x^{a-1}$.
$endgroup$
– nullUser
Nov 14 '13 at 5:14
$begingroup$
Your PDFs are supposed to be $Gamma(a,1),Gamma(b,1)$, correct? You made a typo in the first one writing $x^{x-1}$ instead of $x^{a-1}$.
$endgroup$
– nullUser
Nov 14 '13 at 5:14
$begingroup$
yes that is correct
$endgroup$
– fred
Nov 14 '13 at 5:18
$begingroup$
yes that is correct
$endgroup$
– fred
Nov 14 '13 at 5:18
add a comment |
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