Marginal Density Function, Gamma and Beta distributions












0












$begingroup$



If $Ysimoperatorname{Gamma}(gamma,delta)$ and $Zsimoperatorname{Beta}(alpha,beta)$ then their density functions are, respectively,
$$
f_Y(y)=frac{delta^gamma}{Gamma(gamma)}y^{gamma-1}e^{-delta y},quad y>0,quadgamma>0,quaddelta>0
$$

and
$$
f_Z(z)=frac{Gamma(alpha+beta)}{Gamma(alpha)Gamma(beta)}z^{alpha-1}(1-z)^{beta-1},quad 0leq zleq 1,quadalpha>0,quadbeta>0.
$$

Consider $X_1$ and $X_2$ having $operatorname{Gamma}(a+b,1)$ and $operatorname{Beta}(a,b)$ distributions, respectively, where $a,b>0$. Assume that $X_1$ and $X_2$ are independent.




How do i find the marginal density functions of $Y_1 = X_1X_2$ and $Y_2 = X_1(1-X_2)$?



I know that the marginal density function can be derived from the joint density, but since the joint is not given, how do I create it?



Also how do I manipulate the gamma function? first time I have come across it.










share|cite|improve this question











$endgroup$








  • 3




    $begingroup$
    Perhaps you meant how to find the joint density of $(Y_1,Y_2)$. Do you know change of variables?
    $endgroup$
    – StubbornAtom
    Dec 1 '18 at 10:55










  • $begingroup$
    @StubbornAtom thank you for your response. By change of variables you mean for integration?
    $endgroup$
    – OvermanZarathustra
    Dec 4 '18 at 0:00
















0












$begingroup$



If $Ysimoperatorname{Gamma}(gamma,delta)$ and $Zsimoperatorname{Beta}(alpha,beta)$ then their density functions are, respectively,
$$
f_Y(y)=frac{delta^gamma}{Gamma(gamma)}y^{gamma-1}e^{-delta y},quad y>0,quadgamma>0,quaddelta>0
$$

and
$$
f_Z(z)=frac{Gamma(alpha+beta)}{Gamma(alpha)Gamma(beta)}z^{alpha-1}(1-z)^{beta-1},quad 0leq zleq 1,quadalpha>0,quadbeta>0.
$$

Consider $X_1$ and $X_2$ having $operatorname{Gamma}(a+b,1)$ and $operatorname{Beta}(a,b)$ distributions, respectively, where $a,b>0$. Assume that $X_1$ and $X_2$ are independent.




How do i find the marginal density functions of $Y_1 = X_1X_2$ and $Y_2 = X_1(1-X_2)$?



I know that the marginal density function can be derived from the joint density, but since the joint is not given, how do I create it?



Also how do I manipulate the gamma function? first time I have come across it.










share|cite|improve this question











$endgroup$








  • 3




    $begingroup$
    Perhaps you meant how to find the joint density of $(Y_1,Y_2)$. Do you know change of variables?
    $endgroup$
    – StubbornAtom
    Dec 1 '18 at 10:55










  • $begingroup$
    @StubbornAtom thank you for your response. By change of variables you mean for integration?
    $endgroup$
    – OvermanZarathustra
    Dec 4 '18 at 0:00














0












0








0





$begingroup$



If $Ysimoperatorname{Gamma}(gamma,delta)$ and $Zsimoperatorname{Beta}(alpha,beta)$ then their density functions are, respectively,
$$
f_Y(y)=frac{delta^gamma}{Gamma(gamma)}y^{gamma-1}e^{-delta y},quad y>0,quadgamma>0,quaddelta>0
$$

and
$$
f_Z(z)=frac{Gamma(alpha+beta)}{Gamma(alpha)Gamma(beta)}z^{alpha-1}(1-z)^{beta-1},quad 0leq zleq 1,quadalpha>0,quadbeta>0.
$$

Consider $X_1$ and $X_2$ having $operatorname{Gamma}(a+b,1)$ and $operatorname{Beta}(a,b)$ distributions, respectively, where $a,b>0$. Assume that $X_1$ and $X_2$ are independent.




How do i find the marginal density functions of $Y_1 = X_1X_2$ and $Y_2 = X_1(1-X_2)$?



I know that the marginal density function can be derived from the joint density, but since the joint is not given, how do I create it?



Also how do I manipulate the gamma function? first time I have come across it.










share|cite|improve this question











$endgroup$





If $Ysimoperatorname{Gamma}(gamma,delta)$ and $Zsimoperatorname{Beta}(alpha,beta)$ then their density functions are, respectively,
$$
f_Y(y)=frac{delta^gamma}{Gamma(gamma)}y^{gamma-1}e^{-delta y},quad y>0,quadgamma>0,quaddelta>0
$$

and
$$
f_Z(z)=frac{Gamma(alpha+beta)}{Gamma(alpha)Gamma(beta)}z^{alpha-1}(1-z)^{beta-1},quad 0leq zleq 1,quadalpha>0,quadbeta>0.
$$

Consider $X_1$ and $X_2$ having $operatorname{Gamma}(a+b,1)$ and $operatorname{Beta}(a,b)$ distributions, respectively, where $a,b>0$. Assume that $X_1$ and $X_2$ are independent.




How do i find the marginal density functions of $Y_1 = X_1X_2$ and $Y_2 = X_1(1-X_2)$?



I know that the marginal density function can be derived from the joint density, but since the joint is not given, how do I create it?



Also how do I manipulate the gamma function? first time I have come across it.







probability probability-theory probability-distributions density-function






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share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 1 '18 at 11:57









user10354138

7,3772925




7,3772925










asked Dec 1 '18 at 10:44









OvermanZarathustraOvermanZarathustra

156




156








  • 3




    $begingroup$
    Perhaps you meant how to find the joint density of $(Y_1,Y_2)$. Do you know change of variables?
    $endgroup$
    – StubbornAtom
    Dec 1 '18 at 10:55










  • $begingroup$
    @StubbornAtom thank you for your response. By change of variables you mean for integration?
    $endgroup$
    – OvermanZarathustra
    Dec 4 '18 at 0:00














  • 3




    $begingroup$
    Perhaps you meant how to find the joint density of $(Y_1,Y_2)$. Do you know change of variables?
    $endgroup$
    – StubbornAtom
    Dec 1 '18 at 10:55










  • $begingroup$
    @StubbornAtom thank you for your response. By change of variables you mean for integration?
    $endgroup$
    – OvermanZarathustra
    Dec 4 '18 at 0:00








3




3




$begingroup$
Perhaps you meant how to find the joint density of $(Y_1,Y_2)$. Do you know change of variables?
$endgroup$
– StubbornAtom
Dec 1 '18 at 10:55




$begingroup$
Perhaps you meant how to find the joint density of $(Y_1,Y_2)$. Do you know change of variables?
$endgroup$
– StubbornAtom
Dec 1 '18 at 10:55












$begingroup$
@StubbornAtom thank you for your response. By change of variables you mean for integration?
$endgroup$
– OvermanZarathustra
Dec 4 '18 at 0:00




$begingroup$
@StubbornAtom thank you for your response. By change of variables you mean for integration?
$endgroup$
– OvermanZarathustra
Dec 4 '18 at 0:00










1 Answer
1






active

oldest

votes


















1












$begingroup$

Plugging in the definition, $X_1$ following $operatorname{Gamma}(a+b,1)$ means its density is



$$f_{X_1}(x_1) = frac1{ Gamma(a+b)}, x_1^{a+b-1} e^{-x_1} qquad text{for}~~ 0 < x_1 < infty $$



The density of $X_2$ is



$$f_{X_2}(x_2) = frac{ Gamma(a+b) }{ Gamma(a) Gamma(b) }, x_2^{a-1} (1 - x_2)^{b-1} qquad text{for}~~ 0<x_2<1$$



The fact that $X_1 perp X_2$ means their joint density is just the direct product



$$f_{X_1X_2}(x_1,, x_2) = frac1{ Gamma(a) Gamma(b) }, x_1^{a+b-1} e^{-x_1} , x_2^{a-1} (1 - x_2)^{b-1} qquad text{for}~~ begin{cases}
0<x_1<infty \
0<x_2<1 end{cases} tag*{Eq.(1)}$$



The 2-dim transformation is
$$begin{cases}
Y_1 = X_1 X_2 \
\
Y_2 = X_1 (1 - X_2)
end{cases} Longleftrightarrow begin{cases}
X_1 = Y_1 + Y_2 \
\
X_2 = dfrac{ Y_1 }{ Y_1 + Y_2}
end{cases} qquad text{where}~~ begin{cases}
0<y_1<infty \
0<y_2<infty end{cases}$$

with the Jacobian (of the inverse mapping) as
$$J = left| begin{matrix} dfrac{ partial x_1}{ partial y_1} & dfrac{ partial x_1}{ partial y_2} \
dfrac{ partial x_2}{ partial y_1} & dfrac{ partial x_2}{ partial y_2}end{matrix} right| = left| begin{matrix} 1 & 1 \
dfrac{ y_2 }{ (y_1 +y_2)^2 } & dfrac{ -y_1 }{ (y_1 +y_2)^2 } end{matrix} right| = frac{-1}{ y_1 + y_2 }$$

The transformed joint density for $Y_1$ and $Y_2$ is
begin{align}
f_{Y_1Y_2}( y_1 ,~y_2 ) &= |J| cdot f_{X_1X_2}( x_1,, x_2)Bigg|_{x_1 = y_1+y_2, ,,x_2 = frac{y_1}{y_1 + y_2}} qquad text{, plug in Eq.(1)}\
&= frac1{ y_1 + y_2} cdot frac1{ Gamma(a) Gamma(b) }, (y_1 + y_2)^{a+b-1} e^{-(y_1 + y_2)} , left(frac{y_1}{ y_1 + y_2}right)^{a-1} left(frac{y_2}{ y_1 + y_2} right)^{b-1} \
&= frac1{ Gamma(a) Gamma(b) }, y_1^{a-1} y_2^{b-1} e^{-(y_1 + y_2)} qquad text{for}~~0<y_1<infty ,~0<y_2<infty
end{align}



The marginal density of $Y_1$ can be obtained from the joint as
begin{align}
f_{Y_1}(y_1) &= int_{y_2 = 0}^{infty} f_{Y_1Y_2}( y_1 ,~y_2 ) ,mathrm{d}y_2 \
&= frac1{Gamma(a)} y_1^{a-1} e^{-y_1} int_{y_2 = 0}^{infty} frac1{Gamma(b)} y_2^{b-1} e^{-y_2} ,mathrm{d}y_2 qquad scriptsizetext{integral is just the kernel of Gamma distribution} \
&= frac1{Gamma(a)} y_1^{a-1} e^{-y_1}
end{align}

Thus one identifies the distribution of $Y_1$ as $operatorname{Gamma}(a,1)$.



Similarly, or noting the symmetry in the joint $f_{Y_1Y_2}( y_1 ,~y_2 )$, we have $Y_2$ follows $operatorname{Gamma}(b,1)$.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Thank you very much for your response! Is there another way to solve this question? I don't think i understand the Jacobian method.
    $endgroup$
    – OvermanZarathustra
    Dec 3 '18 at 18:23






  • 1




    $begingroup$
    It is a well-known basic fact that Beta distribution can be viewed as $frac{U}{U+W}$ where $U$ and $W$ are independent and follow Gamma distributions (of the same rate/scale parameter). See this or the 10th item of this. The question is just a symbolic (verbal) argument of the inverse of this definition. However, this way of quoting "known results" is hardly what the question seems to be aiming for.
    $endgroup$
    – Lee David Chung Lin
    Dec 4 '18 at 11:01






  • 1




    $begingroup$
    The question statement basically gives you the joint $f_{X1X2}(x_1,x_2)$ and the intention is most likely asking you to do some calculus one way or the other. For example, you can do the CDF of $Y_1$ as $Pr{Y_1 < y_1 } = Pr{X_1 X_2 < y_1 }$, which is a 2-dim integration across the relevant region (bounded by the hyperbola $x_1 x_2 = y_1$ and $x_2 = 0$, $x_2 = 1$, along with $x_1 = 0$) over the joint density $f_{X_1X_2}(x_1,x_2)$. I doubt you'll find this easier, but if you'd like to see it I can make another answer post.
    $endgroup$
    – Lee David Chung Lin
    Dec 4 '18 at 11:06






  • 1




    $begingroup$
    Meanwhile, originally you posted the question with a snapshot. Which textbook is it? It's rather unlikely that the method of variable transformation (1-dim and 2-dim) is not covered, even if it's only a half-decent material.
    $endgroup$
    – Lee David Chung Lin
    Dec 4 '18 at 11:38











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1 Answer
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1 Answer
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active

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active

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active

oldest

votes









1












$begingroup$

Plugging in the definition, $X_1$ following $operatorname{Gamma}(a+b,1)$ means its density is



$$f_{X_1}(x_1) = frac1{ Gamma(a+b)}, x_1^{a+b-1} e^{-x_1} qquad text{for}~~ 0 < x_1 < infty $$



The density of $X_2$ is



$$f_{X_2}(x_2) = frac{ Gamma(a+b) }{ Gamma(a) Gamma(b) }, x_2^{a-1} (1 - x_2)^{b-1} qquad text{for}~~ 0<x_2<1$$



The fact that $X_1 perp X_2$ means their joint density is just the direct product



$$f_{X_1X_2}(x_1,, x_2) = frac1{ Gamma(a) Gamma(b) }, x_1^{a+b-1} e^{-x_1} , x_2^{a-1} (1 - x_2)^{b-1} qquad text{for}~~ begin{cases}
0<x_1<infty \
0<x_2<1 end{cases} tag*{Eq.(1)}$$



The 2-dim transformation is
$$begin{cases}
Y_1 = X_1 X_2 \
\
Y_2 = X_1 (1 - X_2)
end{cases} Longleftrightarrow begin{cases}
X_1 = Y_1 + Y_2 \
\
X_2 = dfrac{ Y_1 }{ Y_1 + Y_2}
end{cases} qquad text{where}~~ begin{cases}
0<y_1<infty \
0<y_2<infty end{cases}$$

with the Jacobian (of the inverse mapping) as
$$J = left| begin{matrix} dfrac{ partial x_1}{ partial y_1} & dfrac{ partial x_1}{ partial y_2} \
dfrac{ partial x_2}{ partial y_1} & dfrac{ partial x_2}{ partial y_2}end{matrix} right| = left| begin{matrix} 1 & 1 \
dfrac{ y_2 }{ (y_1 +y_2)^2 } & dfrac{ -y_1 }{ (y_1 +y_2)^2 } end{matrix} right| = frac{-1}{ y_1 + y_2 }$$

The transformed joint density for $Y_1$ and $Y_2$ is
begin{align}
f_{Y_1Y_2}( y_1 ,~y_2 ) &= |J| cdot f_{X_1X_2}( x_1,, x_2)Bigg|_{x_1 = y_1+y_2, ,,x_2 = frac{y_1}{y_1 + y_2}} qquad text{, plug in Eq.(1)}\
&= frac1{ y_1 + y_2} cdot frac1{ Gamma(a) Gamma(b) }, (y_1 + y_2)^{a+b-1} e^{-(y_1 + y_2)} , left(frac{y_1}{ y_1 + y_2}right)^{a-1} left(frac{y_2}{ y_1 + y_2} right)^{b-1} \
&= frac1{ Gamma(a) Gamma(b) }, y_1^{a-1} y_2^{b-1} e^{-(y_1 + y_2)} qquad text{for}~~0<y_1<infty ,~0<y_2<infty
end{align}



The marginal density of $Y_1$ can be obtained from the joint as
begin{align}
f_{Y_1}(y_1) &= int_{y_2 = 0}^{infty} f_{Y_1Y_2}( y_1 ,~y_2 ) ,mathrm{d}y_2 \
&= frac1{Gamma(a)} y_1^{a-1} e^{-y_1} int_{y_2 = 0}^{infty} frac1{Gamma(b)} y_2^{b-1} e^{-y_2} ,mathrm{d}y_2 qquad scriptsizetext{integral is just the kernel of Gamma distribution} \
&= frac1{Gamma(a)} y_1^{a-1} e^{-y_1}
end{align}

Thus one identifies the distribution of $Y_1$ as $operatorname{Gamma}(a,1)$.



Similarly, or noting the symmetry in the joint $f_{Y_1Y_2}( y_1 ,~y_2 )$, we have $Y_2$ follows $operatorname{Gamma}(b,1)$.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Thank you very much for your response! Is there another way to solve this question? I don't think i understand the Jacobian method.
    $endgroup$
    – OvermanZarathustra
    Dec 3 '18 at 18:23






  • 1




    $begingroup$
    It is a well-known basic fact that Beta distribution can be viewed as $frac{U}{U+W}$ where $U$ and $W$ are independent and follow Gamma distributions (of the same rate/scale parameter). See this or the 10th item of this. The question is just a symbolic (verbal) argument of the inverse of this definition. However, this way of quoting "known results" is hardly what the question seems to be aiming for.
    $endgroup$
    – Lee David Chung Lin
    Dec 4 '18 at 11:01






  • 1




    $begingroup$
    The question statement basically gives you the joint $f_{X1X2}(x_1,x_2)$ and the intention is most likely asking you to do some calculus one way or the other. For example, you can do the CDF of $Y_1$ as $Pr{Y_1 < y_1 } = Pr{X_1 X_2 < y_1 }$, which is a 2-dim integration across the relevant region (bounded by the hyperbola $x_1 x_2 = y_1$ and $x_2 = 0$, $x_2 = 1$, along with $x_1 = 0$) over the joint density $f_{X_1X_2}(x_1,x_2)$. I doubt you'll find this easier, but if you'd like to see it I can make another answer post.
    $endgroup$
    – Lee David Chung Lin
    Dec 4 '18 at 11:06






  • 1




    $begingroup$
    Meanwhile, originally you posted the question with a snapshot. Which textbook is it? It's rather unlikely that the method of variable transformation (1-dim and 2-dim) is not covered, even if it's only a half-decent material.
    $endgroup$
    – Lee David Chung Lin
    Dec 4 '18 at 11:38
















1












$begingroup$

Plugging in the definition, $X_1$ following $operatorname{Gamma}(a+b,1)$ means its density is



$$f_{X_1}(x_1) = frac1{ Gamma(a+b)}, x_1^{a+b-1} e^{-x_1} qquad text{for}~~ 0 < x_1 < infty $$



The density of $X_2$ is



$$f_{X_2}(x_2) = frac{ Gamma(a+b) }{ Gamma(a) Gamma(b) }, x_2^{a-1} (1 - x_2)^{b-1} qquad text{for}~~ 0<x_2<1$$



The fact that $X_1 perp X_2$ means their joint density is just the direct product



$$f_{X_1X_2}(x_1,, x_2) = frac1{ Gamma(a) Gamma(b) }, x_1^{a+b-1} e^{-x_1} , x_2^{a-1} (1 - x_2)^{b-1} qquad text{for}~~ begin{cases}
0<x_1<infty \
0<x_2<1 end{cases} tag*{Eq.(1)}$$



The 2-dim transformation is
$$begin{cases}
Y_1 = X_1 X_2 \
\
Y_2 = X_1 (1 - X_2)
end{cases} Longleftrightarrow begin{cases}
X_1 = Y_1 + Y_2 \
\
X_2 = dfrac{ Y_1 }{ Y_1 + Y_2}
end{cases} qquad text{where}~~ begin{cases}
0<y_1<infty \
0<y_2<infty end{cases}$$

with the Jacobian (of the inverse mapping) as
$$J = left| begin{matrix} dfrac{ partial x_1}{ partial y_1} & dfrac{ partial x_1}{ partial y_2} \
dfrac{ partial x_2}{ partial y_1} & dfrac{ partial x_2}{ partial y_2}end{matrix} right| = left| begin{matrix} 1 & 1 \
dfrac{ y_2 }{ (y_1 +y_2)^2 } & dfrac{ -y_1 }{ (y_1 +y_2)^2 } end{matrix} right| = frac{-1}{ y_1 + y_2 }$$

The transformed joint density for $Y_1$ and $Y_2$ is
begin{align}
f_{Y_1Y_2}( y_1 ,~y_2 ) &= |J| cdot f_{X_1X_2}( x_1,, x_2)Bigg|_{x_1 = y_1+y_2, ,,x_2 = frac{y_1}{y_1 + y_2}} qquad text{, plug in Eq.(1)}\
&= frac1{ y_1 + y_2} cdot frac1{ Gamma(a) Gamma(b) }, (y_1 + y_2)^{a+b-1} e^{-(y_1 + y_2)} , left(frac{y_1}{ y_1 + y_2}right)^{a-1} left(frac{y_2}{ y_1 + y_2} right)^{b-1} \
&= frac1{ Gamma(a) Gamma(b) }, y_1^{a-1} y_2^{b-1} e^{-(y_1 + y_2)} qquad text{for}~~0<y_1<infty ,~0<y_2<infty
end{align}



The marginal density of $Y_1$ can be obtained from the joint as
begin{align}
f_{Y_1}(y_1) &= int_{y_2 = 0}^{infty} f_{Y_1Y_2}( y_1 ,~y_2 ) ,mathrm{d}y_2 \
&= frac1{Gamma(a)} y_1^{a-1} e^{-y_1} int_{y_2 = 0}^{infty} frac1{Gamma(b)} y_2^{b-1} e^{-y_2} ,mathrm{d}y_2 qquad scriptsizetext{integral is just the kernel of Gamma distribution} \
&= frac1{Gamma(a)} y_1^{a-1} e^{-y_1}
end{align}

Thus one identifies the distribution of $Y_1$ as $operatorname{Gamma}(a,1)$.



Similarly, or noting the symmetry in the joint $f_{Y_1Y_2}( y_1 ,~y_2 )$, we have $Y_2$ follows $operatorname{Gamma}(b,1)$.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Thank you very much for your response! Is there another way to solve this question? I don't think i understand the Jacobian method.
    $endgroup$
    – OvermanZarathustra
    Dec 3 '18 at 18:23






  • 1




    $begingroup$
    It is a well-known basic fact that Beta distribution can be viewed as $frac{U}{U+W}$ where $U$ and $W$ are independent and follow Gamma distributions (of the same rate/scale parameter). See this or the 10th item of this. The question is just a symbolic (verbal) argument of the inverse of this definition. However, this way of quoting "known results" is hardly what the question seems to be aiming for.
    $endgroup$
    – Lee David Chung Lin
    Dec 4 '18 at 11:01






  • 1




    $begingroup$
    The question statement basically gives you the joint $f_{X1X2}(x_1,x_2)$ and the intention is most likely asking you to do some calculus one way or the other. For example, you can do the CDF of $Y_1$ as $Pr{Y_1 < y_1 } = Pr{X_1 X_2 < y_1 }$, which is a 2-dim integration across the relevant region (bounded by the hyperbola $x_1 x_2 = y_1$ and $x_2 = 0$, $x_2 = 1$, along with $x_1 = 0$) over the joint density $f_{X_1X_2}(x_1,x_2)$. I doubt you'll find this easier, but if you'd like to see it I can make another answer post.
    $endgroup$
    – Lee David Chung Lin
    Dec 4 '18 at 11:06






  • 1




    $begingroup$
    Meanwhile, originally you posted the question with a snapshot. Which textbook is it? It's rather unlikely that the method of variable transformation (1-dim and 2-dim) is not covered, even if it's only a half-decent material.
    $endgroup$
    – Lee David Chung Lin
    Dec 4 '18 at 11:38














1












1








1





$begingroup$

Plugging in the definition, $X_1$ following $operatorname{Gamma}(a+b,1)$ means its density is



$$f_{X_1}(x_1) = frac1{ Gamma(a+b)}, x_1^{a+b-1} e^{-x_1} qquad text{for}~~ 0 < x_1 < infty $$



The density of $X_2$ is



$$f_{X_2}(x_2) = frac{ Gamma(a+b) }{ Gamma(a) Gamma(b) }, x_2^{a-1} (1 - x_2)^{b-1} qquad text{for}~~ 0<x_2<1$$



The fact that $X_1 perp X_2$ means their joint density is just the direct product



$$f_{X_1X_2}(x_1,, x_2) = frac1{ Gamma(a) Gamma(b) }, x_1^{a+b-1} e^{-x_1} , x_2^{a-1} (1 - x_2)^{b-1} qquad text{for}~~ begin{cases}
0<x_1<infty \
0<x_2<1 end{cases} tag*{Eq.(1)}$$



The 2-dim transformation is
$$begin{cases}
Y_1 = X_1 X_2 \
\
Y_2 = X_1 (1 - X_2)
end{cases} Longleftrightarrow begin{cases}
X_1 = Y_1 + Y_2 \
\
X_2 = dfrac{ Y_1 }{ Y_1 + Y_2}
end{cases} qquad text{where}~~ begin{cases}
0<y_1<infty \
0<y_2<infty end{cases}$$

with the Jacobian (of the inverse mapping) as
$$J = left| begin{matrix} dfrac{ partial x_1}{ partial y_1} & dfrac{ partial x_1}{ partial y_2} \
dfrac{ partial x_2}{ partial y_1} & dfrac{ partial x_2}{ partial y_2}end{matrix} right| = left| begin{matrix} 1 & 1 \
dfrac{ y_2 }{ (y_1 +y_2)^2 } & dfrac{ -y_1 }{ (y_1 +y_2)^2 } end{matrix} right| = frac{-1}{ y_1 + y_2 }$$

The transformed joint density for $Y_1$ and $Y_2$ is
begin{align}
f_{Y_1Y_2}( y_1 ,~y_2 ) &= |J| cdot f_{X_1X_2}( x_1,, x_2)Bigg|_{x_1 = y_1+y_2, ,,x_2 = frac{y_1}{y_1 + y_2}} qquad text{, plug in Eq.(1)}\
&= frac1{ y_1 + y_2} cdot frac1{ Gamma(a) Gamma(b) }, (y_1 + y_2)^{a+b-1} e^{-(y_1 + y_2)} , left(frac{y_1}{ y_1 + y_2}right)^{a-1} left(frac{y_2}{ y_1 + y_2} right)^{b-1} \
&= frac1{ Gamma(a) Gamma(b) }, y_1^{a-1} y_2^{b-1} e^{-(y_1 + y_2)} qquad text{for}~~0<y_1<infty ,~0<y_2<infty
end{align}



The marginal density of $Y_1$ can be obtained from the joint as
begin{align}
f_{Y_1}(y_1) &= int_{y_2 = 0}^{infty} f_{Y_1Y_2}( y_1 ,~y_2 ) ,mathrm{d}y_2 \
&= frac1{Gamma(a)} y_1^{a-1} e^{-y_1} int_{y_2 = 0}^{infty} frac1{Gamma(b)} y_2^{b-1} e^{-y_2} ,mathrm{d}y_2 qquad scriptsizetext{integral is just the kernel of Gamma distribution} \
&= frac1{Gamma(a)} y_1^{a-1} e^{-y_1}
end{align}

Thus one identifies the distribution of $Y_1$ as $operatorname{Gamma}(a,1)$.



Similarly, or noting the symmetry in the joint $f_{Y_1Y_2}( y_1 ,~y_2 )$, we have $Y_2$ follows $operatorname{Gamma}(b,1)$.






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$endgroup$



Plugging in the definition, $X_1$ following $operatorname{Gamma}(a+b,1)$ means its density is



$$f_{X_1}(x_1) = frac1{ Gamma(a+b)}, x_1^{a+b-1} e^{-x_1} qquad text{for}~~ 0 < x_1 < infty $$



The density of $X_2$ is



$$f_{X_2}(x_2) = frac{ Gamma(a+b) }{ Gamma(a) Gamma(b) }, x_2^{a-1} (1 - x_2)^{b-1} qquad text{for}~~ 0<x_2<1$$



The fact that $X_1 perp X_2$ means their joint density is just the direct product



$$f_{X_1X_2}(x_1,, x_2) = frac1{ Gamma(a) Gamma(b) }, x_1^{a+b-1} e^{-x_1} , x_2^{a-1} (1 - x_2)^{b-1} qquad text{for}~~ begin{cases}
0<x_1<infty \
0<x_2<1 end{cases} tag*{Eq.(1)}$$



The 2-dim transformation is
$$begin{cases}
Y_1 = X_1 X_2 \
\
Y_2 = X_1 (1 - X_2)
end{cases} Longleftrightarrow begin{cases}
X_1 = Y_1 + Y_2 \
\
X_2 = dfrac{ Y_1 }{ Y_1 + Y_2}
end{cases} qquad text{where}~~ begin{cases}
0<y_1<infty \
0<y_2<infty end{cases}$$

with the Jacobian (of the inverse mapping) as
$$J = left| begin{matrix} dfrac{ partial x_1}{ partial y_1} & dfrac{ partial x_1}{ partial y_2} \
dfrac{ partial x_2}{ partial y_1} & dfrac{ partial x_2}{ partial y_2}end{matrix} right| = left| begin{matrix} 1 & 1 \
dfrac{ y_2 }{ (y_1 +y_2)^2 } & dfrac{ -y_1 }{ (y_1 +y_2)^2 } end{matrix} right| = frac{-1}{ y_1 + y_2 }$$

The transformed joint density for $Y_1$ and $Y_2$ is
begin{align}
f_{Y_1Y_2}( y_1 ,~y_2 ) &= |J| cdot f_{X_1X_2}( x_1,, x_2)Bigg|_{x_1 = y_1+y_2, ,,x_2 = frac{y_1}{y_1 + y_2}} qquad text{, plug in Eq.(1)}\
&= frac1{ y_1 + y_2} cdot frac1{ Gamma(a) Gamma(b) }, (y_1 + y_2)^{a+b-1} e^{-(y_1 + y_2)} , left(frac{y_1}{ y_1 + y_2}right)^{a-1} left(frac{y_2}{ y_1 + y_2} right)^{b-1} \
&= frac1{ Gamma(a) Gamma(b) }, y_1^{a-1} y_2^{b-1} e^{-(y_1 + y_2)} qquad text{for}~~0<y_1<infty ,~0<y_2<infty
end{align}



The marginal density of $Y_1$ can be obtained from the joint as
begin{align}
f_{Y_1}(y_1) &= int_{y_2 = 0}^{infty} f_{Y_1Y_2}( y_1 ,~y_2 ) ,mathrm{d}y_2 \
&= frac1{Gamma(a)} y_1^{a-1} e^{-y_1} int_{y_2 = 0}^{infty} frac1{Gamma(b)} y_2^{b-1} e^{-y_2} ,mathrm{d}y_2 qquad scriptsizetext{integral is just the kernel of Gamma distribution} \
&= frac1{Gamma(a)} y_1^{a-1} e^{-y_1}
end{align}

Thus one identifies the distribution of $Y_1$ as $operatorname{Gamma}(a,1)$.



Similarly, or noting the symmetry in the joint $f_{Y_1Y_2}( y_1 ,~y_2 )$, we have $Y_2$ follows $operatorname{Gamma}(b,1)$.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Dec 1 '18 at 15:17









Lee David Chung LinLee David Chung Lin

3,88531140




3,88531140












  • $begingroup$
    Thank you very much for your response! Is there another way to solve this question? I don't think i understand the Jacobian method.
    $endgroup$
    – OvermanZarathustra
    Dec 3 '18 at 18:23






  • 1




    $begingroup$
    It is a well-known basic fact that Beta distribution can be viewed as $frac{U}{U+W}$ where $U$ and $W$ are independent and follow Gamma distributions (of the same rate/scale parameter). See this or the 10th item of this. The question is just a symbolic (verbal) argument of the inverse of this definition. However, this way of quoting "known results" is hardly what the question seems to be aiming for.
    $endgroup$
    – Lee David Chung Lin
    Dec 4 '18 at 11:01






  • 1




    $begingroup$
    The question statement basically gives you the joint $f_{X1X2}(x_1,x_2)$ and the intention is most likely asking you to do some calculus one way or the other. For example, you can do the CDF of $Y_1$ as $Pr{Y_1 < y_1 } = Pr{X_1 X_2 < y_1 }$, which is a 2-dim integration across the relevant region (bounded by the hyperbola $x_1 x_2 = y_1$ and $x_2 = 0$, $x_2 = 1$, along with $x_1 = 0$) over the joint density $f_{X_1X_2}(x_1,x_2)$. I doubt you'll find this easier, but if you'd like to see it I can make another answer post.
    $endgroup$
    – Lee David Chung Lin
    Dec 4 '18 at 11:06






  • 1




    $begingroup$
    Meanwhile, originally you posted the question with a snapshot. Which textbook is it? It's rather unlikely that the method of variable transformation (1-dim and 2-dim) is not covered, even if it's only a half-decent material.
    $endgroup$
    – Lee David Chung Lin
    Dec 4 '18 at 11:38


















  • $begingroup$
    Thank you very much for your response! Is there another way to solve this question? I don't think i understand the Jacobian method.
    $endgroup$
    – OvermanZarathustra
    Dec 3 '18 at 18:23






  • 1




    $begingroup$
    It is a well-known basic fact that Beta distribution can be viewed as $frac{U}{U+W}$ where $U$ and $W$ are independent and follow Gamma distributions (of the same rate/scale parameter). See this or the 10th item of this. The question is just a symbolic (verbal) argument of the inverse of this definition. However, this way of quoting "known results" is hardly what the question seems to be aiming for.
    $endgroup$
    – Lee David Chung Lin
    Dec 4 '18 at 11:01






  • 1




    $begingroup$
    The question statement basically gives you the joint $f_{X1X2}(x_1,x_2)$ and the intention is most likely asking you to do some calculus one way or the other. For example, you can do the CDF of $Y_1$ as $Pr{Y_1 < y_1 } = Pr{X_1 X_2 < y_1 }$, which is a 2-dim integration across the relevant region (bounded by the hyperbola $x_1 x_2 = y_1$ and $x_2 = 0$, $x_2 = 1$, along with $x_1 = 0$) over the joint density $f_{X_1X_2}(x_1,x_2)$. I doubt you'll find this easier, but if you'd like to see it I can make another answer post.
    $endgroup$
    – Lee David Chung Lin
    Dec 4 '18 at 11:06






  • 1




    $begingroup$
    Meanwhile, originally you posted the question with a snapshot. Which textbook is it? It's rather unlikely that the method of variable transformation (1-dim and 2-dim) is not covered, even if it's only a half-decent material.
    $endgroup$
    – Lee David Chung Lin
    Dec 4 '18 at 11:38
















$begingroup$
Thank you very much for your response! Is there another way to solve this question? I don't think i understand the Jacobian method.
$endgroup$
– OvermanZarathustra
Dec 3 '18 at 18:23




$begingroup$
Thank you very much for your response! Is there another way to solve this question? I don't think i understand the Jacobian method.
$endgroup$
– OvermanZarathustra
Dec 3 '18 at 18:23




1




1




$begingroup$
It is a well-known basic fact that Beta distribution can be viewed as $frac{U}{U+W}$ where $U$ and $W$ are independent and follow Gamma distributions (of the same rate/scale parameter). See this or the 10th item of this. The question is just a symbolic (verbal) argument of the inverse of this definition. However, this way of quoting "known results" is hardly what the question seems to be aiming for.
$endgroup$
– Lee David Chung Lin
Dec 4 '18 at 11:01




$begingroup$
It is a well-known basic fact that Beta distribution can be viewed as $frac{U}{U+W}$ where $U$ and $W$ are independent and follow Gamma distributions (of the same rate/scale parameter). See this or the 10th item of this. The question is just a symbolic (verbal) argument of the inverse of this definition. However, this way of quoting "known results" is hardly what the question seems to be aiming for.
$endgroup$
– Lee David Chung Lin
Dec 4 '18 at 11:01




1




1




$begingroup$
The question statement basically gives you the joint $f_{X1X2}(x_1,x_2)$ and the intention is most likely asking you to do some calculus one way or the other. For example, you can do the CDF of $Y_1$ as $Pr{Y_1 < y_1 } = Pr{X_1 X_2 < y_1 }$, which is a 2-dim integration across the relevant region (bounded by the hyperbola $x_1 x_2 = y_1$ and $x_2 = 0$, $x_2 = 1$, along with $x_1 = 0$) over the joint density $f_{X_1X_2}(x_1,x_2)$. I doubt you'll find this easier, but if you'd like to see it I can make another answer post.
$endgroup$
– Lee David Chung Lin
Dec 4 '18 at 11:06




$begingroup$
The question statement basically gives you the joint $f_{X1X2}(x_1,x_2)$ and the intention is most likely asking you to do some calculus one way or the other. For example, you can do the CDF of $Y_1$ as $Pr{Y_1 < y_1 } = Pr{X_1 X_2 < y_1 }$, which is a 2-dim integration across the relevant region (bounded by the hyperbola $x_1 x_2 = y_1$ and $x_2 = 0$, $x_2 = 1$, along with $x_1 = 0$) over the joint density $f_{X_1X_2}(x_1,x_2)$. I doubt you'll find this easier, but if you'd like to see it I can make another answer post.
$endgroup$
– Lee David Chung Lin
Dec 4 '18 at 11:06




1




1




$begingroup$
Meanwhile, originally you posted the question with a snapshot. Which textbook is it? It's rather unlikely that the method of variable transformation (1-dim and 2-dim) is not covered, even if it's only a half-decent material.
$endgroup$
– Lee David Chung Lin
Dec 4 '18 at 11:38




$begingroup$
Meanwhile, originally you posted the question with a snapshot. Which textbook is it? It's rather unlikely that the method of variable transformation (1-dim and 2-dim) is not covered, even if it's only a half-decent material.
$endgroup$
– Lee David Chung Lin
Dec 4 '18 at 11:38


















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