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If $Y\sim\operatorname{Gamma}(\gamma,\delta)$ and $Z\sim\operatorname{Beta}(\alpha,\beta)$ then their density functions are, respectively, $$ f_Y(y)=\frac{\delta^\gamma}{\Gamma(\gamma)}y^{\gamma-1}e^{-\delta y},\quad y>0,\quad\gamma>0,\quad\delta>0 $$ and $$ f_Z(z)=\frac{\Gamma(\alpha+\beta)}{\Gamma(\alpha)\Gamma(\beta)}z^{\alpha-1}(1-z)^{\beta-1},\quad 0\leq z\leq 1,\quad\alpha>0,\quad\beta>0. $$ 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.

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    $\begingroup$ 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
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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 \scriptsize\text{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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  • $\begingroup$ 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
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    $\begingroup$ 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
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    $\begingroup$ 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
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    $\begingroup$ 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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