Question

2. LetX-Gamma(α = 2, β = 4), Y-Gam ma (α = 3, β = 4), X & Y are independent, Z1 = , Z,-X + Y a) (3 pts) State the joint pdf ofX and Y. Simplify the expression, clearing all Г's. b) (9 pts) Find the joint pdf of Zi and Zz, using the two variable transformation method. In addition, clearly write the support for this joint pdf. When done, your answer should include the expression

Answer 1

$(a)~The~joint~distribution~of~X~and~Y~is\\ f_{X,Y}(x,y)=\frac{1}{4^2\Gamma(2)}\exp(-x/4)x^{2-1}\times \frac{1}{4^3\Gamma(3)}\exp(-y/4)y^{3-1},~since~X~and~Y~are~independent\\ = \frac{1}{2048}\exp(-(x&plus;y)/4)xy^2,~0<x,y<\infty\\ =0~otherwise\\ (b)~Transform~(X,Y)\rightarrow (Z_1,Z_2)~where\\ z_1=\frac{x}{x&plus;y},~z_2=x&plus;y\Rightarrow x=z_1z_2,~y=z_2(1-z_1)\\ \begin{vmatrix} J\left(\frac{x,y}{z_1,z_2} \right ) \end{vmatrix}=\begin{vmatrix} \frac{\partial x }{\partial z_1} &\frac{\partial y }{\partial z_1} \\ \frac{\partial x }{\partial z_2} &\frac{\partial y }{\partial z_2} \end{vmatrix}=\begin{vmatrix} z_2 & -z_2\\ z_1&1-z_1 \end{vmatrix}=z_2(1-z_1)&plus;z_1z_2=z_2,\\ ~|Jacobian|=z_2,~x,y>0\Rightarrow 0<z_1<1,~z_2>0\\ The~joint~distribution~of~Z_1~and~Z_2~is\\ f_{Z_1,Z_2}(z_1,z_2)=\frac{1}{2048}z_1(1-z_1)^2z_2^4\exp(-z_2/4),~0<z_1<1,~z_2>0\\ =0~otherwise\\$

$(c)~ f_{Z_1,Z_2}(z_1,z_2)=\frac{1}{2048}z_1(1-z_1)^2 z_2^4\exp(-z_2/4)dz_2\\ =\left(\frac{1}{\frac{\Gamma(2)\Gamma(3)}{\Gamma(5)}}z_1(1-z_1)^2 \right )\times \left(\frac{\Gamma(5)}{4^5}\exp(-z_2/4)z_2^{5-1} \right )\\ =\left(\frac{1}{B(2,3)}z_1(1-z_1)^2 \right ) \left(\frac{\Gamma(5)}{4^5}\exp(-z_2/4)z_2^{5-1} \right );~0<z_1<1,~z_2>0\\ =f_{Z_1}(z_1)f_{Z_2}(z_2)\\ where~marginal~pdf~of~Z_1~is\\ f_{Z_1}(z_1)=\frac{1}{B(2,3)}z_1(1-z_1)^2,~0<z_1<1\\ =0~otherwise\\ Therefore~Z_1\sim Beta(2,3).\\ f_{Z_2}(z_2)=\frac{\Gamma(5)}{4^5}\exp(-z_2/4)z_2^{5-1};~0<z_2<\infty\\ =0~otherwise\\ Hence~Z_2\sim Gamma(\alpha=5,\beta=4).\\ Since~Joint~distribution~of~Z_1~and~Z_2~is~written~as~a~product~of~\\ marginal~distributions~of~Z_1~and~Z_2~so~Z_1~and~Z_2~are~independent.$