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Ted123
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Homework Statement
Let [itex]X_1[/itex] and [itex]X_2[/itex] be independent [itex]\text{Exp}(\lambda)[/itex] random variables.
(i) Find the joint density of [itex]Y_1 = X_1 + X_2[/itex] and [itex]\displaystyle Y_2 = \frac{X_1}{X_2}[/itex]
(ii) Show that [itex]Y_1[/itex] and [itex]Y_2[/itex] are independent.
The Attempt at a Solution
(i) By independence of [itex]X_1[/itex] and [itex]X_2[/itex]
[itex]\begin{displaymath} f_{X_1,X_2} (x_1,x_2) = f_{X_1} (x_1) f_{X_2} (x_2) = \left\{ \begin{array}{lr}
\lambda ^2 e^{-\lambda (x_1 + x_2)} & : (x_1,x_2) \in S \\
0 & : \text{o/w}\;\;\;\;\;\;\;\;\;\;\;\,
\end{array}
\right.
\end{displaymath}[/itex]
where [itex]S=[0,\infty )\times [0,\infty )[/itex]
[itex]y_1 = x_1 +x_2[/itex]
[itex]\displaystyle y_2 = \frac{x_1}{x_2}[/itex]
Inverting the transformation
[itex]\displaystyle x_1 = \frac{y_1 y_2}{y_2 +1}[/itex]
[itex]\displaystyle x_2 = \frac{y_1}{y_2 +1}[/itex]
for [itex](y_1 , y_2)\in T[/itex] where [itex]T=[0,\infty )\times [0,\infty )[/itex]
[itex]J=\begin{vmatrix} \displaystyle \frac{\partial x_1}{\partial y_1} & \displaystyle \frac{\partial x_1}{\partial y_2} \\ \displaystyle \frac{\partial x_2}{\partial y_1} & \displaystyle \frac{\partial x_2}{\partial y_2} \end{vmatrix}[/itex]
[itex]J=\begin{vmatrix} \displaystyle \frac{y_2}{y_2 +1} & \displaystyle \left(\frac{y_1}{y_2 +1}-\frac{y_1 y_2}{(y_2 +1)^2} \right) \\ \displaystyle\frac{1}{y_2 +1} & \displaystyle -\frac{y_1}{(y_2 +1)^2} \end{vmatrix} = \displaystyle -\frac{y_1}{(y_2 +1)^2}[/itex]
[itex]\begin{displaymath} f_{Y_1,Y_2} (y_1,y_2) = \left\{ \begin{array}{lr}
\displaystyle f_{X_1,X_2} \left(\frac{y_1 y_2}{y_2 +1} , \frac{y_1}{y_2 +1}\right) |J| & : (x_1,x_2) \in T \\
0 & : \text{o/w}\;\;\;\;\;\;\;\;\;\;\;\;
\end{array}
\right.
\end{displaymath}[/itex]
[itex]\begin{displaymath} f_{Y_1,Y_2} (y_1,y_2) = \left\{ \begin{array}{lr}
\displaystyle \lambda ^2 e^{-\lambda y_1} \frac{y_1}{(y_2 +1)^2} & : (x_1,x_2) \in T \\
0 & : \text{o/w}\;\;\;\;\;\;\;\;\;\;\;\;
\end{array}
\right.
\end{displaymath}[/itex]
(ii) To determine independence, calculate marginals.
[itex]\begin{displaymath} f_{Y_1} (y_1) = \left\{ \begin{array}{lr}
\displaystyle y_1 \lambda ^2 e^{-\lambda y_1} \int^{\infty}_0 \frac{1}{(y_2 +1)^2}\;dy_2 & : y_1 \geq 0 \\
0 & : \text{o/w}\;\;\;\,
\end{array}
\right.
\end{displaymath}[/itex]
[itex]\begin{displaymath} f_{Y_1} (y_1) = \left\{ \begin{array}{lr}
\displaystyle y_1\lambda ^2 e^{-\lambda y_1} & : y_1 \geq 0 \\
0 & : \text{o/w}\;\;\;\,
\end{array}
\right.
\end{displaymath}[/itex]
[itex]\begin{displaymath} f_{Y_2} (y_2) = \left\{ \begin{array}{lr}
\displaystyle \frac{\lambda ^2}{(y_2 +1)^2} \int^{\infty}_0 y_1 e^{-\lambda y_1} \;dy_1 & : y_2 \geq 0 \\
0 & : \text{o/w}\;\;\;\,
\end{array}
\right.
\end{displaymath}[/itex]
[itex]\begin{displaymath} f_{Y_2} (y_2) = \left\{ \begin{array}{lr}
\displaystyle \frac{1}{(y_2 +1)^2} & : y_2 \geq 0 \\
0 & : \text{o/w}\;\;\;\,
\end{array}
\right.
\end{displaymath}[/itex]
[itex]\therefore[/itex] since [itex]f_{Y_1,Y_2} (y_1,y_2) = f_{Y_1} (y_1) f_{Y_2} (y_2), Y_1[/itex] and [itex]Y_2[/itex] are independent.