Solution Found!
In Exercise 5.8, we established that is a valid joint
Chapter 5, Problem 105E(choose chapter or problem)
In Exercise 5.8, we established that
\(f\left(y_{1}, y_{2}\right)=\left\{\begin{array}{ll}
4 y_{1} y_{2}, & 0 \leq y_{1} \leq 1,0 \leq y_{2} \leq 1, \\
0, & \text { elsewhere }
\end{array}\right.
\)
is a valid joint probability density function. In Exercise , we established that \(Y_{1}\) and \(Y_{2}\) are independent; in Exercise
, we determined that \(E\left(Y_{1}-Y_{2}\right)=0\) and found the value for \(V\left(Y_{1}\right)\). Find \(V\left(Y_{1}-Y_{2}\right)\).
Equation Transcription:
Text Transcription:
f(y_1,y2)={_0, elsewhere ^4y_1y_2, 0</=y_1</=1,0</=y_2</=1,
Y_1
Y_2
E(Y_1-Y_2)=0
V(Y_1)
V(Y_1-Y_2)
Questions & Answers
QUESTION:
In Exercise 5.8, we established that
\(f\left(y_{1}, y_{2}\right)=\left\{\begin{array}{ll}
4 y_{1} y_{2}, & 0 \leq y_{1} \leq 1,0 \leq y_{2} \leq 1, \\
0, & \text { elsewhere }
\end{array}\right.
\)
is a valid joint probability density function. In Exercise , we established that \(Y_{1}\) and \(Y_{2}\) are independent; in Exercise
, we determined that \(E\left(Y_{1}-Y_{2}\right)=0\) and found the value for \(V\left(Y_{1}\right)\). Find \(V\left(Y_{1}-Y_{2}\right)\).
Equation Transcription:
Text Transcription:
f(y_1,y2)={_0, elsewhere ^4y_1y_2, 0</=y_1</=1,0</=y_2</=1,
Y_1
Y_2
E(Y_1-Y_2)=0
V(Y_1)
V(Y_1-Y_2)
ANSWER:
Solution:
Step 1 of 2:
Given the joint probability density function is
