Solution Found!
In Section 5.2, we argued that if Y1 and Y2 have joint
Chapter 5, Problem 67E(choose chapter or problem)
In Section 5.2, we argued that if \(Y_{1}\) and \(Y_{2}\) have joint cumulative distribution function \(F\left(y_1,\ y_2\right)\) then for any \(a<b\) and \(c<d\)
\(P\left(a<Y_1\le b,\ c<Y_2\le d\right)=F(b,\ d)-F(b,\ c)-F(a,\ d)+F(a,\ c).\)
If \(Y_{1}\) and \(Y_{2}\) are independent, show that
\(P\left(a<Y_1\le b,\ c<Y_2\le d\right)=P\left(a<Y_1\le b\right)\times P\left(c<Y_2\le d\right).\)
[Hint: Express \(P\left(a<Y_{1} \leq b\right)\) in terms of \(F_{1}(\cdot)\).]
Equation Transcription:
Text Transcription:
Y_1
Y_2
F(y_1,y_2)
a<b
c<d
P(a<Y_1</=b,c<Y_2</=d)=F(b,d)-F(b,c)-F(a,d)+F(a,c).
Y_1
Y_2
P(a<Y_1</=b,c<Y_2</=d)=P(a<Y_1</=b)P(c<Y_2</=d).
P(a<Y_1</=b)
F_1(cdot)
Questions & Answers
QUESTION:
In Section 5.2, we argued that if \(Y_{1}\) and \(Y_{2}\) have joint cumulative distribution function \(F\left(y_1,\ y_2\right)\) then for any \(a<b\) and \(c<d\)
\(P\left(a<Y_1\le b,\ c<Y_2\le d\right)=F(b,\ d)-F(b,\ c)-F(a,\ d)+F(a,\ c).\)
If \(Y_{1}\) and \(Y_{2}\) are independent, show that
\(P\left(a<Y_1\le b,\ c<Y_2\le d\right)=P\left(a<Y_1\le b\right)\times P\left(c<Y_2\le d\right).\)
[Hint: Express \(P\left(a<Y_{1} \leq b\right)\) in terms of \(F_{1}(\cdot)\).]
Equation Transcription:
Text Transcription:
Y_1
Y_2
F(y_1,y_2)
a<b
c<d
P(a<Y_1</=b,c<Y_2</=d)=F(b,d)-F(b,c)-F(a,d)+F(a,c).
Y_1
Y_2
P(a<Y_1</=b,c<Y_2</=d)=P(a<Y_1</=b)P(c<Y_2</=d).
P(a<Y_1</=b)
F_1(cdot)
ANSWER:
Solution
Step 1 of 1
If Y1 and Y2 are independent we have to show that
Given joint cumulative distribution function is