In Section 5.2, we argued that if Y1 and Y2 have joint

Chapter 5, Problem 67E

(choose chapter or problem)

Get Unlimited 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)

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

Add to cart


Study Tools You Might Need

Not The Solution You Need? Search for Your Answer Here:

×

Login

Login or Sign up for access to all of our study tools and educational content!

Forgot password?
Register Now

×

Register

Sign up for access to all content on our site!

Or login if you already have an account

×

Reset password

If you have an active account we’ll send you an e-mail for password recovery

Or login if you have your password back