When I read the question I interpreted = as this
The circuit Y comes in use when circuit = X fails
so expected lifetime will be -- = expected time circuit X fails .. plus then circuit Y comes in use .. so the expected = time circuit Y fails
 
since Y is not being used when X is in = use
 
But I think the point that the domain = of y is (x to infinity )
so x lifetime is being added already to = the expected Y
So it  will be just Y =
 
Am I right ? Or something is still = missing
----- Original Message -----
From: Adesola, Olasumbo
To: Study Group 1
Sent: Wednesday, April 28, 2004 = 10:50 AM
Subject: [studygroup1] Re: BPP = free Practice question 8.4

Also, you still need to use the joint pdf:

 

E(y) = Integral (y*joint pdf)

 

Olasumbo Adesola

Pricing

Kemper = Autoand Home

Phone: = (904) 596-8420

Fax: (904) = 245-5952

oadesola@kahg.com

<= /DIV>

-----Original Message-----
From: Adesola, Olasumbo = [mailto:OAdesola@kahg.com]
Sent: Wednesday, = April 28, 2004 11:39 AM
To: = Study Group 1
Subject: = [studygroup1] Re: BPP free Practice question 8.4

 

To add more to what Bailey suggested.
Read the question two or three times and pay = attention to details.
It states: ".... the device fails when and only the second circuit = fails.... What is the expected time at which device fails?"

Circuit Y is only used iff X fails but mind you, the device has not failed yet = even though Circuit X failed. The only time the device fails is when = Circuit Y fails.

I hope this clarifies it more.

Olasumbo Adesola
Pricing
Kemper Auto and Home
Phone: (904) = 596-8420
Fax: (904) 245-5952
oadesola@kahg.com

-----Original Message----- =
From: Bailey Ling [mailto:b.ling@myrealbox.com]
Sent: = Tuesday, April 27, 2004 10:13 PM
To: Study Group 1
Subject: [studygroup1] Re: BPP free Practice = question 8.4

If you're doing E[X+Y] then you are counting X twice.  Y can happen only
if X = has already happened.  So finding E[Y] alone will find the expected
number of times the circuit fails.

Hope that helps,
Bailey

Ritu Kumar wrote:

> 
> A device contains two circuits . The = second circuit is a backup for
> the first , so the second is used only = when the first has failed . The
> device fails when and only the second = circuit fails .
> Let X and Y be the times at which the first and second circuit fail ,
> = respectively . X and Y have joint probability density function =
>  =
> f(x,y) = 6 (e^ -x) (e^ = -2y)   for 0<x<y<infinity
>  =
> What is the expected = time at which device fails ?
> 
> a) 0.33
> b) 0.50
> c) 0.67
> d) 0.83
> e) 1.50
> 
> 
> Now according to my understanding , we = should find out the time E[ X +
> Y] , and that should be the expected = time when device fail .
> But in the solution they just find out = E[ Y ] for answer
> Can someone please explain me why
> 
> Thanks in advance =
> Ritu =
>  =
>  =
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