Daniel Ma
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From: DMA@AFLAC (Daniel Ma), on 8/17/98 9:52 AM:
To: studygroup4b@lists.casact.org, dr@g-g-a.com
To CAS 4B study group participants:
This is basically the idea that E[2X] =3D 2E[X]. With conditional =
=0D
expectation, the idea also works. That is,
Let Y=3D X1+X2. Then E[Y|Y<4] =3D 2E[X1|Y<4].
Apparently, the author of the manual must think that E[Y|Y<4] is easier to=
=
=0D
conceptualize since both the expected value and the conditional statement =
=0D
are based on the same variable. Since you have to enumerate all cases where=
=
=0D
X1+X2<4, the solution can be derived directly.
X1 X2
1 1
1 2 =
1 3
2 1
2 2
3 1
=
With the above 6 cases, we are dealing with a reduced sample space, and =
=0D
note that X1 has 3 possible values. Thus,
=
E[X1|X1+X2<4]=3D 1(3/6) + 2(2/6) + 3(1/6) =3D 5/3
Either way requires the same amount of work. =
Daniel Ma
---------
From: dr@g-g-a.com (Dimitra Roidakis), on 8/14/98 3:28 PM:
I have a question about the solution of question 16, p.9. I don't =
understand when answering the question why they equate the following =
things:
E[X1/X1+X2<4]=3DE[X1+X2/X1+X2<4]/2
It's probably simple but I can't figure it out.
Dimitra Roidakis
dr@g-g-a.com
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