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Forum index » Science and Technology » Math » Probability
non-trivial expectation
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porce.lee@gmail.com
science forum beginner


Joined: 13 Sep 2005
Posts: 4

PostPosted: Wed Oct 05, 2005 4:10 am    Post subject: non-trivial expectation Reply with quote
Hi,

This is an non-trivial expectation problem for
non-discrete/non-continuous distribution. Challange this :)

P(x,y) = 2/5 if (x,y)=(3/4, 1/4)
P(x,y) = uniform U[0,1]^2 with total probability 3/5 if
(x,y)!=(3/4,1/4)

Definitely, E[X|Y!=1/4] = 1/2.
Can you calculate E[X|Y=1/4] = ? How can you derive that answer ?
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Mike
science forum addict


Joined: 17 Sep 2005
Posts: 74

PostPosted: Sat Oct 08, 2005 4:28 pm    Post subject: Re: non-trivial expectation Reply with quote
"porce" wrote:
Quote:

This is an non-trivial expectation problem for
non-discrete/non-continuous distribution. Challange this :)

P(x,y) = 2/5 if (x,y)=(3/4, 1/4)
P(x,y) = uniform U[0,1]^2 with total probability 3/5 if
(x,y)!=(3/4,1/4)

Definitely, E[X|Y!=1/4] = 1/2.
Can you calculate E[X|Y=1/4] = ? How can you derive that answer ?

Why not use:

P{X<>3/4}*E[X|X<>3/4,Y=1/4]+P{X=3/4}*E[X|X=3/4,Y=1/4] ( = 3/5)
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