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#1 2012-02-26 12:14:38

abel3
Member
Registered: 2012-02-26
Posts: 4

Bay's Rule

Hi,

I'm having some trouble solving the following problem:

lets say X is normally distributed with mean ux and std sx
Y is normally distributed with mean X and std sy.

What is E(X|Y)?

Thanks!

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#2 2012-02-26 20:38:13

Bob
Administrator
Registered: 2010-06-20
Posts: 10,583

Re: Bay's Rule

hi abel3

Welcome to the forum!  smile

Just to check your question.

What is E(X|Y), the expection for X given Y ?

Bob


Children are not defined by school ...........The Fonz
You cannot teach a man anything;  you can only help him find it within himself..........Galileo Galilei
Sometimes I deliberately make mistakes, just to test you!  …………….Bob smile

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#3 2012-02-27 06:41:33

abel3
Member
Registered: 2012-02-26
Posts: 4

Re: Bay's Rule

Hi Bob,

No, what I meant is that the mean of Y is the realization of the random variable X

Thanks,
Aya

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#4 2012-02-27 07:02:04

Bob
Administrator
Registered: 2010-06-20
Posts: 10,583

Re: Bay's Rule

hi abel3

realization

Sorry, I don't understand this word used in a mathematical context. sad

Bob


Children are not defined by school ...........The Fonz
You cannot teach a man anything;  you can only help him find it within himself..........Galileo Galilei
Sometimes I deliberately make mistakes, just to test you!  …………….Bob smile

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#5 2012-02-27 08:05:40

abel3
Member
Registered: 2012-02-26
Posts: 4

Re: Bay's Rule

What I meant was that Y's mean is also a random variable.
Anyway - thanks. I'll let you know when I figure out the answer smile

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#6 2012-02-28 04:04:47

abel3
Member
Registered: 2012-02-26
Posts: 4

Re: Bay's Rule

Ok, this is the solution:

E(X|Y)=E(X)+cov(X,Y)/var(Y)*(Y-E(Y))

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