Conditional expected value - notation

[Owen90]

Hello,

Expected value of some function g containing 2 random variables X and Y, conditioned on r.v. Y that is equal to some numerical value y, can be written as:



If I now replace value of y on the LHS to some integer, let's say 8, can I leave the RHS as above, or do I need to replace each of the ys there with 8 in order for it to be mathematically correct?

In other words is the following correct?



Thanks in advance for any help.

 

[BigBeachBanana]

I don't believe your first equation is correct to start with:
[math]E(g(X,Y))=\sum_{(x,y)}g(x,y)p(x,y)[/math]The conditional expectation becomes:
[math]E(g(X,Y)|Y=y)=\sum_{(x,y)}g(x,y)p(g(x,y)|y)=\sum_{(x,y)}g(x,y)\frac{p(g(x,y),y)}{p(y)}[/math]

 

[Owen90]

Since Y is fixed to some value y (Y=y), then doesn't it mean that there is no need to sum over ys? If so, I would argue that the first equation I've written is correct.

Now the question I had was whether the notation I've given in the second equation, where Expected value of g(X,Y) is conditioned on Y=8 is still mathematically correct even though on the RHS of equation I have written it with ys and not with 8 in replacement of those ys. In other words, does information with Y=8 on the LHS is in conflict with ys in the RHS - can they be misinterpreted as function of some parameter y or is the information that Y=8 explicit mathematically that those ys are 8.

I hope that I didn't confuse it more with the explanation I've given.

Thanks.

 

[BigBeachBanana]

Since Y is fixed to some value y (Y=y), then doesn't it mean that there is no need to sum over ys? If so, I would argue that the first equation I've written is correct.

You would still need to consider the probability of that value y gets picked over its sample space.
[math]E(g(X,Y)|Y=y) =\sum_{(x,y)}g(x,y)p(g(x,y)|y)\\ =\sum_{(x,y)}g(x,y)\frac{p(g(x,y),y)}{p(y)}\\ =\sum_{(x,y)}g(x,y)\frac{p(g(x,y)\land y)}{p(y)}\\ =\sum_{(x,y)}g(x,y)\frac{p(x,y)}{p(y)}\\ E(g(X,Y)|Y=8)=\sum_{(x,8)}g(x,8)\frac{p(x,8)}{p(8)}\\[/math]

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Unless what you really meant is
[math]E(g(X)|Y=y)=\sum_{x}g(x)p(x|Y=y)=\sum_{x}g(x)\frac{p(x,Y=y)}{p(Y=y)}[/math]

 

[Owen90]

Okay, that makes sense now. Thank you for clarifying!