Stochastic Calculus: Conditional Expectation

[Win_odd Dhamnekar]

 

[blamocur]

In the first problem what is the value of your answer when [imath]Z=1[/imath]?

 

[Win_odd Dhamnekar]

In the first problem what is the value of your answer when [imath]Z=1[/imath]?

My answer when Z=1 is 7.346937. But actual answer must be (3 + 6 + 9 +12 +15 +18 = 63/6) =10.5. Somewhere, something goes wrong.

 

[blamocur]

Somewhere, something goes wrong.

I agree In particular, I have no clue where this formula comes from:

 

[Win_odd Dhamnekar]

I agree In particular, I have no clue where this formula comes from:
View attachment 35131

If the question is Find E[ X + 2Y| Z], then can we bifurcate it into E[X] + E[2Y | Z]?

 

[blamocur]

If the question is Find E[ X + 2Y| Z], then can we bifurcate it into E[X] + E[2Y | Z]?

I don't understand what that means, or how that would address my question.

BTW, what is E[X+2Y|Z] ? A number? A function? A random variable?

 

[Win_odd Dhamnekar]

I don't understand what that means, or how that would address my question.

BTW, what is E[X+2Y|Z] ? A number? A function? A random variable?

Author Professor G.F.Lawler (University of Chicago) of the book from which this problem was taken, replied as

"This problem comes from the exercises in my book on Stochastic Calculus. It is in the first section where the definition of a conditional expectation is given. The intention was just to work it out (a tedious exercise) over all the values that Z can take. The purpose of the exercise is to reinforce the definition of a conditional expectation.

There is no "formula for conditional expectation" of the type you gave --- you are giving a number and the conditional expectation should be a random variable measurable with respect to Z. For a discrete random variable, it is the same as giving for each value Z = z , the conditional expectation given this value."

From his comment, It is clear that E[ X + 2Y| Z ] is random variable measurable w.r.t. all values of Z = z .

 

[BigBeachBanana]

The intention was just to work it out (a tedious exercise) over all the values that Z can take.

Brute force, and list them all out.

Matrix for Z=X/Y


Should be straight forward from here.

 

[blamocur]

the conditional expectation should be a random variable measurable with respect to Z.

I have to admit that I am missing something there. To me [imath]E(X+2Y|Z)[/imath] seems to be a function of [imath]z[/imath]. I also have to admit that I could not find a clear definition of conditional expectation on internet -- can you post the one you are using?
Thanks.

 

[Win_odd Dhamnekar]

Brute force, and list them all out.

Matrix for Z=X/Y
View attachment 35140

Should be straight forward from here.

I prepared the following table for X + 2Y. Now how to proceed further?

 

[BigBeachBanana]

I prepared the following table for X + 2Y. Now how to proceed further?

View attachment 35143

@blamocur brought it up above, the conditional expectation given Z is a function of X and Y. Its value is dependent on what is Z (which is listed above).
By definition:
[math]E(X+2Y|Z) = \sum(X+2Y)\Pr(X+2Y|Z)[/math]

 

[Win_odd Dhamnekar]

@blamocur brought it up above, the conditional expectation given Z is a function of X and Y. Its value is dependent on what is Z (which is listed above).
By definition:
[math]E(X+2Y|Z) = \sum(X+2Y)\Pr(X+2Y|Z)[/math]

Hence answer to this question is

E[ X + 2Y | Z] = [3 + 6 + 9 + 12 + 15+ 18 =63]* 1/6 = 10.5 + [5 +10+15]*1/12= 2.5 + [7+ 14]*1/18 =1.1667 + [9 + 11 + 13 +12 + 11+ 13+ 6 + 10 +14 + 7 +9 + 11 +13 +17 + 8 +16 =179]*1/36 = 4.9722222 +[4 +8 +12=24]*1/12=2 + [ 8 +16=24]*1/18 = 1.33333 +[5 + 10 = 15]*1/18 =0.8333333 + [7 +14=21]*1/18 = 1.1667 = 24.47223