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Simple expectation question

  1. Dec 12, 2007 #1
    I have a random variable Y that represents the size of a population. I know that the expectation E(Y) = a.
    Now suppose, I have another random variable X that represents the number of people in that population that have a certain disease. The expectation is that on average half the population have the disease.

    So, I was wondering which of the following would correctly describe the random variable X

    1) E(X|Y) = 0.5*Y
    2) E(X) = 0.5*E(Y) = 0.5*a
    3) E(X) = 0.5*Y
     
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  3. Dec 12, 2007 #2

    Hurkyl

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    Aren't (1) and (3) nonsensical?

    Of course, you haven't considered things like E(X | Y = y) = 0.5 y...

    Honestly, I would say that the problem is badly stated. My best literal translation is that the problem states your option (3) -- but as I said, that expression is nonsensical; Y is a probability measure and E(X) is a number. They aren't even the same kind of object, so it doesn't even make sense to ask if they are equal!
     
  4. Dec 12, 2007 #3
    Yes, sorry my notation was bad. What I meant by the "Y" in (1) and (3) was the value of the random variable Y. So perhaps, I could restate it like

    1) E(X|Y=y) = 0.5*y
    2) E(X) = 0.5*E(Y) = 0.5*a
    3) E(X) = 0.5*y

    Perhaps that makes a bit more sense??
     
  5. Dec 12, 2007 #4

    Office_Shredder

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    (3) doesn't make any sense if you're not told what y is, if we know y is what Y came out to be, then it's the same as (1) again. Interesting note, if (1) is true, then

    [tex]E(X) = \sum_{y=1}^{ \infty } E(X|Y=y)P(Y=y)[/tex] which comes out to be [tex]\sum_{y=1}^{ \infty } (1/2)y*P(Y=y)=1/2 \sum_{y=1}^{ \infty }y*P(Y=y)[/tex] which by definition is 1/2*E(Y). So (1) implies (2), hence if (1) correctly describes X, so does (2). By the Law of Only One Right Answer, (1) cannot be correct, and hence it must be (2).

    However, if the question was intended to be "which of the following best represents the information given in the problem" I would go with (1)
     
  6. Dec 12, 2007 #5

    Hurkyl

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    Well, this (3) is also nonsensical.

    I would like to make a correction; I no longer think E(X|Y)=0.5*Y is nonsensical. If you set f(y) = E(X|Y=y), then I think it's not unreasonable to read the expression E(X|Y) as denoting the random variable f(Y).

    (it would be nice sometime to see a formal grammar for this stuff, but ah well)


    The problem with the statement of the problem is that it describes a multivariate distribution -- you have a joint distribution on the outcomes of X and Y. When it says "on average" without further qualification, you would naturally assume this means to average over the entire joint outcome space. But since "half the population" is a random variable, it doesn't make sense to equate the two.

    So, the problem is how to reinterpret what is said...

    If forced to guess the author's meaning, I would assume what he really meant is the marginal expectation -- "For each population outcome, the expectation is that on average..." -- in which case your old (1) and your new (1) are both adequate descriptions.
    (And from which (2) is a consequence!)
     
  7. Dec 12, 2007 #6
    Yes, this is exactly what I meant.
    So, given that E(X) = a*E(Y) is the correct expression, am I right in concluding that the variance Var(X) = a*Var(Y)?

    Also from this can I deduce anyting for the covariance Cov(X,Y)?? That is, is there a simple solution for Cov(X,Y)?
     
    Last edited: Dec 12, 2007
  8. Dec 12, 2007 #7

    Hurkyl

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    Absolutely not. Get it out of your head right now that knowing the expectation of a random variable tells you anything at all about it's variance.

    Or its covariance with another random variable.


    In fact, simple counterexamples are extremely easy to produce. Try playing with random variables that have only one or two outcomes.
     
  9. Dec 13, 2007 #8

    EnumaElish

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