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Conditional expectation question

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  • nubie
    replied
    i'm a little perplexed by the problem, too. is Y a function of u, or does u simply represent the "values" that the r.v. Y can have?

    assuming the latter, then we can try this:

    the overall expected value E(Y) can be expressed this way:

    E(Y) = E(Y|Y>1) + E(Y|Y≤1)

    the first and last ones are easier to deal with imho:

    use the pdf to find E(Y), then E(Y|Y≤1) can be found using the cdf.

    you'd just have to differenciate to get the pdf.
    and P(Y≤1) is F(u=1)

    ...i think...but would be nice to KNOW, so somebody please clarify and let me know if/where i went wrong...thanx!

    Leave a comment:


  • managuense
    replied
    That distribution does not make sense. I assume F(10) should equal 1 and F(x) has to be >= 0 for all x. Please check it over and restate it and maybe we can lend some help.

    Leave a comment:


  • audia4
    started a topic Conditional expectation question

    Conditional expectation question

    A group insurance covers the medical claims of the employees of a small company. The value, Y of the claims made in one year has the following cumulative distribution.


    F_Y(u) = 0 u<=0

    = u^2/50 0<u<=5

    = -1/50 u^2 +2/5u -1 5<u<=10

    = 0 u>10

    Find conditional expectation of Y given that Y exceeds 1

    Ie E(Y|Y>1)

    Could someone illustrate this solution by first finding the density function first.

    Thank you.
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