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Thread: P/1 Exam problem

  1. #1
    Actuary.com - Level III Poster edwinjaxfl's Avatar
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    Post P/1 Exam problem

    Sharing a problem for practice. Here it goes:

    Let X be a random variable with mean 3 and variance 2, and let Y be a random variable such that for every x, the conditional distribution of Y given X=x has a mean of x and a variance of x². What is the variance of the marginal distribution of Y? :goofy:

    A) 5
    B) 4
    C) 13
    D) 11
    E) 2
    Edwin

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    Quote Originally Posted by edwinjaxfl View Post
    Sharing a problem for practice. Here it goes:

    Let X be a random variable with mean 3 and variance 2, and let Y be a random variable such that for every x, the conditional distribution of Y given X=x has a mean of x and a variance of x². What is the variance of the marginal distribution of Y? :goofy:

    A) 5
    B) 4
    C) 13
    D) 11
    E) 2
    if you know that Yar(Y)=E(Var(Y|X))+Var(E(Y|X)), you can see that

    Var(Y)=E(X^2)+Var(X)=11+2=13.

    Otherwise, you should at least know that E(g(X))=E(E(g(X))), then
    E(Y)=E(X)=3, E(Y^2)=E(E(Y^2|X))=E(2X^2)=22, and this gives Var(Y)=22-9=13.

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    Quote Originally Posted by ecofreak11 View Post
    if you know that Yar(Y)=E(Var(Y|X))+Var(E(Y|X)), you can see that

    Var(Y)=E(X^2)+Var(X)=11+2=13.

    Otherwise, you should at least know that E(g(X))=E(E(g(X))), then
    E(Y)=E(X)=3, E(Y^2)=E(E(Y^2|X))=E(2X^2)=22, and this gives Var(Y)=22-9=13.
    Yarr, variance for pirates.

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    Actuary.com - Level II Poster
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    That is a really cool problem. I had no idea how to work it.

  5. #5
    Actuary.com - Level III Poster edwinjaxfl's Avatar
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    That's the right answer. The Actex book states that we might not see this type of problem on the P/1 test but that we might on the (MLC) modeling test or the (C) construction test. I thought it was good to work out anyway.
    Edwin

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