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Math Help - E(X) and Var(X)

  1. #1
    Junior Member Ruichan's Avatar
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    E(X) and Var(X)

    1. The random variable X has pdf fX(x) = x^−4for x > 1, and 0, otherwise.
    (a) Compute E(X) and Var(X).
    (b) Let Y =√X. Compute E(Y ) and Var(Y ).
    (c) Let W = 2X + 5. Compute E(W) and Var(W).

    By the way, how do I get the Math symbols to work while pasting directly from my Words? Thanks
    Last edited by Ruichan; October 31st 2006 at 09:38 AM.
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  2. #2
    Grand Panjandrum
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    Quote Originally Posted by Ruichan View Post
    1. The random variable X has pdf fX(x) = x^−4for x > 1, and 0, otherwise.
    This is not a density, as a requirement for a density is that:

    \int_D f(x) dx = 1

    and f(x)=x^{-4} does not meet this requirement if D=(1, \infty).

    RonL
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  3. #3
    Junior Member Ruichan's Avatar
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    Quote Originally Posted by CaptainBlack View Post
    This is not a density, as a requirement for a density is that:

    \int_D f(x) dx = 1

    and f(x)=x^{-4} does not meet this requirement if D=(1, \infty).

    RonL

    That's why I couldn't get the answer at all. I emailed my prof this morning. He just replied saying there's a typo. The correct question is
    1. The random variable X has pdf fX(x) = 3x^−4for x > 1, and 0, otherwise.
    (a) Compute E(X) and Var(X).
    (b) Let Y =√X. Compute E(Y ) and Var(Y ).
    (c) Let W = 2X + 5. Compute E(W) and Var(W).

    Does the 3 make a difference?
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  4. #4
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    \begin{array}{l}<br />
 \int\limits_1^\infty  {\frac{3}{{x^4 }}dx}  = 1 \\ <br />
 E(X) = \int\limits_1^\infty  {x\left( {\frac{3}{{x^4 }}} \right)dx}  = \int\limits_1^\infty  {\left( {\frac{3}{{x^3 }}} \right)dx = ?}  \\ <br />
 E(X^2 ) = \int\limits_1^\infty  {x^2 \left( {\frac{3}{{x^4 }}} \right)dx}  = \int\limits_1^\infty  {\left( {\frac{3}{{x^2 }}} \right)dx = ?}  \\ <br />
 V(X) = E\left( {X^2 } \right) - E^2 (X) \\ <br />
 \end{array}
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  5. #5
    Junior Member Ruichan's Avatar
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    Quote Originally Posted by Ruichan View Post
    That's why I couldn't get the answer at all. I emailed my prof this morning. He just replied saying there's a typo. The correct question is
    1. The random variable X has pdf fX(x) = 3x^−4for x > 1, and 0, otherwise.
    (a) Compute E(X) and Var(X).
    (b) Let Y =√X. Compute E(Y ) and Var(Y ).
    (c) Let W = 2X + 5. Compute E(W) and Var(W).

    Does the 3 make a difference?

    Answers:
    a) E(X) = 3/2, E(X^2)=3, Var(X)=3/4
    b) Can someone help me with this please? All the questions I did before was when X ~Unif(0,1). How do you do it for this when it's not
    unif(0,1)? Thanks

    c) Using the the values I got from a), I got E(W)=3 and Var(W)=3. ----> Can the expected value and variance be the same?
    E(W) = 2E(X) = 2(3/2) = 3. Var(X) = 2^2Var(X) = 4Var(X) = 4(3/4) = 3.
    The answers look so wrong though. Did I do something wrongly? I just integrated with the formula my prof gave.
    Thanks in advance.
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  6. #6
    Junior Member Ruichan's Avatar
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    Quote Originally Posted by Ruichan View Post
    b) Can someone help me with this please? All the questions I did before was when X ~Unif(0,1). How do you do it for this when it's not
    unif(0,1)? Thanks
    (b) Let Y =√X. Compute E(Y ) and Var(Y ).
    E(Y) = integrate (from 0 to 1) x^1/2 dx =2/3
    E(Y^2) = Integrate (0 to 1) x dx = 1/2
    Var(Y) = 1/2 - 2/3^2 = 1/18

    Is it correct if I do it this way?
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  7. #7
    Grand Panjandrum
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    Quote Originally Posted by Ruichan View Post
    (b) Let Y =√X. Compute E(Y ) and Var(Y ).
    E(Y) = integrate (from 0 to 1) x^1/2 dx =2/3
    E(Y^2) = Integrate (0 to 1) x dx = 1/2
    Var(Y) = 1/2 - 2/3^2 = 1/18

    Is it correct if I do it this way?
    Y=\sqrt{X}, where the density of X is f(x)=3x^{-4}.


    Then:

    E(Y)=\int_1^{\infty} \sqrt{x} 3 x^{-4}\ dx=\int_1^{\infty} 3 x^{-3.5}\ dx

    ........................ = \left[ -\frac{3}{2.5}x^{-2.5} \right]_1^{\infty}=3/2.5 = 1.2

    Also:

    E(Y^2)= \int_1^{\infty} (x) 3 x^{-4}\ dx=\int_1^{\infty} 3 x^{-3}\ dx

    and you should be able to take things from there

    RonL
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  8. #8
    Junior Member Ruichan's Avatar
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    Quote Originally Posted by CaptainBlack View Post
    Y=\sqrt{X}, where the density of X is f(x)=3x^{-4}.


    Then:

    E(Y)=\int_1^{\infty} \sqrt{x} 3 x^{-4}\ dx=\int_1^{\infty} 3 x^{-3.5}\ dx

    ........................ = \left[ -\frac{3}{2.5}x^{-2.5} \right]_1^{\infty}=3/2.5 = 1.2

    Also:

    E(Y^2)= \int_1^{\infty} (x) 3 x^{-4}\ dx=\int_1^{\infty} 3 x^{-3}\ dx

    and you should be able to take things from there

    RonL
    Thank you very much.
    Arg.....shows how much i understand probability when I don't even know that I'm supposed to use the pdf from part a). Arg.....

    Question: Therefore it doesn't matter that if X~unif(0,1)? I just need to use the pdf? Then integrate from whatever values that are given? The unif(0,1) is to integrate values from 0 to 1? If for eg, if not uniform, values given are 2<x<7, just need to integrate from 2 to 7?
    Last edited by Ruichan; October 31st 2006 at 09:33 PM.
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  9. #9
    Grand Panjandrum
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    Quote Originally Posted by Ruichan View Post
    Thank you very much.
    Arg.....shows how much i understand probability when I don't even know that I'm supposed to use the pdf from part a). Arg.....

    Question: Therefore it doesn't matter that if X~unif(0,1)? I just need to use the pdf? Then integrate from whatever values that are given? The unif(0,1) is to integrate values from 0 to 1? If for eg, if not uniform, values given are 2<x<7, just need to integrate from 2 to 7?
    Yes, you integrate the function you want the expectation of times the pdf,
    over the range that the pdf is defined on (or rather you integrate over the
    entire line or whatever, but ignore the regions over which the pdf is zero).

    RonL
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  10. #10
    Junior Member Ruichan's Avatar
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    Quote Originally Posted by Plato View Post
    \begin{array}{l}<br />
 \int\limits_1^\infty  {\frac{3}{{x^4 }}dx}  = 1 \\ <br />
 E(X) = \int\limits_1^\infty  {x\left( {\frac{3}{{x^4 }}} \right)dx}  = \int\limits_1^\infty  {\left( {\frac{3}{{x^3 }}} \right)dx = ?}  \\ <br />
 E(X^2 ) = \int\limits_1^\infty  {x^2 \left( {\frac{3}{{x^4 }}} \right)dx}  = \int\limits_1^\infty  {\left( {\frac{3}{{x^2 }}} \right)dx = ?}  \\ <br />
 V(X) = E\left( {X^2 } \right) - E^2 (X) \\ <br />
 \end{array}
    I do have the formula in front of me when trying to do the questions. I just didn't know how to start.
    Thanks for showing me!
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