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Math Help - Var(ax^2 - x )

  1. #1
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    Var(ax^2 - x )

    Var(ax^2 - x )-dsc_0082.jpg Attachment 31478Var(ax^2 - x )-dsc_0083.jpg

    can anyone please help me with part iii) ? i know that VAR (x) = E(X^2)- ( E(X) )^2 .... But I get my E(B^2) = 6875/4 , whereas my E(B) = 5125/12 , where my B= 3X^2 - X, i take E(B^2)- ( E(B) )^2, MY ANS IS NEGATIVE . WHICH IS WRONG! i have attached my working in the photo .
    Last edited by delso; August 27th 2014 at 11:00 PM.
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  2. #2
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    Re: Var(ax^2 - x )

    $\displaystyle \begin{align*} \mathrm{Var}\,\left( 3X^2 - X \right) &= \mathrm{E}\,\left[ \left( 3X^2 - X \right) ^2 \right] - \left[ \mathrm{E}\,\left( 3X^2 - X \right) \right] ^2 \\ &= \mathrm{E}\,\left( 9X^4 - 6X^3 + X^2 \right) - \left[ 3\, \mathrm{E}\,\left( X^2 \right) - \mathrm{E}\,\left( X \right) \right] ^2 \\ &= 9\,\mathrm{E}\,\left( X^4 \right) - 6\,\mathrm{E}\,\left( X^3 \right) + \mathrm{E}\,\left( X^2 \right) - \left\{ 9 \, \left[ \mathrm{E}\,\left( X^2 \right) \right] ^2 - 6 \, \mathrm{E}\,\left( X^2 \right) \, \mathrm{E}\, \left( X \right) + \left[ \mathrm{E}\,\left( X \right) \right] ^2 \right\} \\ &= 9\,\mathrm{E}\,\left( X^4 \right) - 6\,\mathrm{E}\,\left( X^3 \right) + \mathrm{E}\,\left( X^2 \right) - 9 \, \left[ \mathrm{E}\,\left( X^2 \right) \right] ^2 + 6 \, \mathrm{E}\,\left( X^2 \right) \, \mathrm{E}\,\left( X \right) - \left[ \mathrm{E}\,\left( X \right) \right] ^2 \end{align*}$

    These expectations should be easy to calculate.
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    Re: Var(ax^2 - x )

    Quote Originally Posted by Prove It View Post
    $\displaystyle \begin{align*} \mathrm{Var}\,\left( 3X^2 - X \right) &= \mathrm{E}\,\left[ \left( 3X^2 - X \right) ^2 \right] - \left[ \mathrm{E}\,\left( 3X^2 - X \right) \right] ^2 \\ &= \mathrm{E}\,\left( 9X^4 - 6X^3 + X^2 \right) - \left[ 3\, \mathrm{E}\,\left( X^2 \right) - \mathrm{E}\,\left( X \right) \right] ^2 \\ &= 9\,\mathrm{E}\,\left( X^4 \right) - 6\,\mathrm{E}\,\left( X^3 \right) + \mathrm{E}\,\left( X^2 \right) - \left\{ 9 \, \left[ \mathrm{E}\,\left( X^2 \right) \right] ^2 - 6 \, \mathrm{E}\,\left( X^2 \right) \, \mathrm{E}\, \left( X \right) + \left[ \mathrm{E}\,\left( X \right) \right] ^2 \right\} \\ &= 9\,\mathrm{E}\,\left( X^4 \right) - 6\,\mathrm{E}\,\left( X^3 \right) + \mathrm{E}\,\left( X^2 \right) - 9 \, \left[ \mathrm{E}\,\left( X^2 \right) \right] ^2 + 6 \, \mathrm{E}\,\left( X^2 \right) \, \mathrm{E}\,\left( X \right) - \left[ \mathrm{E}\,\left( X \right) \right] ^2 \end{align*}$

    These expectations should be easy to calculate.
    any simpler way beside this ? i know that VAR (a X) = ( a^2 )X , can i apply this method to the question above ? how to do it ? since this is VAR (aX^2) , which is different from VAR (a X)
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    Re: Var(ax^2 - x )

    Quote Originally Posted by Prove It View Post
    $\displaystyle \begin{align*} \mathrm{Var}\,\left( 3X^2 - X \right) &= \mathrm{E}\,\left[ \left( 3X^2 - X \right) ^2 \right] - \left[ \mathrm{E}\,\left( 3X^2 - X \right) \right] ^2 \\ &= \mathrm{E}\,\left( 9X^4 - 6X^3 + X^2 \right) - \left[ 3\, \mathrm{E}\,\left( X^2 \right) - \mathrm{E}\,\left( X \right) \right] ^2 \\ &= 9\,\mathrm{E}\,\left( X^4 \right) - 6\,\mathrm{E}\,\left( X^3 \right) + \mathrm{E}\,\left( X^2 \right) - \left\{ 9 \, \left[ \mathrm{E}\,\left( X^2 \right) \right] ^2 - 6 \, \mathrm{E}\,\left( X^2 \right) \, \mathrm{E}\, \left( X \right) + \left[ \mathrm{E}\,\left( X \right) \right] ^2 \right\} \\ &= 9\,\mathrm{E}\,\left( X^4 \right) - 6\,\mathrm{E}\,\left( X^3 \right) + \mathrm{E}\,\left( X^2 \right) - 9 \, \left[ \mathrm{E}\,\left( X^2 \right) \right] ^2 + 6 \, \mathrm{E}\,\left( X^2 \right) \, \mathrm{E}\,\left( X \right) - \left[ \mathrm{E}\,\left( X \right) \right] ^2 \end{align*}$

    These expectations should be easy to calculate.
    my E(X^4) = integration of hx^ 4 from 0 to 5 , where h = 0.2 , so my E(X^4) = 125
    by using the same method , i find E(X^3 ) , E(X^3)= integration of hx^ 3 from 0 to 5 , so my E(X^3)=125/4

    by substituiting E(X^4) = 125 , E(X^3)=125/4 , E(X^2)=25/3 , E(X) = 5/2 , my ans if VAR(3X^2 - X) = 5275/12 , but the ans form book is 6025/12 . What's wrong with my working?
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  5. #5
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    Re: Var(ax^2 - x )

    I come up with your answer of 5275/12.

    This way of computing the answer works just fine; I actually prefer to just calculate

    E(3x^2 - x) directly = integral from 0 to 5 of .2(3x^2 - x) dx = 22.5, and
    E[(3x^2 - x)^2] directly = integral from 0 to 5 of .2[(3x^2 - x)^2] dx = 11350/12, and then
    Var(3x^2 - x) = 11350/12 - 22.5^2 = 5275/12

    Maybe book is wrong?
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