# Thread: Var(ax^2 - x )

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

Attachment 31478

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 .

2. ## 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.

3. ## Re: Var(ax^2 - x )

Originally Posted by Prove It
\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)

4. ## Re: Var(ax^2 - x )

Originally Posted by Prove It
\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?