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Math Help - variance of uniform distribution

  1. #1
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    variance of uniform distribution

    Let b>a and let X-uniform(a,b) . Prove Var(X) =  (a+b)^2/12

    Var(X)= E(X^2)-E(X)^2

    E(X^2)=integral from a to b of x^2/(b-a) = (b^3-a^3)/b-a

    I know E(X)= (b-a)/2 so E(X^2) must be wrong as when I subtract I do not get required Var(X).
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  2. #2
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    Re: variance of uniform distribution

    Quote Originally Posted by Duke View Post
    Let b>a and let X-uniform(a,b) . Prove Var(X) =  (a+b)^2/12

    Var(X)= E(X^2)-E(X)^2

    E(X^2)=integral from a to b of x^2/(b-a) = (b^3-a^3)/b-a

    I know E(X)= (b-a)/2 so E(X^2) must be wrong as when I subtract I do not get required Var(X).
    Consider this: \int_a^b {\frac{{x^2 }}{{b - a}}dx}  = \left. {\frac{{x^3 }}{3(b-a)}} \right|_a^b
    Last edited by Plato; December 3rd 2011 at 12:17 PM.
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    Re: variance of uniform distribution

    but surely there is a missing factor of 1/(b-a) on the RHS
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    Re: variance of uniform distribution

    Quote Originally Posted by Duke View Post
    but surely there is a missing factor of 1/(b-a) on the RHS
    Yes. I edited it.
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    Re: variance of uniform distribution

    So does it work out correctly for you ?
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    Re: variance of uniform distribution

    Quote Originally Posted by Duke View Post
    So does it work out correctly for you ?
    \begin{align*}\frac{b^3-a^3}{3(b-a)}-\frac{(b+a)^2}{4}&=\frac{4(b^2+ab+a^2)-3(b^2+2ab+a^2)}{12}\\&=\frac{b^2-2ab+a^2}{12} \end{align*}
    Last edited by Plato; December 3rd 2011 at 01:03 PM.
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    Re: variance of uniform distribution

    so you are saying b^3-a^3=(b-a)(b+a)^2
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    Re: variance of uniform distribution

    Quote Originally Posted by Duke View Post
    so you are saying b^3-a^3=(b-a)(b+a)^2
    NO. You have two mistakes in the OP.

    E(X)=\frac{a+b}{2}~\&~\text{Var}(X)=\frac{(b-a)^2}{12}
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    Re: variance of uniform distribution

    Quote Originally Posted by Duke View Post
    Let b>a and let X-uniform(a,b) . Prove Var(X) =  (a-b)^2/12
    The pdf is f_X=\frac{x}{b-a}

    The expectation is E(X)=\int_a^b {x \cdot f_X dx}  = \frac{1}{{b - a}}\int_a^b {xdx}  = \frac{1}{{b - a}}\frac{{b^2  - a^2 }}{2} = \frac{{b + a}}{2}.

    Then E(X^2)= \int_a^b {x^2 \cdot f_X dx}  = \frac{1}{{b - a}}\int_a^b {x^2dx}  = \frac{1}{{b - a}}\frac{{b^3 - a^3 }}{3} = \frac{{b^2 + ba+ a^2}}{3} .

    Now what is \text{Var}(X)=E(X^2)-(E(X))^2~?
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