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Math Help - Variance in Univariate Normal

  1. #1
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    Variance in Univariate Normal

    In Univariate Normal, Show
    Var(x)=\sigma^2
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  2. #2
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    What have you tried?
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  3. #3
    Rhymes with Orange Chris L T521's Avatar
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    Quote Originally Posted by ynotidas View Post
    In Univariate Normal, Show
    Var(x)=\sigma^2
    Well, we need to use the fact that E\left[X\right]=\int_{-\infty}^{\infty}\frac{1}{\sigma\sqrt{2\pi}}xe^{-\tfrac{1}{2}\left(\frac{x-\mu}{\sigma}\right)^2}\,dx=\mu

    Thus, E\left[X^2\right]=\frac{1}{\sigma\sqrt{2\pi}}\int_{-\infty}^{\infty}x^2e^{-\frac{1}{2}\left(\frac{x-\mu}{\sigma}\right)^2}\,dx

    Let u=\frac{x-\mu}{\sigma}\implies x=\sigma u+\mu\implies \,dx=\sigma\,du

    Therefore, \frac{1}{\sigma\sqrt{2\pi}}\int_{-\infty}^{\infty}x^2e^{-\frac{1}{2}\left(\frac{x-\mu}{\sigma}\right)^2}\,dx\xrightarrow{u=(x-\mu)/\sigma}{}\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}\left(\sigma u+\mu\right)^2e^{-\tfrac{1}{2}u^2}\,du =\frac{1}{\sqrt{2\pi}}\left[\sigma^2\int_{-\infty}^{\infty}u^2e^{-\tfrac{1}{2}u^2}\,du-2\mu\sigma\int_{-\infty}^{\infty}ue^{-\frac{1}{2}u^2}\,du+\mu^2\int_{-\infty}^{\infty}e^{-\frac{1}{2}u^2}\,du\right]

    Analyze each integral individually:

    \sigma^2\int_{-\infty}^{\infty}u^2e^{-\tfrac{1}{2}u^2}\,du=2\sigma^2\int_0^{\infty}u^2e^  {-\tfrac{1}{2}u^2}\,du

    Let z=\tfrac{1}{2}u^2\implies \,dz=u\,du

    Thus, 2\sigma^2\int_0^{\infty}u^2e^{-\tfrac{1}{2}u^2}\,du\xrightarrow{z=\tfrac{1}{2}u^2  }=2\sqrt{2}\sigma^2\int_{0}^{\infty}z^{\frac{1}{2}  }e^{-z}\,dz=2\sqrt{2}\sigma^2\Gamma\!\left(\tfrac{3}{2}  \right) =\sqrt{2}\sigma^2\Gamma\!\left(\tfrac{1}{2}\right)  =\sqrt{2\pi}\sigma^2

    2\mu\sigma\int_{-\infty}^{\infty}ue^{-\frac{1}{2}u^2}\,du

    Let z=\tfrac{1}{2}u^2\implies \,dz=u\,du

    Thus,
    2\mu\sigma\int_{-\infty}^{\infty}ue^{-\frac{1}{2}u^2}\,du\xrightarrow{z=\tfrac{1}{2}u^2}  {}2\mu\sigma\int_{\infty}^{\infty}e^{-z}\,dz=0

    \mu^2\int_{-\infty}^{\infty}e^{-\frac{1}{2}u^2}\,du=\mu^2\sqrt{2\pi}

    Thus, E\left[X^2\right]=\frac{1}{\sqrt{2\pi}}\cdot\left[\sigma^2\sqrt{2\pi}+\mu^2\sqrt{2\pi}\right]=\sigma^2+\mu^2

    Thus, \text{Var}\left(X\right)=E\left[X^2\right]-\left(E\left[X\right]\right)^2=\sigma^2+\mu^2-\mu^2=\sigma^2

    Does this make sense?
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  4. #4
    MHF Contributor matheagle's Avatar
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    Chris, I mean anteater (my bad)
    I thought you did this last week.
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  5. #5
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    last week Chris did the expected value proof
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