Results 1 to 5 of 5

Thread: Variance in Univariate Normal

  1. #1
    Junior Member
    Joined
    Jan 2009
    Posts
    30

    Variance in Univariate Normal

    In Univariate Normal, Show
    $\displaystyle Var(x)=\sigma^2$
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Junior Member
    Joined
    Mar 2009
    Posts
    64
    What have you tried?
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Rhymes with Orange Chris L T521's Avatar
    Joined
    May 2008
    From
    Chicago, IL
    Posts
    2,844
    Thanks
    5
    Quote Originally Posted by ynotidas View Post
    In Univariate Normal, Show
    $\displaystyle Var(x)=\sigma^2$
    Well, we need to use the fact that $\displaystyle E\left[X\right]=\int_{-\infty}^{\infty}\frac{1}{\sigma\sqrt{2\pi}}xe^{-\tfrac{1}{2}\left(\frac{x-\mu}{\sigma}\right)^2}\,dx=\mu$

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

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

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

    Analyze each integral individually:

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

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

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

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

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

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

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

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

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

    Does this make sense?
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor matheagle's Avatar
    Joined
    Feb 2009
    Posts
    2,763
    Thanks
    5
    Chris, I mean anteater (my bad)
    I thought you did this last week.
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Junior Member
    Joined
    Jan 2009
    Posts
    30
    last week Chris did the expected value proof
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. LRT for variance of normal distribution
    Posted in the Advanced Statistics Forum
    Replies: 2
    Last Post: Feb 24th 2011, 11:34 AM
  2. How do you derive Mean and Variance of Log-Normal Distribution?
    Posted in the Advanced Statistics Forum
    Replies: 4
    Last Post: Nov 10th 2009, 04:39 PM
  3. Proof within Univariate Normal
    Posted in the Advanced Statistics Forum
    Replies: 10
    Last Post: Aug 21st 2009, 11:18 AM
  4. Normal distribution- cheak for mean and variance
    Posted in the Advanced Statistics Forum
    Replies: 4
    Last Post: Feb 2nd 2009, 04:55 AM
  5. MSE of Normal Variance Estimate
    Posted in the Advanced Statistics Forum
    Replies: 2
    Last Post: Jan 28th 2009, 10:17 PM

Search Tags


/mathhelpforum @mathhelpforum