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Thread: Proof within Univariate Normal

  1. #1
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    Proof within Univariate Normal

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    E(X)=μ<br><br>thats all the information i've been given.<br>I dont know where to start.<br>
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  2. #2
    Rhymes with Orange Chris L T521's Avatar
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    Quote Originally Posted by ynotidas View Post
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    E(X)=μ<br><br>thats all the information i've been given.<br>I dont know where to start.<br>
    First, note that the pdf of $\displaystyle \mathcal{N}\left(\mu,\sigma^2\right)$ is $\displaystyle f\!\left(x\right)=\frac{1}{\sqrt{2\pi}\sigma}e^{-\frac{1}{2}\left(\frac{x-\mu}{\sigma}\right)^2}$

    Now, $\displaystyle E\left(x\right)=\int_{-\infty}^{\infty}xf\!\left(x\right)\,dx=\frac{1}{\s qrt{2\pi}\sigma}\int_{-\infty}^{\infty}xe^{-\frac{1}{2}\left(\frac{x-\mu}{\sigma}\right)^2}\,dx$

    Making the substitution $\displaystyle u=\frac{x-\mu}{\sigma}\implies \sigma\,du=\,dx$.

    Thus, we have $\displaystyle \frac{1}{\sqrt{2\pi}\sigma}\int_{-\infty}^{\infty}xe^{-\frac{1}{2}\left(\frac{x-\mu}{\sigma}\right)^2}\,dx\xrightarrow{u=\left(x-\mu\right)/\sigma}{}\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}\left(\sigma u+\mu\right)e^{-\frac{1}{2}u^2}\,du$ $\displaystyle =\frac{\sigma}{\sqrt{2\pi}}\int_{-\infty}^{\infty}ue^{-\frac{1}{2}u^2}\,du+\frac{\mu}{\sqrt{2\pi}}\int_{-\infty}^{\infty}e^{-\frac{1}{2}u^2}\,du$

    For the first integral:

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

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

    Thus, $\displaystyle \frac{\sigma}{\sqrt{2\pi}}\int_{-\infty}^{\infty}ue^{-\frac{1}{2}u^2}\,du\xrightarrow{z=\tfrac{1}{2}u^2} {}\frac{\sigma}{\sqrt{2\pi}}\int_{\infty}^{\infty} e^{-z}\,dz=\frac{\sigma}{\sqrt{2\pi}}\cdot0=0$.

    Now for the second integral:

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

    Recall that the Gaussian integral $\displaystyle \int_{0}^{\infty}e^{-ax^2}\,dx=\frac{1}{2}\sqrt{\frac{\pi}{a}}\implies\ int_{-\infty}^{\infty}e^{-ax^2}\,dx=\sqrt{\frac{\pi}{a}} $.

    So it follows that

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

    Therefore, $\displaystyle E\left(x\right)=\frac{1}{\sqrt{2\pi}\sigma}\int_{-\infty}^{\infty}xe^{-\frac{1}{2}\left(\frac{x-\mu}{\sigma}\right)^2}\,dx=\mu$.

    Does this make sense?
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  3. #3
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    Thanks Chris,

    I'm going to have to look over it again and again, and analyse everystep to see if i understand all of it, i dont want to waste your time by asking a stupid question ehheh...

    THANKS AGAIN!
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  4. #4
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    BUT, in saying that,

    In the beginning, why did you integrate the E[x]??

    also do you know any websites that can help me in understanding more on this? thanx!!
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  5. #5
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    Quote Originally Posted by ynotidas View Post
    BUT, in saying that,

    In the beginning, why did you integrate the E[x]??

    also do you know any websites that can help me in understanding more on this? thanx!!
    $\displaystyle E(x)$ was not integrated, it is defined to be the integral of $\displaystyle x\times f(x)$ where $\displaystyle f(x) $ is the pdf of $\displaystyle x$, and that is what was done.

    CB
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  6. #6
    Rhymes with Orange Chris L T521's Avatar
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    Quote Originally Posted by ynotidas View Post
    BUT, in saying that,

    In the beginning, why did you integrate the E[x]??

    also do you know any websites that can help me in understanding more on this? thanx!!
    I didn't integrate the expected value. That's how the expected value is defined for continuous distributions (like the normal):

    $\displaystyle E\left(X\right)=\int_{-\infty}^{\infty}xf\!\left(x\right)\,dx$, where $\displaystyle f\!\left(x\right)$ is the pdf of the distribution.

    Here's a site that evaluated the integral the way I did: The Normal Distribution
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  7. #7
    MHF Contributor matheagle's Avatar
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    Note that
    Thus, $\displaystyle \frac{\sigma}{\sqrt{2\pi}}\int_{-\infty}^{\infty}ue^{-\frac{1}{2}u^2}\,du\xrightarrow{z=\tfrac{1}{2}u^2} {}\frac{\sigma}{\sqrt{2\pi}}\int_{\infty}^{\infty} e^{-z}\,dz=\frac{\sigma}{\sqrt{2\pi}}\cdot0=0$.
    is a lot of work.
    The integrand is an odd function so the integral must be zero.
    I just commented on that in my review of that calc book.
    Noticing that a function is even or odd can save some work.

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

    I first show that the normal is a valid density with that polar coordinate trick (multiplying it by itself)
    and then we have

    $\displaystyle \frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}e^{-\frac{1}{2}u^2}\,du=1$
    since this is a valid density of a N(0,1).
    Last edited by matheagle; Aug 21st 2009 at 12:12 AM.
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  8. #8
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    OH okkkkkkkkkkkkkk i'll keep reading down..

    thanx everyone!
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  9. #9
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    How do you get the integration of e^-z from -infinity to +infinity to be equal to zero?


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  10. #10
    MHF Contributor matheagle's Avatar
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    Just integrate and let the bounds go to infinity, calc 2.
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  11. #11
    Rhymes with Orange Chris L T521's Avatar
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    Quote Originally Posted by ynotidas View Post
    How do you get the integration of e^-z from -infinity to +infinity to be equal to zero?


    After the substitution $\displaystyle z=\tfrac{1}{2}u^2$, the limits of integration became positive (and the same)!

    So it follows from integration properties that $\displaystyle \int_a^a f\!\left(x\right)\,dx=0$.

    That's why $\displaystyle \int_{\infty}^{\infty}e^{-z}\,dz=0$
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