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Math Help - 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 \mathcal{N}\left(\mu,\sigma^2\right) is f\!\left(x\right)=\frac{1}{\sqrt{2\pi}\sigma}e^{-\frac{1}{2}\left(\frac{x-\mu}{\sigma}\right)^2}

    Now, 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 u=\frac{x-\mu}{\sigma}\implies \sigma\,du=\,dx.

    Thus, we have \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 =\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:

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

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

    Thus, \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:

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

    Recall that the Gaussian integral \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

    \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, 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!!
    E(x) was not integrated, it is defined to be the integral of x\times  f(x) where f(x) is the pdf of 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):

    E\left(X\right)=\int_{-\infty}^{\infty}xf\!\left(x\right)\,dx, where 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, \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
    \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

    \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; August 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 z=\tfrac{1}{2}u^2, the limits of integration became positive (and the same)!

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

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